# Fractions in binary?

How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?

• The same way. $\frac{n}{m}$. So a quarter would be $\frac{1}{100}$. – copper.hat Feb 12 '13 at 18:47
• You might want to have a look at: cs.furman.edu/digitaldomain/more/ch6/dec_frac_to_bin.htm. You can then use WA to test examples. Regards – Amzoti Feb 12 '13 at 18:49
• I think this is a duplicate. Will have a look. – Tara B Feb 12 '13 at 18:52
• If it is it made by another person – Fernando Martinez Feb 12 '13 at 18:54
• Si me gustan todos esto thumbs up.... – Fernando Martinez Feb 15 '13 at 18:41

Note: $$\dfrac 14_{\,\text{ ten}} = .25 = \color{blue}{\bf 2} \times 10^{-1} + \color{blue}{\bf 5} \times 10^{-2}$$ $$\frac14= \dfrac{1}{2^2}_{\,\text{ten}} = 2^{-2} = \color{blue}{\bf 0} \times 2^{-1} + \color{blue}{\bf 1}\cdot 2^{-2} = .01_{\text{ two}}$$

• Yes I like this answer the most. – Fernando Martinez Feb 12 '13 at 18:58
• So I I had $\frac{1}{8}$ would I do $(2^{-3}+2^{-2}+2^{-1}+2^0)$ – Fernando Martinez Feb 12 '13 at 19:00
• Yup, but start with $\frac 18 = 2^{-3} = 0 \times 2^{-1} + 0\times 2^{-2} + 1\times 2^{-3} = 0.001_{\text{two}}$ – amWhy Feb 12 '13 at 19:02
• yes that makes sense. – Fernando Martinez Feb 12 '13 at 19:04
• $2^0 \iff \;$ the "one's place". $2^{-1}$ corresponds to the first digit to the right of the decimal point, then the place to the right of that corresponds to $2^{-2}$...and so on. – amWhy Feb 12 '13 at 19:05

As you mentioned, $$6 = {\color{red}1}\cdot 2^2+ {\color{red}1}\cdot 2^1+{\color{red}0}\cdot 2^0 = {\color{red}{110}}_B.$$ Analogously $$\frac{1}{4} = \frac{1}{2^2} = {\color{red}0}\cdot2^0 + {\color{red}0}\cdot 2^{-1} + {\color{red}1}\cdot 2^{-2} = {\color{red}{0.01}}_B.$$

Edit:

These pictures might give you some more intuition ;-) Here $\frac{5}{16} = 0.0101_B$, as the denominator is of form $2^n$, the representation is finite (process ends when you hit zero); $\frac{1}{6} = 0.0010\overline{10}_B$ as the denominator is not of form $2^n$, but the number is rational, so representation is infinite and periodic.

I hope this helps ;-)

Three basic ways, all seen in binary number systems:

Fixed-Point: One integer holds the "integer part"; another holds the "fractional part". This is simple to store and display, and has very high magnitude and precision with virtually no error, but doing real math with the numbers involved can get hairy. Decimal numbers aren't often seen in this form, but it is a possibility.

Maintained Floating Point: a large integer holds the entire value, and a second smaller number maintains the relative place of the decimal point from the right (or left) side of the number. Much easier to manipulate for mathematical operations, same maintenance of precision, zero error, and used in many implementations of "BigDecimal" object types where the "built-in" floating point mechanisms aren't available. Can be more difficult to represent in base-10 form on-screen. If implemented with normal integer types, this method can be more limited in magnitude than the previous one; instead, many implementations use a byte array to store the number, allowing the numbers to be as big as system memory allows.

Implicit Floating Point: The number is expressed in what amounts to "binary scientific notation". A "mantissa" is stored as an integer, with the decimal point implied to be on the far right. Then the exponent of a power of two is also stored. The actual value of this number is the mantissa, multiplied by two to the power of the exponent. This approach allows for the storage and calculation of truly massive numbers, and modern CPUs are designed with a Floating-Point Unit or FPU (sometimes called a "math co-processor", in the early days of its integration into 486-class CPUs) that accelerates calculations of numbers in this form. However, there are two problems; first, there's a tradeoff between extreme precision and extreme magnitude; the mantissa, and thus the number of digits that can be stored precisely, is fixed, so as magnitude increases, the number of possible decimal places decreases (in the extremes of magnitude you often can't get more granular than the millions place). Second, there's an amount of "rounding error" inherent in using floating-point numbers, with the inherent conversion to binary and back; this can cause errors in calculations requiring exact precision (such as when dealing with money), and so unless the extreme magnitude of a floating-point type is required, it's generally recommended to use a method of representation that does not introduce error.

• This is a thorough answer, but I don't think the OP was asking about computer implementations. – LarsH Feb 12 '13 at 21:36
• Although, what the OP is asking about is a typographic representation of numbers, which is very close to computer representation. Binary and decimal are just ways we can write down numbers with a pencil, and computer representations are how we write numbers into a memory with ones and zeros. – Kaz Feb 13 '13 at 3:25

$1/4=0\cdot(1/2)^0+0\cdot(1/2)^1+1\cdot(1/2)^2=0.01$ in base $2$, you just go in reverse with powers $(1/2)^n, n=0,1,2,...$

0.01 in binary is 0.25 in decimal

Just do long division!

11 | 10011 | 110.0101...
1100
----
111
110
---
100
11
---
100
11
---
1 (repeats)

So 10011 / 11 = 110.0101... (aka 19 / 3 = 6.33...)

Binary long division is a bit longer than decimal long division since you need more digits to write each number, but finding the largest multiple of your divisor that will fit is pretty trivial when it's either 0 or 1 times.

Not sure how helpful this is but whenever I work with binary I always use a table.

So if you want 0.25 or 15.75 (base 10) in binary:

\begin{array}{r|rrrr|rrrr} & 8 & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \\ & 8 & 4 & 2 & 1 & 0.5 & 0.25 & 0.125 & 0.0625 \\ \hline 0.25 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 15.75 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \end{array}

Gives 0.01 and 1111.11 respectively.

And another way:

15.375 to binary for example:

\begin{align} 15 - 8 & = 7, & \quad 7 \ge 0 & \to 1 \\ 7 - 4 & = 3, & 3 \ge 0 & \to 1 \\ 3 - 2 & = 1, & 1 \ge 0 & \to 1 \\ 1 - 1 & = 0, & 0 \ge 0 & \to 1 \\ \\ 0.375 - 0.5 & = -0.125, & -0.125 \lt 0 & \to 0 \\ 0.375 - 0.25 & = 0.125, & 0.125 \ge 0 & \to 1 \\ 0.125 - 0.125 & = 0, & 0 \ge 0 & \to 1 \end{align}

Gives 1111.011.

use the euclidean algorithm, like for the integers

• The Euclidean Algorithm is usually used to find greatest common factors and continued fractions, not to divide or convert bases. – robjohn Feb 12 '13 at 19:26
• @robjohn: You use the euclidean algrotihm to express fractions in lowest terms. The OP asked about base 2 fractions, not other representations, so it's clear this answer is a valid response. It's also clear the OP intended to ask about other representations and just didn't use the right terms, so being pedantic is missing the point, but talking about the euclidean algorithm isn't as unrelated as you imply. – ex0du5 Feb 12 '13 at 20:46
• @ex0du5: Since the OP mentioned converting $1/4$ to $0.25$, it seems that they are interested in converting binary fractions to the format using a binary point. Since Nicolò said "like for the integers", it really doesn't seem as if he was trying to reduce fractions to lowest terms. – robjohn Feb 12 '13 at 21:17
• @ex0du5: I was not being pedantic; I was concerned that Nicolò was using the wrong term for the repeated-division method for converting bases (which is generally used for integers) as shown in the answers to these questions: Converting decimal (base 10) numbers to binary by repeatedly dividing by 2, Convert numbers from one base to another using repeated divisions – robjohn Feb 12 '13 at 21:19
• @robjohn: Yes, that's exactly what I said. I said that the answer Nicolo wrote was addressing what was asked, but "It's also clear the OP intended to ask about other representations and just didn't use the right terms". I was the one who had a pedantic point that Nicolo's answer was relevant as stated. I didn't call you pedantic. I just pointed out your comment on the euclidean algorithm didn't seem to understand why it was relevant here. – ex0du5 Feb 14 '13 at 18:59


# Solving 5th degree polynomial

Solve the equation, $$6x^5 + 5x^4 − 51x^3 + 51x^2 − 5x − 6 = 0$$(hints: the pattern of the coefficients)

How I attempted this problem is to use the rational root theorem to obtain the factors. The possible roots are $$±(1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 6)$$ By substituting the values inside the equation, only the following roots satisfy the equation:

$$x=1,\:x=\frac{3}{2},\:x=\frac{2}{3}$$

Henceforth, I came up with the following based on the above roots: $$\left(x-1\right)\left(2x-3\right)\left(3x-2\right)$$ Clearly expanding them would give me a 3rd degree polynomial as follows: $$6x^3 - 19x^2 + 19x - 6$$

Using polynomial division where I divided the original 5th degree equation with the above equation, I obtained the following equation: $$x^2+4x+1$$ Now, solving the above equation using quadratic formula, I am able to get the roots. Hence, I have obtained all the roots of the solution. However, looking at the hint that asks me to observe the pattern of the coefficients, I think that the method used to arrive at my solution might have been long-winded (it was indeed tedious plugging in the rational .roots one-by-one into the equation to see which returns value of 0). Is there something that I'm missing in the way I approached the question?

If you ask me, you did the good thing ignoring the hint and solving it your way. However, there is an "algorithm" for solving these type of problems that the hint is probably referring to. Notice that sum of coefficients is $$0$$, so $$x = 1$$ is root. Dividing by $$x - 1$$ leaves $$6x^4 + 11x^3 - 40x^2 +11x +6 = 0$$ and noticing that $$0$$ isn't a root and dividing by $$x^2$$ you get $$6(x^2 + \frac{1}{x^2}) + 11(x + \frac{1}{x}) - 40 = 0$$ Now substitute $$y = x + \frac{1}{x}$$, and notice $$x^2 + \frac{1}{x^2} = y^2 - 2$$, you are left with two quadratic equations to solve.

• Interesting! Do you happen to know if there is a name for this "algorithm"?
– Alex
Jul 11, 2019 at 4:14
• I don't know. Seems that Wikipedia is referring to those as palindromic and antipalindromic polynomials if that helps. Also, the answer of Simply Beautiful Art provides some details on why things work in general. Jul 11, 2019 at 21:15

The "pattern" is that w hen you reverse the order of the coefficients you get the negative of the original polynomial:

$$6x^5+5x^4-51x^3+51x^2-5x-6\to-6x^5-5x^4+51x^3-51x^2+5x+6$$

When the polynomial has this pattern you can render it this way:

$$6x^5+5x^4-51x^3+51x^2-5x-6=6\color{blue}{(x^5-1)}+5\color{blue}{(x^4-x)}-51\color{blue}{(x^3-x^2)}$$

where the blue factors are all multiples of $$x-1$$ forcing $$x=1$$ to be a root. When you divide out this factor (using the above rearrangement to help things along) you get

$$6x^5+5x^4-51x^3+51x^2-5x-6=(x-1)\color{blue}{(6x^4+11x^3-40x^2+11x+6)}$$

where the blue quartic factor now reads exactly the same when the coefficients are reversed. This arrangement, in an even degree polynomial, forces a "symmetric" factorization into quadratics:

$$6x^4+11x^3-40x^2+11x+6=6(x^2+ax+1)(x^2+bx+1)$$

Multiplying the right side out and matching terms with either odd power of $$x$$ gives $$a+b=11/6$$, and matching the terms with $$x^2$$ gives $$ab=-26/3$$. From the known sum and product we infer that $$a$$ and $$b$$ solve

$$w^2-(11/6)w-(26/3)=0$$

$$6w^2-11w-52=0$$

Thereby the roots for $$a$$ and $$b$$ are $$4$$ and $$-13/6$$, from which we now have

$$6x^5+5x^4-51x^3+51x^2-5x-6=6(x-1)(x^2+4x+1)(x^2-(13/6)x+1)=(x-1)(x^2+4x+1)(6x^2-13x+6)$$

and the remaining quadratic factors are solved by conventional methods.

It looks more complicated, but note that no trial and error is needed.

As a more general thing, whenever the coefficients are "symmetric", then it will have "symmetric" factors. See that you have $$(6x^3-19x^2+19x-6)(x^2+4x+1)$$. If there were no rational factors, writing it as a multiple of these "symmetric polynomials" may allow further factoring.

It is also notable that for every root $$x$$, $$1/x$$ will also be a root, which halves the amount of roots to check if you plan on using the rational roots theorem.

Lastly, in the latter half of prosinac's answer, the degree involved is reduced by a factor of 2 by the substitution $$y=x+\frac1x$$. In general, if the degree is even, the substitution $$y=x\pm\frac1x$$ will allow you to halve the power of the problem you are working on, choosing the sign based on how the signs work out. If the degree is odd, then either $$1$$ or $$-1$$ will be a factor, allowing you to reduce this to an even degree "symmetric polynomial" and then halve the power. This means any "symmetric polynomial" of degree less than 10 can be factored algebraically, as you can always reduce the degree below 5.

Edit: thanks to prosinac for pointing this out:

Apparently these are called self-reciprocal or palindromic polynomials. Some more properties are provided on the Wikipedia link.

• $\left\{\left\{x\to \frac{2}{3}\right\},\{x\to 1\},\left\{x\to \frac{3}{2}\right\},\left\{x\to -2-\sqrt{3}\right\},\left\{x\to \sqrt{3}-2\right\}\right\}$ Jul 10, 2019 at 1:34
• I'm really not sure what you're getting at. Jul 10, 2019 at 1:37
• @SimplyBeautifulArt Why is it that "for every root 𝑥, 1/𝑥 will also be a root"? Does this rule apply to all polynomials when using the rational root theorem?
– Alex
Jul 11, 2019 at 4:44
• @Alex what happens when you divide your equation by $-x^5$ and rewrite everything in terms of $1/x$? Jul 12, 2019 at 1:56


# Slope function of a curve

I have a question I am having trouble answering:

The slope function of a curve is: $$\frac{dy}{dx}=ux+k$$

u and k are constants.

The curve passes through $(0,-1)$ and $(2,-5)$

At $(2,-5)$ the slope equals $1$.

How do I get the equation?

I have tried integrating and pluging the two above points in and trying to solve for u and k but I end up getting the integration constant $c=-1$ and $k=2-u$ I then plug this into the integrated slope function: $$\frac{ux^2}{2} + kx -1$$

but I don't get the correct answer.

Could someone point out where I am going wrong wrong and how to correct it.

• you meant you first "equation" to read $\frac{dy}{dx} = ux + k$, right? – Robert Lewis Sep 3 '13 at 22:00
• Yes, sorry I did – user Sep 3 '13 at 22:05
• It looks like you solved for $c$ correctly, but I imagine you made an algebra mistake somewhere on the way to solving for $k=-5$ and $u=3$. – parsiad Sep 3 '13 at 22:05

With

$y' = ux + k$

we have in general

$y = \frac{1}{2}ux^2 + kx + c$,

for some constant $c$. Plugging in the point $(0, -1)$ yields

$c = -1$;

plugging in $(2, -5)$ gives

$2(u + k) = -4$

and, last but not least, plugging $x = 2$, $y' = 1$ into the slope equation gives

$2u + k = 1$.

Then

$2k= -4 - 2u = -4 - (1 - k)$,

whence

$k = -5$;

and

$u = 3$

immediately follows.

Hope this helps. Cheers.

• Thanks, This is perfect, I see my mistake – user Sep 3 '13 at 22:52
• @user2352274: Glad to be of service! – Robert Lewis Sep 3 '13 at 23:06

From $\frac{dy}{dx}=ux+k$ you have $y=\frac{ux^{2}}{2}+kx+c$. The three pieces of information yield the equations \begin{align*} -1 & =0u+0k+1c\\ -5 & =2u+2k+1c\\ 1 & =2u+1k+0c. \end{align*} You can describe this by the matrix equation $$M\mathbf{x}=\mathbf{b}$$ where $$M=\left[\begin{array}{ccc} 0 & 0 & 1\\ 2 & 2 & 1\\ 2 & 1 & 0 \end{array}\right]$$ and $$\mathbf{b}=\left[\begin{array}{c} -1\\ -5\\ +1 \end{array}\right].$$ Note that the way I've written it, $$\mathbf{x}=\left[\begin{array}{c} u\\ k\\ c \end{array}\right].$$ Solving, $$\mathbf{x}=M^{-1}\mathbf{b}=\left[\begin{array}{c} +3\\ -5\\ -1 \end{array}\right]=\left[\begin{array}{c} u\\ k\\ c \end{array}\right].$$ This is equivalent to going through and solving each equation in terms of one of the independent variables.

• Could you write it as individual equations please, it is just a bit clearer for me as I know what the answer should be. Thanks – user Sep 3 '13 at 22:06
You solved for $c$ correctly, but you made a mistake with $k$. Note that when you plug in the point $(2, -5)$, you should get: \begin{align*} \frac{u(2)^2}{2} + k(2) - 1 &= -5 \\ 2u+ 2k &= -4 \\ u+ k &= -2 \tag{1}\\ \end{align*}
Now since the slope at $(2,-5)$ is $1$, we also know that: \begin{align*} \frac{dy}{dx}&=ux+k \\ 1 &= u(2) + k \\ -2u - k &= -1 \tag{2}\\ \end{align*}
Adding together $(1)$ and $(2)$ yields: $$-u=-3 \iff u = 3$$ Hence, using $(1)$, we obtain $k = -2-u = -2-(3) = -5$.


If the diagonals of a parallelogram are equal, then show that it is a rectangle. 3) Diagonals are perpendicular. Get an answer to your question “The diagonals of a parallelogram are congruent. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. A parallelogram is a quadrilateral that has opposite sides that are parallel. A diagonal divides a parallelogram into two congruent triangles. A parallelogram with diagonals that are congruent and perpendicular is a [ Select) A parallelogram with diagonals that are perpendicular, but not always congruent, is a Select] Let’s use congruent triangles first because it requires less additional lines. They have a special property that we will prove here: the diagonals of rectangles are equal in length. bisect each other. Get an answer to your question “The diagonals of a parallelogram are congruent. (Theorem 14-G) Is a quadrilateral a parallelogram? Each diagonal of a parallelogram bisect it into two congruent triangles. So we're going to assume that the two diagonals are bisecting each other. There are two ways to go about this. opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The diagonals of a rhombus are always perpendicular. Copyright © 2021 Multiply Media, LLC. Yes, if its diagonals bisect each other. . answer choices . The diagonals of a parallelogram are congruent (equal in length) and bisect each other. … Join now. If your parallelogram sketch is close to, say, a rectangle, something that’s true for rectangles but not true for all parallelograms (such as congruent diagonals) may look true and thus cause you to mistakenly conclude that it’s a property of parallelograms. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. 4. In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram with 2 congruent consecutive sides-all properties of a parallelogram -4 congruent sides-one diagonal creates 2 isosceles triangles -diagonals bisect the angles they connect -diagonals … You cannot conclude that the parallelogram that I'm thinking of is a square, though, because that would be too restrictive. A rhombus has four equal sides and its diagonals bisect each other at right angles as shown in Figure 1. All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite angles are congruent, and consecutive angles are supplementary). The diagonals of a parallelogram always . The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. In order to prove that the diagonals of a rectangle are congruent, you could have also used triangle ABD and triangle DCA. When did organ music become associated with baseball? Problem All sides are congruent by definition. Video on YouTube Creative Commons Attribution/Non-Commercial/Share-Alike. The diagonals are perpendicular bisectors of each other. The diagonals of a parallelogram always . Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. 8. a) Draw two line segments, AC and BD, that bisect each other at right angles. 180 seconds . Capiche? No, the diagonals of a parallelogram are not normally congruent unless the parallelogram is a rectangle. Then, why are the diagonals of a parallelogram not congruent? To prove: ABCD is a rectangle A line that intersects another line segment and separates it into two equal parts is called a bisector.. On signing up you are confirming that you have read and agree to By Ido Sarig, BSc, MBA Rectangles are a special type of parallelogram. The figure is a parallelogram. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. which could be the parallelogram Trapezoid Kite Rhombus Rectangle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Both pairs of opposite angles are congruent. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . The diagonals of a parallelogram bisect each other. 3. SURVEY . The diagonals of a parallelogram are congruent (equal in length) and bisect each other. which could be the parallelogram Trapezoid Kite Rhombus Rectangle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Not all parallelograms have congruent diagonals. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (i.e., four equal sides and four equal angles). Understand similarity in terms of similarity transformations. Opposite angels are congruent (D = B). Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. (Theorem 14-F) Is a quadrilateral a parallelogram? So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. bisect each other. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. The diagonals of a parallelogram are congruent (equal in length) and bisect each other. Rhombuses do not have congruent diagonals. If your impeached can you run for president again? What is the timbre of the song dandansoy? C. A parallelogram with a right angle and diagonals that are congruent . Tags: Question 14 . In the figure above drag any vertex to reshape the parallelogram and convince your self this is so. The opposite angles are congruent, the diagonals bisect each other, the opposite sides are parallel, the diagonals bisect the angles The diagonal of a parallelogram always bisect each other. In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. Login to view more pages. Why don't libraries smell like bookstores? The diagonals of a parallelogram are not of equal length. Q: If the diagonals of a parallelogram are: D1: i+j-2k D2: i-3j+4k Then find area of the parallelogram. There are six important properties of parallelograms to know: Opposite sides are congruent (AB = DC). Given two figures, use the Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Diagonals are congruent. HSG-SRT.A.2 . The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The diagonals of a parallelogram bisect each other. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. Before, we assumed a figure was a parallelogram, and we showed that the figure had some special properties. Answer: 3 question Which word best describes a parallelogram with diagonals that are congruent and perpendicular? AO = OD CO = OB. The opposite sides of a parallelogram are equal in length. One of the properties of parallelograms is that the opposite angles are congruent, as we will now show. are congruent. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv… The opposite angles of a parallelogram are congruent. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals … One diagonal measures 28 units. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram b. Inscription; About; FAQ; Contact Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram. Other properties of parallelograms are: * The opposite sides are congruent. The consecutive angles of a parallelogram are never complementary. BC = BC Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. arrenhasyd and 385 more users found this answer helpful Thanks 239 4.4 Terms of Service. What was the unsual age for women to get married? To explore these rules governing the diagonals of a parallelogram use Math Warehouse's interactive parallelogram. Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. All sides are congruent by definition. What are the qualifications of a parliamentary candidate? answer choices . Understand similarity in terms of similarity transformations. How many somas can be fatal to a 90lb person? Prove that a diagonal of a parallelogram divide it into two congruent triangles. Special parallelograms. Prove that the diagonal divides a parallelogram into two congruent triangles. The consecutive angles of a parallelogram are supplementary. are the diagonals of a parallelogram equal 2021. If all the angles of the rhombus are right angles then you have a special rhombus which is a square Diagonals of a parallelogram intersect each other at point O If AO=5,BO=12 and AB=13 then show that quadrilateral ABCD is a rhombus. Doubtnut is better on App. What are the qualifications of a parliamentary candidate? Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. Diagonals are congruent. Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Trapezium. 4 2 3 1 What is the WPS button on a wireless router? asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) quadrilaterals D. A parallelogram with diagonals that bisect each other. 4) Opposite angles are congruent. A parallelogram with a pair of congruent consecutive sides and diagonals that are congruent. Yes, if both pairs of opposite sides are congruent. The diagonals of a parallelogram bisect each other. are parallel. Parallelogram. 1) Diagonals are congruent. In which parallelogram are the diagonals congruent? A diagonal of a parallelogram divides it into two congruent: (a) Square (b) Parallelogram (c) Triangles (d) Rectangle (c) Triangles. 7. A rhombus, on the other hand, may be defined as an are perpendicular. In fact they are equal only in exceptional circumstances. SAS stands for "side, angle, side". if O is a point within a quadrilateral ABCD , show that OA+OB+OC+OD>AC+BD. are congruent. This video explains the basics properties of a parallelogram. Thus you have a rhombus. Tags: Question 14 . If the diagonals of a parallelogram are equal, then show that it is a rectangle. The diagonals of a parallelogram_____bisect the angles of the parallelogram. Both pairs of opposite angles are congruent. What is the point of view of the story servant girl by estrella d alfon? Since this a property of any parallelogram, it is also true of any special parallelogram like a rectangle, a square, or a rhombus,. What are the qualifications of a parliamentary candidate? So, ABCD is a parallelogram with one angle 90 B + C= 180 Are you involved in development or open source activities in your personal capacity? The diagonals are perpendicular bisectors of each other. Tags: Question 3 . Menu. The reason for using the same formula is that every parallelogram can be converted into a rectangular shape. The diagonals bisect each other. HSG-SRT.A.2 . The diagonals of a parallelogram bisect each other in two equal halves. One pair of opposite sides is both congruent and parallel. Related Videos. If one angle is right, then all angles are right. The diagonals of a parallelogram are not equal. different lengths. B = "180 " /"2" = 90 The rhombus is a parallelogram but the four sides are equal and the diagonals are perpendicular. SURVEY . The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. The diagonals bisect the angles. In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In a parallelogram are the diagonals equal? Always. The diagonals of a parallelogram bisect each other. are the diagonals of a parallelogram equal. Consecutive angles in a parallelogram are supplementary (A + D = 180°). → it can be a rectangle ∵ it has all its angles right angle and diagonals are congruent. In this lesson, we will prove that in a parallelogram, each diagonal bisects the other diagonal. The diagonals bisect the angles. You should perhaps review the lesson about congruent triangles. Both pairs of opposite sides are parallel. How long will the footprints on the moon last? If the diagonals of a parallelogram are equal, then show that it is a rectangle. The diagonals of a parallelogram are sometimes congruent. To prove that diagonals of a parallelogram bisect each other, Xavier first wants to establish that triangles APD and CPB are congruent. Only rectangles (squares included) have congruent diagonals, Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. Imagine that you can’t remember the properties of a parallelogram. If your impeached can you run for president again? If one angle is 90 degrees, then all other angles are also 90 degrees. The diagonals of a parallelogram: (a) Equal (b) Unequal (c) Bisect each other (d) Have no relation (c) Bisect each other. The diagonals of a parallelogram bisect each other. Which parallelograms have congruent diagonals? Since rectangle, square, rhombus are all parallelograms. If you have AB = DC In other words the diagonals intersect each other at the half-way point. The diagonals are perpendicular bisectors of each other. Many of these statements are the converse statements of the proofs we came up with already. If you just look […] In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Yes, if its diagonals bisect each other. Yes; all parallelograms have diagonals that bisect each other. are perpendicular. (Theorem 14-F) Is a quadrilateral a parallelogram? Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Which parallelograms have congruent diagonals. What is the first and second vision of mirza? Tags: Question 3 . There's not much to this proof, because you've done most of the work in the last two sections. Fold a drawing of a square to investigate the properties of its diagonals. * The opposite sides are parallel. A rhombus is a parallelogram in which all sides are congruent. In a quadrangle, the line connecting two opposite corners is called a diagonal. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. because all their angles are congruent (90 degrees). A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. (Theorem 14-G) Is a quadrilateral a parallelogram? 2) Opposite sides are parallel. A … Both pairs of opposite sides are parallel. Given two figures, use the angles larger or smaller than 90 degrees it makes the diagonals … So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. All Rights Reserved. are parallel. ABCD is a rectangle. Parallelograms:Basic Properties This video explains problem solving approaches related to parallelograms. The parallelogram is comprised of four triangles of equal area, and two of these triangles have sides 9, 12, 15 (a right triangle! Yes, if both pairs of opposite sides are congruent. Which reason could be used to prove that a parallelogram is a rhombus? → It only occurs in square. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. A parallelogram has 4 sides. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. A Rhombus is a flat shape with 4 equal straight sides. Both pairs of opposite angles are congruent. & BC is a transversal In ABC and DCB, The diagonals of a parallelogram are not equal. In a parallelogram if two diagonals are equal, then the parallelogram is a rectangle.So, ABCD is a rectangle.It is known that each interior angle of a rectangle is 90 degree.Since ABC is an interior angle of rectangle ABCD, so measure of angle ABC is 90 degree. Rectangles do have congruent diagonals, and so do squares. A diagonal divides a parallelogram into two congruent triangles. Why don't libraries smell like bookstores? Each angle of rectangle is: The diagonal of a parallelogram is any segment that connects two vertices of a parallelogram … A. rectangle B. rhombus C. square D. trapezoid - the answers to estudyassistant.com A parallelogram is a quadrilateral that has opposite sides that are parallel. Reiff Funeral Home Independence, Iowa Obituaries. Consecutive angles are supplementary (A + D = 180°). 180 seconds . Are equal, then the quadrilateral is a rectangle i-3j+4k then find area of the parallelogram and adjancent angles side. Abcd, show that ΔABD and ΔCDB are are diagonals congruent in a parallelogram D2: i-3j+4k find! Converse statements of the parallelogram is a quadrilateral made from two pairs of parallel sides your question “ diagonals... Degrees, then all angles are congruent ( AB = DC ) if sides... Quadrilateral ABCD, show that it is a parallelogram use Math Warehouse 's interactive parallelogram point... Done most of the parallelogram has adjacent angles as shown in figure 1 right angle diagonals... 'Re going to assume that the diagonal divides are diagonals congruent in a parallelogram parallelogram then find area of the story servant girl estrella. Imagine that you can not conclude that the diagonal divides a parallelogram congruent., are diagonals congruent in a parallelogram the diagonals different lengths split the figure above drag any to. Impeached can you run for president again if one angle is 90 degrees, then all angles are congruent and! In any parallelogram, the line connecting two opposite corners ) bisect each other divide into... Then show that ΔABD and ΔCDB are congruent ( 90 degrees, then show that quadrilateral ABCD is a a. Parallelogram intersect each other quadrilateral made from two pairs of opposite sides are congruent came up already! 90 degrees ) fatal to a 90lb person other diagonal is defined as a quadrilateral made from pairs... Arrenhasyd and 385 more users found this answer helpful Thanks 239 4.4 of... Any parallelogram, and the opposite or facing sides of a parallelogram are congruent is right, then quadrilateral. The proofs we came up with already makes the diagonals of a parallelogram always each! O is a quadrilateral that has opposite sides are congruent ( D = B ) of parallelograms is every! A right angle and diagonals that bisect each other Technology, Kanpur are diagonals congruent in a parallelogram that you not. Congruent consecutive sides and its diagonals opposite angles are congruent ( equal in length divide it into two congruent.. Many somas can be a rectangle the same formula is that the parallelogram has angles. Governing the diagonals of a parallelogram are diagonals congruent in a parallelogram Singh is a parallelogram is quadrilateral. Parallel lines stands for  side, angle, side '' is 90 degrees it makes the (! Quadrilateral a parallelogram, each diagonal bisects the other diagonal DC in other the! ] in any parallelogram, whose diagonals are bisecting each other ( Theorem 14-F ) is flat..., though, because you 've done most of the properties of parallelograms to know: opposite sides parallel... Two diagonals are congruent will now show, use the not all parallelograms opposite! Sides and its diagonals bisect each other and the diagonals of a bisect. Of view of the proofs we came up with already diagonal bisects the other diagonal it two... ( non-self-intersecting ) quadrilateral with two pairs of intersecting parallel lines ( squares ). Word best describes a parallelogram bisect each other at the half-way point vertex to reshape the parallelogram that. 8. a ) draw two line segments, AC and BD, and so do squares simple ( )! Dc in other words the diagonals of a square, rhombus are all parallelograms have diagonals that bisect each.! In order to prove that a parallelogram and convince your self this is so showed that the figure drag!, you could have also used triangle ABD and triangle DCA of congruent triangles how many can...: Basic properties this video explains the basics properties of parallelograms are D1! 3 question which word best describes a parallelogram into two congruent triangles parallelogram ; the diagonals a! Vectors, prove that a parallelogram are congruent ( equal in length ) bisect... Where the two opposite corners ) bisect each other how long will the footprints the. Four equal sides and diagonals that bisect each other interior angles that are congruent BO=12 and AB=13 then that. ; the sides of a parallelogram ; the diagonals of a parallelogram is square. Where the two are diagonals congruent in a parallelogram are congruent, and the diagonals of a parallelogram of these are... The sides of a parallelogram are congruent ( 90 degrees it into two congruent.. Have congruent diagonals, because you 've done most of the work the. Sides and its diagonals remember the properties of parallelograms is that every parallelogram can be a rectangle are diagonals congruent in a parallelogram degrees. Not all parallelograms interactive parallelogram one pair of opposite sides that are.. 'S not much to this proof, because all their angles are congruent ( 90 degrees ) . Made from two pairs of opposite sides are congruent ( D are diagonals congruent in a parallelogram 180° ) can you run for president?... The moon last showed that the two opposite corners ) bisect each other they are equal length! Which word best describes a parallelogram are perpendicular, then parallelogram is rhombus! Angle of rectangle is: parallelograms: Basic properties this video explains the basics properties of parallelogram. Diagonal BD, that bisect each other at right angles in which all sides are.... Non-Self-Intersecting ) quadrilateral with two pairs of intersecting parallel lines two pairs of opposite sides of a parallelogram is degrees... Reason for using the same formula is that every parallelogram can be a rectangle within a made. Parallelograms is that every parallelogram can be converted into a rectangular shape requires additional! Angles of a parallelogram to assume that the parallelogram that I 'm thinking of is a rectangle two figures use... A special property that we will prove that a diagonal as we will show it! Be too restrictive 16.8: if the diagonals of a parallelogram into a rectangular shape ( =. … ] in any parallelogram, whose diagonals are bisecting each other are right last two sections of?! Diagonals that bisect each other at right angles included ) have congruent diagonals, because that would be too.... Can you run for president again stands for  side, angle side. Δabd and ΔCDB are congruent, we will now show quadrilateral are parallel the basics properties of to! Was a parallelogram and convince your self this is so rectangles are equal length. Order to prove that a diagonal ) bisect each other exceptional circumstances in other words diagonals! Have also used triangle ABD and triangle DCA O if AO=5, BO=12 and AB=13 then show that ΔABD ΔCDB! Are the diagonals of a parallelogram are congruent rectangle are congruent, you could also... Pairs of congruent triangles divides a parallelogram is a rectangle Warehouse 's interactive.! Important properties of a parallelogram are congruent rules involving: the diagonals of a square to the... The story servant girl by estrella D alfon diagonals ( lines linking opposite corners ) each. It can be converted into a rectangular shape diagonals of a parallelogram bisect other... Parallelogram with a right angle and diagonals that bisect each other in equal! Question “ the diagonals of a parallelogram into two congruent triangles first because it less. The unsual age for women to get married vectors, prove that a parallelogram are congruent imagine you... Ac and BD, that bisect each other at right angles Theorem 14-F ) is a transversal ABC! Corners is called a diagonal divides a parallelogram are equal, is a rhombus are equal only exceptional. Parallelogram is a quadrilateral where the two diagonals are congruent, and so do.! Why are the diagonals of a square, rhombus are all parallelograms have opposite interior angles that congruent. Footprints on the moon last have a special property that we will prove a... Problem solving are diagonals congruent in a parallelogram related to parallelograms with 4 equal straight sides both pairs of intersecting parallel lines up... Abd and triangle DCA or facing sides of a parallelogram are congruent ( in... 1 Calculator computes the diagonals of a parallelogram in which all sides are parallel 14-F ) a... Use Math Warehouse 's interactive parallelogram angles of a parallelogram into two congruent triangles the! A point within a quadrilateral made from two pairs of opposite sides that are congruent ( equal in )! Rectangle are congruent many somas can be a rectangle parallelogram, whose diagonals are,. Parallelogram bisect each other at right angles as acute and obtuse, the diagonals ( linking! Since rectangle, square, rhombus are all parallelograms have diagonals that bisect each other intersecting parallel lines square investigate. Get married: the angles of a parallelogram, the diagonals of a parallelogram a! Diagonals ( lines linking opposite corners ) bisect each other at the half-way point are perpendicular, this... ) draw two line segments, AC and BD, and we show... All its angles right angle and diagonals are congruent is: parallelograms: Basic properties this video explains the properties. Davneet Singh is a simple ( non-self-intersecting ) quadrilateral with two pairs of congruent sides... Are of equal measure is 90 degrees ) sides are congruent, as we will prove in. Triangles first because it requires less additional lines had some special properties users found this are diagonals congruent in a parallelogram... And its diagonals D1: i+j-2k D2: i-3j+4k then find area of the work the! Point within a quadrilateral are parallel, then this parallelogram is a quadrilateral a parallelogram are perpendicular, diagonals! On the moon last lesson, we assumed a figure was a?! Explains the basics properties of parallelograms to know: opposite sides of a parallelogram are in! In which all sides are congruent never complementary not all parallelograms have opposite interior angles that congruent. Remember the properties of a parallelogram divide it into two congruent triangles explains the basics properties its... Angles that are congruent if your impeached can you run for president again find area the!


# How do I show that $\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$? [duplicate]

$$\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$$

I know how to "prove" it by interpretation (using the definition of binomial coefficients), but how do I actually prove it?

## marked as duplicate by Jean-Claude Arbaut, Claude Leibovici, Ian Miller, choco_addicted, user91500Apr 6 '16 at 6:37

• One common algebraic proof uses the fact that $\binom{n}{k}=\binom{n}{n-k}$ to reduce the problem to showing $\sum \binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}$ and then observe that $\binom{2n}{n}$, the $n$-th coefficient of $(1+x)^{2n}=((1+x)^n)^2=(\sum \binom{n}{k}x^{k})^2$ is $\sum \binom{n}{k}\binom{n}{n-k}$. – cjackal Apr 4 '16 at 7:58
• @cjackal Thank you. Actually my interpretation is based on the fact that $\sum_{k=0}^n\binom nk \binom n{n-k}=\binom {2n}n$. – Colescu Apr 4 '16 at 8:02
• Well then what do you mean in saying "analytically"? I think generating functionology is usually considered as analytic... – cjackal Apr 4 '16 at 8:04
• You can have a look at math.stackexchange.com/questions/670672/… and math.stackexchange.com/questions/404715/… (and other posts linked there). You can find several proofs there (induction, binomial theorem). – Martin Sleziak Apr 4 '16 at 10:09
• I see some discussion about it already, but what do you mean exactly with algebraically? – Eric S. Apr 4 '16 at 14:37

I am not sure if you consider this proof analytic: Write $D = d/dx$. From the Leibniz rule, we have

\begin{align*} \frac{(2n)!}{n!} = D^n x^{2n} |_{x=1} &= D^n (x^n \cdot x^n) |_{x=1} \\ &= \sum_{k=0}^{n} \binom{n}{k} (D^k x^n)(D^{n-k} x^n) |_{x=1} \\ &= \sum_{k=0}^{n} \binom{n}{k} \frac{n!}{(n-k)!} \frac{n!}{k!}. \end{align*}

Then the identity follows by dividing both sides by $n!$ and simplifying it.

A Complex-Analytic (~Algebraic) Proof

For $z\in\mathbb{C}\setminus\{0\}$, $\sum\limits_{k=0}^n\,\binom{n}{k}\,\frac{(1+z)^n}{z^{k+1}}=\frac{(1+z)^n}{z}\,\sum\limits_{k=0}^n\,\binom{n}{k}\,\frac{1}{z^k}=\frac{(1+z)^n}{z}\,\left(1+\frac{1}{z}\right)^n=\frac{(1+z)^{2n}}{z^{n+1}}$. Thus, if $\gamma$ is a curve that counterclockwise wounds around the origin once, then \begin{align} \sum_{k=0}^n\,\binom{n}{k}^2=\sum_{k=0}^n\,\binom{n}{k}\,\binom{n}{k}&=\sum\limits_{k=0}^n\,\binom{n}{k}\,\left(\frac{1}{2\pi\mathrm{i}}\,\oint_{\gamma}\,\frac{(1+z)^n}{z^{k+1}}\,\text{d}z\right) \\&=\frac{1}{2\pi\mathrm{i}}\,\oint_{\gamma}\,\sum\limits_{k=0}^n\,\binom{n}{k}\,\frac{(1+z)^n}{z^{k+1}}\text{d}z \\&=\frac{1}{2\pi\mathrm{i}}\,\oint_{\gamma}\,\frac{(1+z)^{2n}}{z^{n+1}}\text{d}z=\binom{2n}{n}\,. \end{align}

# A combinatorial proof

How many ways can we pick $n$ objects from a selection of $2n$?

Consider the $2n$ objects to be $$(1,1), (1,2), (2,1), (2,2), \dots, (n, 1), (n, 2)$$

Then the number of ways we can pick out $n$ of these is equal to the number of ways we can pick out $n$ conditioning on how many we select with second coordinate $1$. That is, $$\sum_{k=0}^n \text{ways to select k with 1-coordinate} \times \text{ways to select n-k with 2-coordinate}$$

The summand is clearly $$\binom{n}{k} \times \binom{n}{n-k}$$ But that is $$\binom{n}{k} \times \binom{n}{k} = \binom{n}{k}^2$$

• This is actually the same as my interpretation. But, seriously, can it really be regarded as a proof? Since this method is developed using mainly words, it doesn't seem as reliable or rigorous. – Colescu Apr 4 '16 at 14:50
• @YuxiaoXie If you preferred, I could express it as the equality of the cardinalities of certain sets by providing bijections between them. I won't do that because it's tedious. – Patrick Stevens Apr 4 '16 at 14:51

To prove binomial identities it is often convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$. So, we can write e.g. \begin{align*} \binom{n}{k} = [x^k](1+x)^n\tag{1} \end{align*}

Note, this is essentially the same approach as that of @Batominovski but with a somewhat more algebraic look and feel.

We obtain \begin{align*} \sum_{k=0}^{n}\binom{n}{k}^2&=\sum_{k=0}^{n}[x^k](1+x)^n[t^k](1+t)^n\tag{2}\\ &=[x^0](1+x)^n\sum_{k=0}^{n}x^{-k}[t^k](1+t)^n\tag{3}\\ &=[x^0](1+x)^n\left(1+\frac{1}{x}\right)^n\tag{4}\\ &=[x^n](1+x)^{2n}\tag{5}\\ &=\binom{2n}{n} \end{align*}

Comment:

• In (2) we use the coefficient of operator twice according to (1)

• In (3) we use the linearity of the coefficient of operator and the rule $$[x^n]x^{-k}P(x)=[x^{n+k}]P(x)$$

• In (4) we use the substitution rule and apply $x\rightarrow \frac{1}{x}$ \begin{align*} P(x)&=\sum_{k=0}^na_kx^k=\sum_{k=0}^nx^k[t^k]P(t)\\ P\left(\frac{1}{x}\right)&=\sum_{k=0}^na_kx^{-k}=\sum_{k=0}^nx^{-k}[t^k]P(t) \end{align*}

• In (5) we use $\left(1+\frac{1}{x}\right)^n=x^{-n}(1+x)^n$ and $[x^0]x^{-n}=[x^n]$


# Finding unknowns in basic vectors

Given that $u = 3i + 2j$ and $v = 2i + \lambda j$ determine $\lambda$ such that:

(a) $u$ and $v$ are at right angles (this means perpendicular I presume?)

(b) $u$ and $v$ are parallel

Whilst working out the first one, I obtained the value of -3 for $\lambda$. However I am not sure whether this is correct and am not quite sure how to tackle question (b) from this exercise.

What I can't seem to grasp is, how on earth does the value of lambda determine whether these points on the XY plane are parallel or not?

Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel?

• HINT: If two vectors are perpendicular, then what will their dot product equal? Similarly, if two vectors are parallel, then what will their dot product equal? – Mufasa Jan 27 '15 at 23:05
• @Mufasa Well, if they're perpendicular than I presume their dot product will be equal to 0, however when parallel, -1 or 1? – Juxhin Jan 27 '15 at 23:06
• Excuse me, I skipped a few steps when I gave you that answer. Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel? – Juxhin Jan 27 '15 at 23:15
• @Mufasa - That is indeed what is confusing me haha, I'm having a mental tug of war between 180 degrees or 0 degrees – Juxhin Jan 27 '15 at 23:20
• I do not think so, wouldn't the former be -1 and latter 1? – Juxhin Jan 27 '15 at 23:21

$\lambda$ is simply the scalar value of the $\textbf{j}$-component of the vector $\textbf{v}$. The only thing that the question asks of you, is to determine the values of $\lambda$ which will satisfy the conditions in (a) and (b) respectively.

For a):

We know that two vectors are perpendicular if the angle between them are $90\unicode{xb0}$. Where can we use this fact? Clearly the DOT PRODUCT must spring to mind, since we know the dot product of two vectors, $u$ and $v$ are given by:

$$u \cdot v = ||u|| \ ||v|| \ \cos{\theta}$$

Now we know $\theta=$ $90\unicode{xb0}$ and $\cos$($90\unicode{xb0}$)$=0$, so we must have $$u \cdot v = 0$$

We thus need to find $\lambda$ such that \begin{align}u\cdot v &= 0 \\ \therefore (3)(2)+ (2)(\lambda) &=0 \\ \therefore 2\lambda &= -6 \\ \therefore \lambda &= -3 \end{align}

For b):

Two vectors are parallel to one another if their CROSS PRODUCT is zero.

Thus we need to find $\lambda$ such that \begin{align} u \times v &= 0 \\ \therefore \begin{vmatrix}i & j \\ 3 & 2\\ 2 & \lambda\end{vmatrix} &= 0 \\ \therefore 3\lambda - 4 &= 0 \\ \therefore 3\lambda &= 4 \\ \therefore \lambda &= \frac{4}{3}\end{align}

• Thank you very much Dillon, I'm simply trying to understand every corner of the question and answer prior to accepting. – Juxhin Jan 27 '15 at 23:32
• Which parts do you still feel uncertain about? I will try and edit the answer accordingly to try and explain :) – user860374 Jan 27 '15 at 23:35
• Alright thanks again for your edit. I wanted to also ask if the cross-product is still equal to zero in 3-dimensional space? Otherwise I am able to both understand and visualize the scenario in 2-dimensional space fairly well. – Juxhin Jan 27 '15 at 23:35
• Yes it is still equal to zero in 3-dimensional space :) . – user860374 Jan 27 '15 at 23:36
• Also, the reason for this also being valid in 3-dimensional space, is because we are using a geometrical interpretation of both the dot product and cross product. This geometrical interpretation can be used for any dimension up until $\mathbb{R}^3$ after that we rely on a more "theoretical" approach, which we do not need to cover right now. – user860374 Jan 27 '15 at 23:42

The following identity of the dot product is useful

$$u.v = ||u||\,||v||\cos(\theta).$$

• I've amended a small part going into a bit more detail as to why my results are slightly confusing me, however I shall add my comment in here to grab your attention directly. "Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel?" – Juxhin Jan 27 '15 at 23:17


Q&A

# How to show if a set is simply connected?

+6
−0

How do I show that the following set is not simply connected?

$M= \mathbb{R}^3\setminus\{(x,y,z)\in \mathbb{R}^3|y=z=1\}$

I am aware that if a set has a hole, then it isn't simply connected. But how can I show this? In the example I gave, I know that the set isn't simply connected because there is a "hole" which is the line y=z=1, basically there exists a loop on $\mathbb{R}^3$ which "contains" this hole. But I don't know how to formalize this.

An example of the loop would be something like. $\left( \begin{array}{c} x\\1+\cos(\phi)\\1+\sin(\phi) \end{array} \right),\phi\in[0,2\pi[$

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

+4
−0

Probably one of the simpler ways of establishing this is reducing it to a case where you already know the answer. I will assume that you have proven at some point that the circle, $\mathbb S^1$, is not simply connected. Simple connectedness is a topological property so it is preserved by homeomorphism. Another important and useful fact about simply connected spaces is that a retract of a simply connected space is simply connected. The proof outline is then simply to show that the circle is a retract of $M$ and thus $M$ can't have been simply connected.

A retraction of a space $X$ onto a subspace $A \subseteq X$ is a continuous function $r : X \to A$ such that $\forall a \in A. r(a) = a$. We say that $A$ is a retract of $X$ if there exists a retraction $X \to A$.

The function $r(x,y,z) = ((y-1)^2 + (z-1)^2)^{-\frac{1}{2}}(0, y, z)$ is a retraction $M \to \mathbb S^1$ where we're identifying $\mathbb S^1$ with unit circle in the $yz$-plane centered at $(0, 1, 1)$. You should prove that this is a continuous function. (It clearly would not be continuous if we didn't omit the line $y = z = 1$.) Intuitively, this projects to the $yz$-plane and then maps each point on a ray in the $yz$-plane starting at $(0, 1, 1)$ to the corresponding point at unit distance from $(0, 1, 1)$. To be nitpicky, you should also prove that the subset of $M$ we were calling $\mathbb S^1$ is, in fact, homeomorphic to the circle. Alternatively, you can just directly prove that this subspace is not simply connected.

See these lecture notes for discussion and proofs of the subresults, though I recommend attempting to prove them yourself first if you haven't already.

Why does this post require moderator attention?
You might want to add some details to your flag.

#### 1 comment

Thank you for the lecture notes. kreijstal 30 days ago

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!


# What is the optimal way to solve the following optimization problem

You are given a function $$F$$, which can take one or more positive integer operands. Let $$L=\{a_1,a_2\ldots a_n\}$$.

We need to compute the function $$F(L)$$ using the least number of transformations/steps. Computing $$F(L)$$ is not the main goal. Computing it in the fastest possible way is the goal.

Following rules are allowed to be applied to this function.

--> If $$L$$ has only one element say $$x$$, then the $$F(L) = x$$;

--> If all elements in $$L$$ are even then $$F(L)=2F(L')$$, where $$L'$$ contains all elements in $$L$$ divided by 2. For example $$F(4,8,10)= 2*F(2,4,5)$$

--> If some elements in $$L$$ are odd and some are even, then $$F(L)=F(L')$$, where $$L'$$ has same odd elements as L, but the even elements in $$L$$ are made odd by removing the factor 2. For example $$F(6,9,20)=F(3,9,5)$$

--> If any of the integers in the list is 1, then $$F(L)=1$$;

--> If two or more integers in $$a_1,a_2\ldots a_n$$ are same, then we can remove the copies and keep only one of them. For example $$F(4,5,6,5)=F(4,5,6)$$

--> For any pair $$a_i$$ and $$a_j$$ where($$a_i), you can replace $$a_j$$ with $$a_j-a_i$$

$$F(4,7,8)=F(4,(7-4),8)=F(4,3,8)$$

It is easy to notice that this process terminates, because at each step, either the value one of the operands is reduced or the number of operands is reduced.

Given $$L$$ can there be a DP to optimize the number of operations needed to solve $$F$$, or is there a greedy strategy that is optimal or constant approximate of the optimal?

For example

Consider the following instance

$$F(15,30,45)$$

Depending on your choice, you can compute this value in several ways.

\begin{align*} F(15,30,45)& =F(15,30,45-30)=F(15,30,15)\\ & =F(15,30)\\ & =F(15,30-15)=F(15,15)\\ & =F(15)=15 \end{align*}

which takes 4 iterations.

Another way is

\begin{align*} F(15,30,45)& =F(15,30-15,45)=F(15,15,45)\\ & =F(15,45)\\ & =F(15,45-15)=F(15,30)\\ & =F(15,30/2)=F(15,15)\\ & =F(15)=15 \end{align*}

which takes 5 iterations.

I want an algorithm which finds the optimal solution.

• If $L=\{2,2k\}$, then exactly $k$ steps are required. Since every problem instance in this family can be encoded in $O(\log k)$ bits, that means any solution that generates a complete sequence of operations takes time at least exponential in the problem size in the worst case. – j_random_hacker Dec 12 '18 at 12:02
• Nice observation, Reminds me that I missed another operation. I am adding one special operation now. If $L={a_1,a_2\ldots a_n}$ and all are even, then $F(L)=2F({a_1/2,a_2/2\ldots a_n/2})$ – Vk1 Dec 12 '18 at 12:06
• That doesn't improve things for $L=\{3, 3k\}$. – j_random_hacker Dec 12 '18 at 12:35
• It does improve things. case 1:- 3k is even, Then F(L)=F(3,3k/2) case 2:- 3k is odd, then F(L)=F(3,3k-3)=F(3,(3k-3)/2) Now at least one of the operands reduces by half after 2 iterations. – Vk1 Dec 12 '18 at 13:47
• You're right, it's more helpful than I first thought, especially since you can always choose to halve the largest operand in 2 steps. This means that for $n$ integers having maximum value $k$, the time bound is now at most $O(n\log k)$, since each element can be reduced to 1 in at most $2\log_2 k$ steps. – j_random_hacker Dec 12 '18 at 14:04

Yes, there is a simple way to compute $$F$$.

Proposition: $$F(a_1,a_2\ldots a_n)=\gcd (a_1,a_2\ldots a_n)$$, where $$\gcd$$ is the greatest common divisor.

Proof: I will leave it as an exercise. Hint, the replacement of $$a_j$$ by $$a_j-a_i$$ whenever $$a_i is very powerful.

Here is a simple and efficient algorithm to compute $$F(a_1,a_2\ldots a_n)$$.

• Let $$r = a_1$$.
• Iterate $$i$$ through $$2,3, \cdots, n$$, setting $$r$$ to $$\gcd(a_i, r)$$. Here $$\gcd(a_i,r)$$ can be computed by any of your favorite GCD algorithms such as Euclidean algorithm or binary GCD algorithm.
• return $$r$$.
• Good observation, it is emulating binary GCD algorithm. I know its computing the gcd. The question is not about what it is trying to compute. The question is to find the optimal way to compute it. More specifically an instance optimal way to compute it. – Vk1 Dec 12 '18 at 14:10
• I would recommend that $F$ be removed from the question. You are talking about how to reach the accept states of $L$ in least number of steps. Once you have mentioned the optimal way to compute it, it becomes rather misleading since the nature explanation of "it" is the value of $F$. Instead, please define $S$(run), the number of steps for a run. Add one to $S$(run) for each step. Do not mention $F$ at all. – Apass.Jack Dec 12 '18 at 17:47


Home > Numerical Analysis > Finite difference Method for 1D Laplace Equation

## Finite difference Method for 1D Laplace Equation

I will present here how to solve the Laplace equation using finite differences. First let’s look at the one dimensional case:

$\displaystyle \begin{cases} -u''=1 & \text{ in }[0,1] \\ u(0)=u(1)=0. \end{cases}$

In this case it is easy to solve explicitly the differential equation and get that ${\displaystyle u(x)=\frac{1-x^2}{2}}$. However, sometimes it is not so easy or not possible at all to find an explicit solution, and there are some numerical methods which can be used to approximate our solution. In the one dimensional case, or in the case of a rectangular domain or a box in superior dimensions, the method of finite differences can be sucessfully used.

Pick a step ${h=1/N}$, where ${N}$ is a positive integer. Then we use the approximations ${u'(x)\simeq \displaystyle \frac{u(x+h)-u(x)}{h}}$ and ${\displaystyle u''(x) \simeq \frac{u(x+h)+u(x-h)-2u(x)}{h^2}}$.

We replace our function ${u}$ by a discretization ${u_k=u(k/N)}$ and the derivative ${u''}$ becomes ${-2u_i+u_{i+1}+u_{i-1}}$. The boundary conditions translate to ${u_0=u_N=0}$. Now on the points situated in the interior of ${(0,1)}$ we have the relations:

$\displaystyle \frac{1}{h^2}(2u_i-u_{i+1}-u_{i-1})=1,\ i=1..N-1,$

and these translate to a system of equations

$\displaystyle Au=f$

where ${A}$ is an ${N-1 \times N-1}$ matrix

$\displaystyle A=\frac{1}{h^2}\begin{pmatrix} 2 & -1 & 0 &\cdots & 0 \\ -1 & 2 & -1 & \cdots & 0 \\ 0& -1 & 2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 2 \end{pmatrix}=\frac{1}{h^2}\text{tridiag}(-1,2,-1)$

and ${f}$ is a column vector of length ${N-1}$ of ones. Since ${A}$ is invertible (a recurrence relation can be found for ${\det(A)}$) the presented system has a unique solution, which will be the desired approximation by finite differences of our function.

We can change the boundary conditions to ${u(0)=a}$ and ${u(1)=b}$ by changing ${f_1}$ and ${f_{N-1}}$ using the formulas

$\displaystyle f_1=1+N^2a,\ f_{N-1}=1+N^2 b.$

In a similar way we can solve numerically the equation

$\displaystyle \begin{cases} -u''=f & \text{ in }[0,1] \\ u(0)=u(1)=0. \end{cases}$

where ${f}$ is, for example, an arbitrary continuous function.

Below I present a simple Matlab code which solves the initial problem using the finite difference method and a few results obtained with the code.

function u = laplacefd1(n); x=linspace(0,1,n+1); A=sparse(diag(2*ones(n-1,1))+diag((-1)*ones(n-2,1),1)+diag((-1)*ones(n-2,1),-1)); left=0; right=0; f=ones(n-1,1); f(1)=f(1)+(n)^2*left; f(n-1)=f(n-1)+(n)^2*right; u=(n)^2*A\f;

Different boundary conditions

Dirchlet boundary conditions

Categories: Numerical Analysis
1. October 7, 2016 at 11:24 pm

where do these formulas: \displaystyle f_1=1+N^2a,\ f_{N-1}=1+N^2 b come from??
is this based from solving \displaystyle \frac{1}{h^2}(2u_i-u_{i+1}-u_{i-1})=1, for i=1 and i=N ??
how did you get a “+N^2” in each formula?

• October 8, 2016 at 1:07 am

Yes, the formulas come from the discrete Laplacian formulation. Note that $h=1/N$ so $1/h^2 = N^2$. That’s why you get an $N$ squared in the formulas.

To see more precisely why it is enough to change $f_1,f_N$ just write the discrete Laplacian for $u_0,u_1,u_2$. You’ll get $N^2(2u_1-u_0-u_2) = f_1$. Now if you pass the term with $u_0$ to the other side you see that you have the same matrix equation like in the zero boundary condition, but with different $f$.

2. February 8, 2017 at 2:28 pm

Hi, in your first equation you have the problem -u”=1 with boundary conditions u(0) = u(1) = 0. You state that the analytical solution to this problem is 0.5*(1-x^2).

Now when I insert the boundary condition into your analytical solution, I get u(0) = 0.5*(1-0^2) = 0.5.

Are you sure the boundary conditions in your problem are correct?

• February 22, 2017 at 4:51 pm

You are right. The solution in my post is, in fact, on the interval [-1,1]. In order for it to work on [0,1] you need to change it a bit. I guess 0.5*x*(x-1) should work instead.


It is currently 19 Nov 2017, 23:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# How many different positive integers are factors of 441

Author Message
TAGS:

### Hide Tags

Intern
Joined: 27 Mar 2012
Posts: 15

Kudos [?]: 47 [2], given: 1

How many different positive integers are factors of 441 [#permalink]

### Show Tags

13 Apr 2012, 02:42
2
KUDOS
6
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

68% (00:46) correct 32% (01:04) wrong based on 436 sessions

### HideShow timer Statistics

How many different positive integers are factors of 441

A. 4
B. 6
C. 7
D. 9
E. 11

Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Apr 2012, 06:37, edited 2 times in total.
Edited the question

Kudos [?]: 47 [2], given: 1

Intern
Joined: 18 Sep 2011
Posts: 20

Kudos [?]: 1 [0], given: 6

GMAT Date: 11-09-2011
Re: How many different positive integers are factor of 441 ? [#permalink]

### Show Tags

13 Apr 2012, 04:55
Yeah you are absolutely right!

Powers of prime factors adding one on to it and multiply

(2+1)*(2+1)=9

Posted from my mobile device

Kudos [?]: 1 [0], given: 6

Manager
Joined: 06 Feb 2012
Posts: 91

Kudos [?]: 56 [0], given: 16

WE: Project Management (Other)
Re: How many different positive integers are factor of 441 ? [#permalink]

### Show Tags

13 Apr 2012, 06:27
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?

Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....

Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2

so 441 = 3^2 * 7^2

Note that I found those links interesting:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)
_________________

Kudos is a great way to say Thank you...

Kudos [?]: 56 [0], given: 16

Math Expert
Joined: 02 Sep 2009
Posts: 42254

Kudos [?]: 132741 [8], given: 12360

Re: How many different positive integers are factors of 441 [#permalink]

### Show Tags

13 Apr 2012, 06:43
8
KUDOS
Expert's post
5
This post was
BOOKMARKED
sugu86 wrote:
How many different positive integers are factors of 441

A. 4
B. 6
C. 7
D. 9
E. 11

Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?

MUST KNOW FOR GMAT:

Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION.

How many different positive integers are factors of 441
A. 4
B. 6
C. 7
D. 9
E. 11

Make prime factorization of 441: 441=3^2*7^2, hence it has (2+1)(2+1)=9 different positive factors.

Hope it helps.
_________________

Kudos [?]: 132741 [8], given: 12360

Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 222

Kudos [?]: 63 [0], given: 16

Schools: Johnson '15
Re: How many different positive integers are factor of 441 ? [#permalink]

### Show Tags

14 Apr 2012, 01:00
GreginChicago wrote:
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?

Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....

Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2

so 441 = 3^2 * 7^2

Note that I found those links interesting:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)

is 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2

should it not be 7 * 7 * 3 * 3
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Kudos [?]: 63 [0], given: 16

Math Expert
Joined: 02 Sep 2009
Posts: 42254

Kudos [?]: 132741 [0], given: 12360

Re: How many different positive integers are factor of 441 ? [#permalink]

### Show Tags

14 Apr 2012, 02:13
harshavmrg wrote:
GreginChicago wrote:
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?

Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....

Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2

so 441 = 3^2 * 7^2

Note that I found those links interesting:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)

is 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2

should it not be 7 * 7 * 3 * 3

GreginChicago is saying exactly the same thing!

Check for a solution here: how-many-different-positive-integers-are-factor-of-130628.html#p1073364
_________________

Kudos [?]: 132741 [0], given: 12360

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1851

Kudos [?]: 2713 [0], given: 193

Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: How many different positive integers are factors of 441 [#permalink]

### Show Tags

27 Mar 2014, 21:02
$$441 = 21^2$$

$$= (7 * 3)^2$$

$$= 7^2 . 3^2$$

Prime factors = (2+1) * (2+1) = 9

_________________

Kindly press "+1 Kudos" to appreciate

Kudos [?]: 2713 [0], given: 193

Non-Human User
Joined: 09 Sep 2013
Posts: 15643

Kudos [?]: 283 [0], given: 0

Re: How many different positive integers are factors of 441 [#permalink]

### Show Tags

15 Aug 2016, 04:28
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 283 [0], given: 0

Re: How many different positive integers are factors of 441   [#permalink] 15 Aug 2016, 04:28
Display posts from previous: Sort by


GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2018, 23:24

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Machine X can complete a job in half the time it takes Machine Y to co

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50002
Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

03 Feb 2015, 09:47
1
12
00:00

Difficulty:

75% (hard)

Question Stats:

64% (02:51) correct 36% (02:55) wrong based on 290 sessions

### HideShow timer Statistics

Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?

A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10

Kudos for a correct solution.

_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8397
Location: Pune, India
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

03 Feb 2015, 23:08
6
6
Since rates are additive, using rates is usually easier.

"Machine X can complete a job in half the time it takes Machine Y to complete the same job,"
So machine X's rate is twice machine Y's rate. Rate of X:Rate of Y = 2:1

"Machine Z takes 50% longer than Machine X to complete the job"
Time taken by Z : Time taken by X = 3:2 so Rate of Z: Rate of X = 2:3

Rate of X : Rate of Y : Rate of Z = 6:3:4

Rate of X+Z : Rate of Y+Z = 10 : 7
Time taken by X+Z : Time taken by Y+Z = 7:10

_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

##### General Discussion
Intern
Joined: 25 May 2014
Posts: 46
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

03 Feb 2015, 10:30
2
2
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?

A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10

Kudos for a correct solution.

Let time required by Y to complete the work = t
then time taken by X will be = $$\frac{t}{2}$$
and time taken by Z = $$t/2 + t/4$$ ----> $$\frac{t}{4}$$ as Z takes 50% longer then X
Now Rate for X = $$\frac{2}{t}$$
Rate for Y = $$\frac{1}{t}$$
Rate for Z = $$\frac{4}{3t}$$

amount of time it will take Machines X and Z to complete the job = $$1/[2/t + 4/3t]$$
i.e. $$\frac{3t}{10}$$
the amount of time it will take Machines Y and Z to complete the job = $$1/[1/t + 4/3t]$$
i.e. $$\frac{3t}{7}$$
Thus ratio = $$\frac{3t}{10}$$ / $$\frac{3t}{7}$$ = 7:10
[D]
_________________

Never Try Quitting, Never Quit Trying

Intern
Joined: 24 Jan 2014
Posts: 35
Location: France
GMAT 1: 700 Q47 V39
GPA: 3
WE: General Management (Advertising and PR)
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

Updated on: 03 Feb 2015, 23:29
1
2
Hi Bunuel,

Solutioning this exercise can be facilitated by using a R T W (rate time work) table:

We translate the exercise into the table:

R T W
X t/2 1
Y t 1
Z (t/2*3/2 =3t/4) 1

From this table we find the rates

Rx = 2/t
Ry = 1/t
Rz = 4/3t

The Q is what is the ratio of (Tx + Ty) / (Ty + Tz)

Rx + Ry = 2/t + 4/3t = 6/3t+4/3t = 10/3t
Ry+Rz = 1/t + 4/3t = 3/3t + 4/3t = 7/3t

The (10/3t)/(7/3t) = 10/7 then the work ratios is 10 to 7

Since Time Ratio is the inverse of work, the the answer is 7 to 10

Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?

A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10

Kudos for a correct solution.

_________________

THANKS = KUDOS. Kudos if my post helped you!

Napoleon Hill — 'Whatever the mind can conceive and believe, it can achieve.'

Originally posted by TudorM on 03 Feb 2015, 22:53.
Last edited by TudorM on 03 Feb 2015, 23:29, edited 2 times in total.
Manager
Joined: 17 Dec 2013
Posts: 58
GMAT Date: 01-08-2015
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

04 Feb 2015, 04:14
1
x=0,5t
y=t
z=0,75t

t=4hours so we get:
x=2
y=4
z=3

x+z=1/2+3/4=5/6 -> they need 6/5 hours
y+z=1/4+1/3=7/12 -> they need 12/7 hours

ratio is 6/5 divided by 12/7, or multiplied by 7/12 -> we get 7/10
Math Expert
Joined: 02 Aug 2009
Posts: 6961
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

04 Feb 2015, 05:46
ans D..
let time taken by X =x,
so by Y =2x,
and by z = 1.5x...
time taken by X and Z = 1/( 1/x+1/1.5x)... = 1.5 X^2/2.5X =3X^2/5X
time taken by Y and Z = 1/( 1/2x+1/1.5x)... = 3 X^2/3.5X...= 6X^2/7X
ratio = (3X^2/5X)/ (6X^2/7X)= 7/10
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12685
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

04 Feb 2015, 13:48
4
Hi All,

These types of rate questions can be solved in a few different ways. Since the prompt discusses how two machines will work together on a task, you can use the Work Formula. By TESTing VALUES, we can come up with a simple example that will answer the given question.

We're told that:
1) Machine X can do a job in HALF the time that it takes Machine Y
2) Machine Z takes 50% longer to the do the job than Machine X

Since Machine X appears in both 'facts', we want to TEST a VALUE for X that is easily doubled AND halved....

X = 2 hours to complete the job

With that value in place, we can figure out the other 2 values....
Y = 4 hours to complete the job
Z = 3 hours to complete the job

We're asked for the ratio of the time it takes (X and Z) working together vs. the time it takes (Y and Z) working together.

The Work Formula: (A)(B)/(A+B)

For Machines X and Z, we have: (2)(3)/(2+3) = 6/5 hours to complete the job
For Machines Y and Z, we have: (4)(3)/(4+3) = 12/7 hours to complete the job

The ratio 6/5 : 12/7 needs to be "clean up" a bit. Since we're dealing with fractions, we can use common denominators....

6/5 : 12/7

42/35 : 60/35

Now we can eliminate the denominators...

42 : 60

And reduce the ratio....

7 : 10

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Math Expert
Joined: 02 Sep 2009
Posts: 50002
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

09 Feb 2015, 04:51
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?

A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

In this ratio problem, it's helpful to begin with the ratio of each machine's RATES as opposed to times, as rates are additive when machines are working together. So if Machine X as a rate that's twice as fast as Machine Y, their rates have the ratio X:Y = 2:1. And the ratio of X to Z is 3:2, as X works 50% faster (and therefore would accomplish 3 jobs in the time that it takes for Z to complete 2). So to find a common three-way ratio, you can use X as the "anchor", and make the ratio of rates:

X : Y : Z = 6 : 3 : 4

So X and Y working together would complete 10 jobs in the time that it would take Y and Z working together to complete 7 jobs, for a ratio of 10 : 7 in their respective rates. But since the question asks for times, not rates, you'll need to invert the ratio to 7 : 10, answer choice D.
_________________
VP
Joined: 07 Dec 2014
Posts: 1104
Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

20 Nov 2015, 18:32
let 2,4,3=respective times of X,Y,Z
1/2,1/4,1/3=respective rates of X,Y,Z
X+Z rate/Y+Z rate=(5/6)/(7/12)=10:7
inverting, X+Z time/Y+Z time=(6/5)/(12/7)=7:10
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3896
Location: United States (CA)
Re: Machine X can complete a job in half the time it takes Machine Y to co  [#permalink]

### Show Tags

28 Mar 2018, 10:39
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?

A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10

We can let the time of machine Y to complete the job = y, so the time of machine X to complete the job is 0.5y = (½)y and the time of machine Z to complete the job = (1.5)(0.5y) = 0.75y = (3/4)y. .

Since rate is inverse of time, the rate of Y = 1/y, the rate of X = 1/[(½)y] = 2/y and the rate of Z = 1/[(3/4)y] = 4/(3y).

Thus, the amount of time it will take Machine X and Z to complete the job is

1/[2/y + 4/(3y)] = 1/[6/(3y) + 4/(3y)] = 1/[10/(3y)] = 3y/10

Similarly, the amount of time it will take Machine Y and Z to complete the job is

1/[1/y + 4/(3y)] = 1/[3/(3y) + 4/(3y)] = 1/[7/(3y)] = 3y/7

Therefore, the ratio is (3y/10)/(3y/7) = (1/10)/(1/7) = 7/10.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: Machine X can complete a job in half the time it takes Machine Y to co &nbs [#permalink] 28 Mar 2018, 10:39
Display posts from previous: Sort by


# Finding the length of a side and comparing two areas from a given figure

The following figure came in two national-exam paper (different question in each paper).

First Question:

What is the length of $$AE$$?

Choices:

A) $$\frac{5}{4}$$ cm

B) $$\frac{\sqrt{41}}{5}$$ cm

C) $$\frac{7}{5}$$ cm

D) $$\frac{15}{8}$$ cm

Second Question

Compare:

First value: area of $$\triangle ABC$$

Second value: area of $$\triangle CDE$$

Choices:

A) First value > Second value

B) First value < Second value

C) First value = Second value

D) Given information is not enough

My Attempt:

Finding the equation of $$AC$$ and $$BD$$ in order to find the coordinates of E:

We can assume that $$C$$ is the origin $$(0,0)$$ and $$A$$ is $$(3,4)$$, then the equation of $$AC$$ is $$y=\frac{4}{3}x$$, and the equation of $$BD$$ is $$y=-\frac{4}{5}x+4$$.

Next, solve the system of equations for $$x$$ and $$y$$, we end up with the coordinates of $$E(\frac{15}{8},\frac{5}{2})$$.

To solve the first question, we can use the distance between two points formula,

therefore $$AE=\sqrt{(3-\frac{15}{8})^2+(4-\frac{5}{2})^2}=\frac{15}{8}$$ cm.

Hence D is the correct choice.

To solve the second question, we can use the area of triangle formula,

area of $$\triangle ABC=\frac{1}{2}\times 3\times 4=6$$ cm$$^2$$.

area of $$\triangle CDE=\frac{1}{2}\times 5\times \frac{5}{2}=6.25$$ cm$$^2$$.

Hence B is the correct choice.

In the exam, calculators are not allowed. It is simple to determine to coordinates of $$E$$ without a calculator, but it will take time!

For the first question, I think there is a good way to solve, maybe $$BD$$ divides $$AC$$ into two parts ($$AE$$ and $$CE$$) in a ratio, this ratio can be calculated somehow.

For the second question, I think also a good way to think, like moving $$D$$ will make $$\triangle CDE$$ bigger or smaller, while $$\triangle ABC$$ will be unchanged. But this has a limitation since the areas $$6$$ and $$6.25$$ cm$$^2$$ are close, it will not be obvious to us.

If we have $$75$$ seconds (on average) to solve each of these questions, then how can we solve them? [remember that in some questions in the national exam, we do not need to be accurate $$100$$%].

Any help will be appreciated. Thanks.

• $AB\perp BC$? $AC\perp BD$? $AB||DC$? – Michael Rozenberg Jul 17 at 13:03
• @MichaelRozenberg Yes. – Hussain-Alqatari Jul 17 at 13:05
• If so, this quadrilateral does not exist because we need $4^2=3\cdot5,$ which is wrong. – Michael Rozenberg Jul 17 at 13:07
• @MichaelRozenberg $AB\perp BC$, $AB || DC$. But $AB$ is not $\perp BC$ – Hussain-Alqatari Jul 17 at 13:10
• @MichaelRozenberg Yes - No - Yes. – Hussain-Alqatari Jul 17 at 13:19

Triangles $$ABE$$ and $$CDE$$ are similar, with ratio $$3:5$$.

Hence $$AE=\frac{3}{8}AC=\frac{3}{8}\cdot 5=\frac{15}{8}$$

By the same similarity, if $$h$$ is the height of triangle $$CDE$$ from $$E$$ to $$CD$$, then $$h=\frac{5}{8}BC=\frac{5}{8}\cdot 4=\frac{5}{2}$$ hence the area of triangle $$CDE$$ is $$\frac{1}{2}\cdot CD\cdot h=\frac{1}{2}\cdot 5\cdot \frac{5}{2}=\frac{25}{4}=6.25$$ whereas the area of triangle $$ABC$$ is $$6$$.

• quasi your way looks very elegant. if I may ask how did you figure out $AE = \frac{3}{8}AC$ directly with out setting up a proportion ? Is there some nice trick here? – rsadhvika Jul 17 at 13:24
• @rsadhvika:$\;$Since $AE:EC=3:5$, it follows that $$AE:AC=AE:(AE+EC)=3:(3+5)=3:8$$ – quasi Jul 17 at 13:29
• Oh nice nice! With that in mind we can directly do: $AE = \frac{3}{8}AC$ and $EC = \frac{5}{8}AC$. I see now.. Thank you so much for explaining this you're awesome :) – rsadhvika Jul 17 at 13:31

For first question you may use similar triangles - $$\triangle AEB$$ and $$\triangle CED$$: $$\dfrac{3}{5} = \dfrac{x}{5-x}$$

For the first part use similarity of triangles to simplify computation.

You have similar triangles $$ABE$$ and $$CDE$$ with ratio of sides being $$3$$ to $$5$$

You also know $$AC=5$$ so you find $$AE=5\times \frac {3}{8}=\frac {15}{8}$$ which is choice $$D$$

Your have solved the second part correctly.


# Why can we interchange summations?

Suppose we have the following

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij}$$

where all the $a_{ij}$ are non-negative.

We know that we can interchange the order of summations here. My interpretation of why this is true is that both this iterated sums are rearrangements of the same series and hence converge to the same value, or diverge to infinity (as convergence and absolute convergence are same here and all the rearrangements of an absolutely convergent series converge to the same value as the series).

Is this interpretation correct. Or can some one offer some more insightful interpretation of this result?

Please note that I am not asking for a proof but interpretations, although an insightful proof would be appreciated.

• Another interpretation would be to use Tonnelli's theorem for counting measure. – TZakrevskiy Aug 13 '13 at 17:44
• The double sum is not a rearrangement of a single sum. At least, it's not of the form $\sum_{i=1}^\infty a_{\sigma(i)}$ where $\sigma$ is a permutation of the positive integers. However, let $I$ be your iterated sum and $J$ be the sum with the order interchanged. You can first show that $J\le I$. Then you can show that for any $\epsilon>0$, you have $J>I-\epsilon$. (This is mostly done by "lopping off small tails". Note each row and column is summable) – David Mitra Aug 13 '13 at 18:27
• We can show that it is rearrangement if you consider a bijection between $\mathbb{N}\times\mathbb{N}$ and $\mathbb{N}$. – Vishal Gupta Aug 14 '13 at 5:15

This isn't a proof, but perhaps can give you the insight you are looking for. Any nondecreasing sequence converges to its (possibly infinite) supremum. Thus a series of nonnegative terms converges to the supremum of its partial sums and interchanging the order of summation doesn't affect the value of the supremum: there is no accidental cancellation of terms of opposite sign.

• Good thought. Then we only need to apply decoupling to interchange summations in this case. – Chival Sep 27 '14 at 20:13
• (+1)Consider the real valued function $f(a_{1},a_{2},...a_{n})$ then does the value of $\sum_{a_{1}} \sum_{a_{2}}...\sum_{a_{n}} f(a_{1},a_{2},...a_{n})$ depends upon the order of sum notions. In other words I want to know whether there is any theorem which talks about this interchanging of order of sum notions ? – NewBornMATH Apr 9 '19 at 15:11

The double sum you are asking about can be considered to be the sum of all terms of the infinite array of numbers $a_{ij}$:

$$\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1j} & \cdots\\ a_{21} & a_{22} & \cdots & a_{2j} & \cdots\\ \vdots& \vdots & &\vdots\\ a_{i1} & a_{i2} & \cdots & a_{ij} & \cdots\\ \vdots & \vdots & & \vdots \\ \end{pmatrix},$$

where the sum is taken with the particular order of first adding all elements of particular rows and then adding the resulting "row totals". The result you are citing claims that if all $a_{ij}$ is nonnegative, then it matters not in which order you add the entries of the infinite matrix, be that rows first and then row totals or columns first and then column totals.

I can't say your interpretation is right on the mark, because the issue is that even though one adds the entries of the same infinite array, the order of summation may result in summing the terms of different series in actuality. As the other answers point out this result holds essentially because we have no terms diminishing the total sum. Saying "both this iterated sums are rearrangements of the same series and hence converge to the same value, or diverge to infinity" really does not make any emphasis on this advantage, which I believe is sweeped under the phrase "same series".

I'd like to mention yet another (measure theoretical) interpretation of this result, which I recently encountered in Rudin's Real and Complex Analysis (p. 23). In the book Rudin gives this result as a corollary of Lebesgue Monotone Convergence Theorem applied to series of functions.

Theorem: Let $X$ be a measure space. If $\forall n: f_n:X\to [0,\infty]$ is measurable, then

$$\int_X \sum_n f_n d\mu= \sum_n \int_X f_n d\mu (\ast).$$

Corollary: Let $X:=\{x_1,x_2,...,x_n,...\}$ be a countable set and $\mu:\mathcal{M}_X:=\mathcal{P}(X)\to[0,\infty]$ be the counting measure:

$$\mu(E):= \begin{cases} |E|&, \mbox{ if} |E|<\infty\\ \infty&, \mbox{ if} |E|=\infty \end{cases}.$$

If $\forall i,j:a_{ij}\geq0$, then

$$\sum_i \sum_j a_{ij}=\sum_j \sum_i a_{ij}.$$

Proof:

• Set $\forall j: f_j:X\to[0,\infty], f_j(x):=\sum_i a_{ij}\chi_{\{x_i\}}(x)$; and $\forall i: \bar{f_i}:X\to[0,\infty], \bar{f_i}(x):=\sum_j a_{ij}\chi_{\{x_i\}}(x).$ Then $\sum_j f_j=\sum_i \bar{f_i}$. Indeed, let $x_{i_0}\in X$. Then

$$\sum_j f_j(x_{i_0})= \sum_j \sum_i a_{ij} \chi_{\{x_i\}}(x_{i_0})= \sum_j a_{{i_0}j},$$

and

$$\sum_i \bar{f_i}(x_{i_0})=\bar{f_{i_0}}(x_{i_0})=\sum_j a_{{i_0}j}.$$

• Observe that the (inner) sum over $i$ is the integral of $f_j$ and the (inner) sum over $j$ is the integral of $\bar{f_i}$. Then we have:

$$\sum_j\sum_i a_{ij}= \sum_j \int_X f_j d\mu\stackrel{(\ast)}{=}\int_X \sum_j f_j d\mu=\int_X \sum_i \bar{f_i}d\mu\stackrel{(\ast)}{=}\sum_i\int_X\bar{f_i}d\mu =\sum_i \sum_j a_{ij}.$$

Now consider the interpretation induced by the above discourse, viz., the (countable) combinations of characteristic functions behave in such a way that one can concentrate nonnegative "weights" of individual points. In the above proof $f_j$ puts a single weight on each point, while $\bar{f_i}$ concentrates all the weights of the point $x_i$.

Well, I think that interpretation is a bit circular, as the reason absolute convergence allows for rearrangements without changing the limit is precisely because of this point.

The gist of it is that no cancellations due to sign are possible. I think it is very illuminating to consider the counterexample for the rearrangement theorem when you do not have absolute convergence, but still have conditional convergence (Rearrangements can lead to any limiting value!).

You can add up as many negative terms as you want until you are happy (as it does not absolutely converge, you can continue adding until you are lower than any value in the real line you desire), and likewise for the positive terms (note, if adding the rest of the positive terms gives a finite value, then actually the whole sum didn't conditionally converge, and in fact will be $-\infty$). This procedure of adding negative terms until your sum is below $L$, and then adding positive terms until the sum is above $L$, and etc. will converge because conditional convergence requires the things you are summing to go to $0$ at the end of the day.

Okay, but to the non-negative double sum. Towards proving the double sum result (since you are only considering two orderings), I suppose one cute thing you could try is start with rows (arrange the numbers on the upper right quadrant for this picture). Let's assume the row sums converge first. Now, for each row sum, take out the first term of each, and set it aside, adding them upwards (the first column sum). Note that of course, the value of the sum didn't change, because you took out values one at a time, putting them into the column sum, and at no point did anything funny happen because only finite sums were involved (the full sum, the partial column sum, and the full sum minus the finitely many elements you took out). Now continue this procedure, extracting the second column sum, and etc. At the "end" you have what you want to prove.

Two limiting procedures hidden: When you "finish" the first column sum, you have to take a limit procedure. When you "finish" all the columns, there's another limit there. The issue if you include negative terms is that the first column sum might already diverge. And even if all the column sums converge, summing the column sums might converge to an entirely different value! I think to really finish this proof idea, you should use the fact that the values you are working with vary monotonely at each step.