# $x_1 \cdot \dots \cdot x_n=1 \implies x_1 + \dots + x_n \ge n$

Suppose $x_1, \dots, x_n$ are positive real numbers such that $x_1 \cdot \ldots \cdot x_n = 1$. Prove $x_1 + \dots + x_n \ge n$.

I don't want to use the AM-GM inequality to prove this (from which this statement would follow).

I suppose induction is the way to go. The base case is true. The case for $n=2$ is true by $$x_1 + x+2 - 2 \sqrt{x_1 x_2} = (\sqrt{x_1} - \sqrt{x_n})^2 \ge 0.$$

Now suppose the statement holds for $n$, then $$x_1 \cdot \ldots \cdot x_n = 1 \implies x_1 + \dots + x_n \ge n.$$

So if $x_1 \cdot \ldots \cdot (x_n x_{n+1}) = 1$, then $$x_1 + \dots + x_n x_{n+1} \ge n.$$

Then we also have $$x_1 + \dots + x_{n-1} + x_n + x_{n+1} \ge n - x_{n}x_{n+1} + x_n + x_{n+1}.$$

Now I'm not sure where else to go from here.

• Seem similar in proving AM-GM by induction I think – Azlif Dec 21 '17 at 23:25
• @MathematicianByMistake Induction certainly works, since this is just a special case of AM-GM which is proven by induction. – B. Mehta Dec 21 '17 at 23:29

Thus, it's enough to prove that $$n-x_nx_{n+1}+x_n+x_{n+1}\geq n+1$$ or $$(x_n-1)(x_{n+1}-1)\leq0,$$ which you can assume before.

Indeed, since $\prod\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\leq0.$

• It's not entirely clear to me why this can be assumed. – B. Mehta Dec 21 '17 at 23:31
• @B. Mehta Since $\prod\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\leq0.$ – Michael Rozenberg Dec 21 '17 at 23:33
• Ah, so there's a relabeling step added earlier. – B. Mehta Dec 21 '17 at 23:34
• @B.Mehta I believe the logic is: there has to be at least one value greater-than-or-equal-to 1 and at least one less-than-or-equal-to 1. Rearrange so that these two values come last. – Addem Dec 21 '17 at 23:34
• @MichaelRozenberg Thanks, very helpful. I would like to verify this last part with there needing to be a pair with one greater and one and one less. By contradiction, suppose for every $x_i$ and $x_j$, $(x_i -1)(x_j-1) > 0$, then either all pairs are greater than one or less than one which in either case would result in a product of the $x_k$'s being greater than 1 or less than one, respectively. – user330531 Dec 21 '17 at 23:42

The inequality follows from AM-GM, indeed: $\;x_1 + \dots + x_n \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_n}\,$.
Below is a proof of AM-GM by induction, along a different idea than the ones posted already.

Lemma.   $\;(n-1)t^n+1 \ge n t^{n-1}\;$ for all $\;t \ge 1\,$.

Let $f\,(t)=(n-1)t^n - n t^{n-1} + 1\,$, then $\,f(1) = 0\,$ and $\,f'(t)=n(n-1)t^{n-2}(t-1) \ge 0\,$, so $\,f\,$ is increasing from $\,f(1)=0\,$ for $\,t \ge 1\,$ and therefore $\,f(t) \ge 0\,$ for $\,t \ge 1\,$.

(The stronger statement is true that $\,f(t) \ge 0\,$ for all $\,t \ge 0\,$. That follows from the identity $\,(n-1)t^n - n t^{n-1} + 1 = (t-1)^2 \cdot \sum_{k=1}^{n-1} k \cdot t^{k-1}\,$ which has an elementary no-calculus proof.)

Induction step.   The AM-GM inequality is homogeneous, so it can be assumed WLOG that the smallest $x_j=1\,$. Assume again WLOG that $\,j=n\,$, then the inequality reduces to:

$$x_1+x_2+ \cdots + x_{n-1} + 1 \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}} \quad\style{font-family:inherit}{\text{with}}\quad x_i \ge 1, \;i=1,2,\ldots,n-1$$

By the induction hypothesis:

$$\,x_1+\cdots+x_{n-1} \ge (n-1) \sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} \tag{1}$$

Also, it follows from the lemma with $\,t = \sqrt[n(n-1)]{x_1\cdots x_{n-1}}\ge 1\,$ that: $$\,(n-1)\cdot\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \ge n \cdot \sqrt[n]{x_1\cdot \ldots \cdot x_{n-1}} \tag{2}$$

Therefore:

$$x_1+ \ldots +x_{n-1}+1 \;\stackrel{(1)}{\ge}\; (n-1)\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \;\stackrel{(2)}{\ge}\; n \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}}$$


# What does it mean to divide a complex number by another complex number?

Suppose I have: $w=2+3i$ and $x=1+2i$. What does it really mean to divide $w$ by $x$?

EDIT: I am sorry that I did not tell my question precisely. (What you all told me turned out to be already known facts!) I was trying to ask the geometric intuition behind the division of complex numbers.

• Jul 2, 2013 at 6:23
• suitcaseofdreams.net/Geometric_division.htm Jul 2, 2013 at 6:25
• Perhaps a polar form might be a little more 'intuitive'? Jul 2, 2013 at 6:33
• While multiplication/division of complex numbers can be interpreted geometrically, I don't think it is meant to be interpreted that way. Jul 2, 2013 at 6:40
• @user1551 au contraire it is meant to be interpreted geometrically. Jul 2, 2013 at 16:40

Complex numbers' multiplication is better understood if you forget the "cartesian vector in the complex plane" analogy: $$z = a + b i \quad z \in \mathbb{C}, a, b \in \mathbb{R}$$ And in stead think in polar coordinates: $$z = r \angle \theta = r e^{i \theta} \quad z \in \mathbb{C}, r \in \mathbb{R}^+, \theta \in [0, 2 \pi)$$ Wherein $r$ is the magnitude, $\theta$ is the angle.

When multiplying it is easy: $$z w = (r_z r_w) \angle (\theta _z + \theta _w)$$ You add the angles and multiply the magnitudes.

When dividing you do what comes naturally: $$\frac z w = \left(\! \frac{r_z}{r_w} \!\right) \angle (\theta _z - \theta _w)$$ To divide means to find the difference in angles and the factor in magnitude.

It means to find another complex number $y$, such that $xy=w$. (Just as it does for real numbers.)

When you divide $a$ by $b$, you're asking "What do I multiply $b$ by in order to get $a$?".

Multiplying two complex numbers multiplies their magnitudes and adds their phases:

So when you divide a complex $a$ by a complex $b$ you are asking: "How much do I need to scale $b$ and rotate $b$ in order to get $a$? Please give me a complex number with a magnitude equivalent to how much I must scale and a phase equivalent to how much I must rotate.".

Example

Consider $\frac{1 + i}{1 - i}$. How much do we need to scale and rotate $1-i$ in order to make it the same as $1+i$?

Well, when graphed on the complex plane you can see that $1-i$ has a 45 degree clockwise rotation and a magnitude of $\sqrt{2}$. $1+i$, on the other hand, has a 45 degree counter-clockwise rotation and the same magnitude of $\sqrt{2}$.

Since the magnitudes are the same, we don't need any scaling. Our result's magnitude will be 1.

To rotate from 45 degrees clockwise to 45 degrees counter-clockwise, we must rotate 90 degrees counter-clockwise. Thus our result will have a phase of 90 degrees counter-clockwise (which is upwards along the imaginary Y axis).

Move a distance of 1 up the imaginary Y axis and you get the answer... $\frac{1 + i}{1 - i} = i$. We can confirm this by doing the multiplication: $(1-i) \cdot i = i+1$.

• wow for the gif Jul 2, 2013 at 16:39
• @Harsh I'm glad you like it. I made it (and a bunch of others) myself as part of an explanation of Grover's Quantum Search. Jul 3, 2013 at 21:32
• man... I was thinking you were very resourceful, and had taken it from somewhere... Now, to learn that you MADE IT! Salute Jul 3, 2013 at 22:01

Since multiplication can be nicely visualized as a rotation in the complex plane, you may find it helpful to think of division as a form of multiplication: $$\frac{2+3i}{1+2i} = \frac{2+3i}{1+2i}\cdot \frac{1-2i}{1-2i} = -\frac{1}{3}\cdot(2+3i)(1-2i)$$ So, instead of thinking about division, you can think of it as multiplying by the conjugate.

Geometrically, it means that the magnitudes get divided, and the angles get subtracted.

That is, imagine the complex numbers plotted in polar form.

ie: if

$w = r_1 e ^ {i \theta_1 }$

$x = r_2 e ^ {i \theta_2 }$

then

$w/x = (r_1 / r_2)e^{i(\theta_1 - \theta_2)}$

There are multiple ways by which you can describe this relation;

Since you have asked to give geometric representation so phasor form must be better to understand.

Dividing two complex numbers means to take the complex number that is of the magnitude equal to that of the division of amplitudes of the $$X$$ and $$Y$$ complex number and the phase of the new generated complex number is actually the difference of the phase between them.

Eg. $$X = 2+3i \ , \ Y = 9+3i$$

so the $$|Z|= \frac{ |X|}{|Y|} = \dfrac{\sqrt{2^2+3^2}}{\sqrt{9^2+3^2}}$$ and $$\arg(Z)= \tan^{-1}\left(\frac32\right) - \tan^{-1}\left(\frac 39 \right)$$

This $$Z$$ is also a complex number with amplitude and phase in geometric way.. but you to visualize the complex imaginary axis and real axis :D

Since $\frac wx = \frac{w\cdot x^*}{|x|^2}$, complex division is basically equivalent to multiplication of $w$ and the complex conjugate of $x$, but rescaled by the squared absolute value $|x|^2$.


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# Rapid Math

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23 Oct 2017, 01:32
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Hi Everyone,

Although GMAT doesn't test your ability to perform complicated calculations, it would be beneficial to know some calculation tricks that would save you some time.

Trick 7: Rapidly Square Any Number Ending in 5

Multiply the tens digit by the next whole number then affix the number 25
$$65^2: 6*7=42 ==> 'affix' 25 ==> 4225$$

Trick 8: Rapidly Multiply Any Two-Digit Number by 11

Write the number leaving some space between the two digits, then insert the sum of the number's two digit in between.

24*11 ==> 2 ? 4 ==> 2+4=6 ==> 264

Trick 9: Rapidly Multiply by 25 (or 0.25, 2.5, 250, etc.)

Divide the number by 4 and affix or insert Zeroes or decimal points
Example: 28*25 ==> 28/4 = 7 ==> 700 (add 2 Zeroes)

Trick 10: Rapidly Divide by 25 (or 0.25, 2.5, 250, etc.)

Remove Zeroes or decimal points then multiply the number by 4
Example: 700/25 ==> 7/25 ==> 7*4=28

Trick 24: Rapidly Square Any Two-Digit Number ending in 1

$$21^2 ==> 20^2=400 ==> Add: 20+21=41 ==> Add: 400+41= 441 ==> 21^2 = 441$$

Trick 59: Rapidly Multiply Any Three-Digit Number by 11 (advanced variation of trick 8)

Write the number leaving some space between the two digits, then insert the summations of the numbers in between (summations of 2 digits everytime; as shown below).

Example:

342*11=??
342*11= 3--2 (write the first & the last number "3"& "2")
342*11=3-62 (the sum of 4+2)
342*11=3762 (the sum of 3+4)

Source: Rapid Math Tricks & Tips. EDWARD H. JULIUS
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23 Oct 2017, 01:47
Thanx. It will help

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01 Nov 2017, 07:43
Updated with Tricks 9 & 10
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If we chase perfection we can catch excellence Vince Lombardi

GMAT Club Live: 5 Principles for Fast Math: https://gmatclub.com/forum/gmat-club-live-5-principles-for-fast-math-251028.html#p1940045
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Re: Rapid Math   [#permalink] 01 Nov 2017, 07:43
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# Find the parametric equation to the curve

Find the parametric equation for the curve.

$$x^{2}+y^{2}=10$$

I haven't learned parametric equations fully yet, so I wanted to check with you guys and see if you can confirm if I'm doing this correctly and possibly go more in depth on the problem if you can?

Because it's centered at (0,0) the origin, and it has a radius of sqrt(10) then the answer is this, right?

$$(x(t),y(t)) = (\sqrt{10}\cos t\,,\, \sqrt{10}\sin t)$$

• Yes, it is...and you should probably remark that $\;0\le t\le 2\pi\;$ – DonAntonio May 4 '14 at 21:15

taking $x=a\sin t,y=a\cos t$ from $x^2+y^2=10$ we get $$a^2\sin^2t+a^2\cos^2t=a^2(\sin^2t+\cos^2t)=a^2=10$$ from above $a=\sqrt{10}$
You are correct. Here's an easy check: note that the position vector at any point is given by $\langle\sqrt{10}\cos(t), \sqrt{10}\sin(t) \rangle$. If we take the magnitude of this vector, we get:
$$\sqrt{10\cos^2{t} + 10\sin^2{t}} = \sqrt{10}$$
And therefore the curve is a constant $\sqrt{10}$ distance from the origin, so it must be at least a part of a circle centered at the origin with radius $\sqrt{10}$. To make the argument that it's a complete circle, you can use a bit of calculus to verify that the curve never turns back on itself. To do this, take a derivative of the position function to yield the velocity function, and note that it is never $0$.


# Proving an expression is composite

I am trying to prove that $n^4 + 4^n$ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.

This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.

I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.

• Write it as a difference of squares. – Adam Hughes Jul 1 '14 at 19:54
• How? It's $n^4 + 4^n$ not $n^4 - 4^n$ – Mathmo123 Jul 1 '14 at 19:55
• @Mathmo123: see my answer below. – Adam Hughes Jul 1 '14 at 20:03

$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...

• Ah. Very neat!! – Mathmo123 Jul 1 '14 at 20:05
• Originally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. – pidude Jul 1 '14 at 20:12

Hint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?

Let me know if you need further hints.

• I originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. – pidude Jul 1 '14 at 19:59
• Ah... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor – Mathmo123 Jul 1 '14 at 20:01
• Ok, so I was able to prove that. Now I'm working numbers which are multiples of 5. – pidude Jul 1 '14 at 20:09

I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.

• We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication
3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression
4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression
5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2
6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential
• We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again
1. Treat $n^2+2^n$ as $a$
2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer
3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$
4. $2^{2m}n^2=(n2^m)^2$ associative law
5. Treat $n2^m$ as $b$
• It implies that both $a$ and $b$ are both positive integer
• From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$
• Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$

Some interesting factorizations of a polynomial of type $x^4+\text{const}$: $$x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$

$$x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$

So one can ask, how to select the coefficients $a,b,c,d$ in

$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$

such that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.

If we expand $(3)$ we get

$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$

And the coefficients disappear, if

$$\begin{eqnarray} c+a &=& 0 \\ d+ ac +b &=& 0 \\ ad+bc &=& 0 \end{eqnarray}$$

When solving for $b,c,d$ we get

$$\begin{eqnarray} c &=& -a \\ b &=& \frac{a^2}{2} \\ d &=& \frac{a^2}{2} \end{eqnarray}$$

and therefore

$$x^4+\frac{a^4}{4} = (x^2+a x+\frac{a^2}{2})(x^2-ax+\frac{a^2}{2})$$

For $a=2$ this gives $(1)$, $a=\sqrt[]{2}$ this gives $82)$ . Substituting $a=2^{t+1}$ we get

$$x^4+4^{2t+1} = (x^2+2\cdot 2^t x+2^{2t+1})(x^2-2\cdot 2^tx+2^{2t+1})$$

Substituting $x=n=2t+1$ gives the required result for odd $n$.


Write a function that computes Kendall rank coefficient Tau (see lecture notes for additional details on Kendall Tau).

Kendall Coefficient {as quoted from Wikipedia):

Let (x1, y1), (x2, y2), ., (xn, yn) be a set of observations of the joint random variables X and Y respectively, such that all the values of xi and yi are unique. Any pair of observations (xi and yi) are said to be concordant if the ranks for both elements (more precisely, the sort order by x and by y) agree: that is, if both xi > xj and yi > yi or if both xi < xj and yi < yj. They are said to be discordant, if xi > xj and yi < yj or if xi < xj and yi > yj. If xi = xj or yi = yj, the pair is neither concordant nor discordant.

Equation 1

TAU=(number of concordant pairs-number of discordant pairs)/(0.5 x n x (n - 1))

The above equations ignore ties (sets of pairs where either xi = xj or yi = yj). One way of correcting for ties is to consider how many ties (or non-ties) are present in the data. This variant known as Kendall’s tau_b is computed by modifying the denominator 0.5n(n-1) in Equation 1 to consider ties:

sqrt(mx) x sqrt(my)

Where mx is number of non-ties for x and my is number of non-ties for y.

Note: The R function {cor} returns tau_b.

YOUR TASKS: 1. Write a custom function that computes Kendall ignoring ties (Equation 1) and correcting for ties (as explained above).

1. Your function should return a single value (Kendall’s coefficient)

2. Your function should allow the user to choose to ignore ties or not and thus produce either tau or tau_b.

3. Apply the function to two vectors provided below and check the results provided by your function against those obtained using the R function {cor}.

4. Check how fast is your function relative to {cor} function using microbenchmark function from package ‘microbenchmark’.

5. Finally, generate large vectors x and y (n = 1000) from normal distribution using rnorm function. Check again how fast is your function relative to {cor} function using microbenchmark function from package ‘microbenchmark’.

To carry out this exercise, please start by uploading the library ‘microbenchmark’ and please use x and y variables as defined below. This will ensure that all results are the same for all of us. It will also help you to evaluate if your outcomes agree with those provided in solution functions or alternative R functions.

library(microbenchmark)
x <- c(-30, 0, 14, 5, 6, 7)
y <- c(-5, 0, 12, 16, 16, 8)

Once your function is ready, you can continue to the next step. The function solutions included in this tutorial are hidden here, but can be reviewed in the .rmd file or in the answer key file (.html). Please try to solve this on your own before checking the solution.

my.kend <- function(x, y, ties=T) {
n <- length(x)
nzero <- 0.5*n*(n-1)
s1 <- sign(x - t(matrix(x, n, n, byrow=F)))
s2 <- sign(y - t(matrix(y, n, n, byrow=F)))
t1 <- sum(s1[lower.tri(s1)]!=0)
t2 <- sum(s2[lower.tri(s2)]!=0)
n2 <- t1^0.5*t2^0.5
if (!ties) tau <- sum(s1[lower.tri(s1)]*s2[lower.tri(s2)])/nzero
if (ties) tau <- sum(s1[lower.tri(s1)]*s2[lower.tri(s2)])/n2
return(tau)
}

my.kend2 <- function(x, y, ties=T) {
n <- length(x)
nzero <- 0.5*n*(n-1)
k <- 0
s1 <- 0
s2 <- 0
out <- vector(mode='numeric', length=nzero)
for (i in 1:length(x)) {
for (j in 1:length(x))  {
if (j > i) {
k <- k + 1
out[k] <- sign(x[i] - x[j])*sign(y[i] - y[j])
if(x[i] - x[j]==0) s1 = s1 + 1
if(y[i] - y[j]==0) s2 = s2 + 1
}
}
}
if (!ties) tau <- sum(out)/nzero
if (ties) tau <- sum(out)/((nzero-s1)^0.5*sum(nzero-s2)^0.5)
tau
}

Now check if your function yields comparable results to those provided by standard R functions or custom functions developed by your instructor.

x <- c(-30, 0, 14, 5, 6, 7)
y <- c(-5, 0, 12, 16, 16, 8)

# compute kendall tau using generic function {cor}
cor(x, y, method='kendall')
## [1] 0.4140393
# check if your function is consistent with standard cor function
my.kend(x, y, ties=T)
## [1] 0.4140393
# if you developed a second solution check if that solution works as well.
my.kend2(x, y, ties=T)      # check if your function works as well
## [1] 0.4140393
# now check if your function(s) also compute(s) uncorrected tau correctly
my.kend(x, y, ties=F)
## [1] 0.4
my.kend2(x, y, ties=F)
## [1] 0.4

Once you established that your function works, you should investigate if your function is faster or slower than cor function. Keep in mind, however, that the performance of the function may depend on sample size, so your function may work better than ‘cor’ function for small sample sizes, but perform poorly for large sample sizes (or vice versa).

microbenchmark(cor(x, y, method='kendall'), my.kend(x, y, ties=T),
my.kend2(x, y, ties=T))  # which is faster for a much larger dataset?
## Unit: microseconds
##                           expr    min      lq     mean  median       uq
##  cor(x, y, method = "kendall") 65.700 80.0480 95.31707 95.1510 109.6875
##        my.kend(x, y, ties = T) 23.033 26.2425 36.04066 35.6815  42.2900
##       my.kend2(x, y, ties = T) 10.195 10.9500 15.45499 12.0830  24.5435
##      max neval
##  215.976   100
##   70.986   100
##   40.780   100

You can see that my custom functions my.kend and my.kend2 are much faster than cor function when analyzed data are tiny.

Let’s now check how the functions perform for much larger datasets.

x <- rnorm(500,1,1)
y <- rnorm(500,1,3) + x
cor(x, y, method='kendall')
## [1] 0.19301
my.kend(x, y, ties=T)                                               # check if your function agrees
## [1] 0.19301
my.kend2(x, y, ties=T)                                               # check if your function agrees
## [1] 0.19301
microbenchmark(cor(x, y, method='kendall'), my.kend(x, y, ties=T),
my.kend2(x, y, ties=T))  # which is faster for a much larger dataset?
## Unit: milliseconds
##                           expr       min        lq      mean    median
##  cor(x, y, method = "kendall")  3.233603  3.529627  4.097588  3.924766
##        my.kend(x, y, ties = T) 13.880630 18.497122 21.550797 20.513782
##       my.kend2(x, y, ties = T) 50.971912 58.680038 64.747797 62.303116
##         uq      max neval
##   4.657272  5.69241   100
##  23.238596 52.78090   100
##  69.767135 94.52662   100

As clear from the report, the custom-written solutions that I wrote (you may come up with better algorithms) perform well at small sample size, but underperform at larger sample sizes. That is, compiled, professionally written R functions are faster for large samples than custom written solutions written by amatuers. This means that when we iterate small datasets, custom functions can be faster. When huge datasets are processed, R functions are likley to be superior to custom solutions.

Let us evaluate this issue here by using ‘cor’ and our custom function to carry out the same analysis.

r1 <- as.data.frame(matrix(rnorm(20000), 10, 2000))
# NOTE r1 is a dataframe of many (s=2000) tiny samples (n=10 observations). Each column is a sample.
r2 <- as.data.frame(matrix(rnorm(20000), 2000, 10))
# NOTE r2 is a dataframe of s=10 huge samples (n=2000 observations). Each column is a sample.

system.time(sapply(r1, function(x) cor(x, r1[,1], method='kendall')))
##    user  system elapsed
##    0.16    0.03    0.19
system.time(sapply(r1, function(x) my.kend(x, r1[,1])))
##    user  system elapsed
##    0.09    0.00    0.09
system.time(sapply(r1, function(x) my.kend2(x, r1[,1])))
##    user  system elapsed
##    0.09    0.00    0.09
system.time(sapply(r2, function(x) cor(x, r2[,1], method='kendall')))
##    user  system elapsed
##    0.59    0.00    0.60
system.time(sapply(r2, function(x) my.kend(x, r2[,1])))
##    user  system elapsed
##    3.92    1.01    4.93
system.time(sapply(r2, function(x) my.kend2(x, r2[,1])))
##    user  system elapsed
##    9.39    0.03    9.42

Clearly, my functions worked faster when we processed many, many short vectors (data frame r1), but underperformed badly when a few very large vectors were evaluated (data frame r2).

Always cite packages.

citation('microbenchmark')
##
## To cite package 'microbenchmark' in publications use:
##
##   Olaf Mersmann (2018). microbenchmark: Accurate Timing Functions.
##   R package version 1.4-4.
##   https://CRAN.R-project.org/package=microbenchmark
##
## A BibTeX entry for LaTeX users is
##
##   @Manual{,
##     title = {microbenchmark: Accurate Timing Functions},
##     author = {Olaf Mersmann},
##     year = {2018},
##     note = {R package version 1.4-4},
##     url = {https://CRAN.R-project.org/package=microbenchmark},
##   }


# Thread: Conditional Probability Brain Teaser

1. ## Conditional Probability Brain Teaser

A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.

What is the probability that the marble Tom put in the sock was red?

I've been trying to approach this with Bayes' Theorem but not making any progress:

P(A l B) = P(B l A) * P(A) / P(B)

I'm confused as to what A and B would represent in the problem.

Appreciate the help, thanks!

2. Originally Posted by Penguins
A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.

What is the probability that the marble Tom put in the sock was red?

I've been trying to approach this with Bayes' Theorem but not making any progress:

P(A l B) = P(B l A) * P(A) / P(B)

I'm confused as to what A and B would represent in the problem.

Appreciate the help, thanks!
Let the number of blue marbles in the sock be x.

Draw a tree diagram. The first two branches are the events Tom putting in a blue marble or a red marble. The next branches are the events of pulling out a red marble or a blue marble.

I get $\frac{\frac{3}{2(3 + x)}}{\frac{3}{2(3 + x)} + \frac{1}{3 + x}} = \frac{3}{5}$.

3. I don't understand how you set up the problem like you did. What do the terms in your numerator and denominator represent?

Here is what I get:

P(A l B) = (P(B l A) * P(A)) / P(B)

A is Tom putting a red marble in the sock.
B is Jason taking a red marble out of the sock.

P (B l A) = 3 / (x+3)
P (A) = unknown, let it be p
P (B) = p*(3/(x+3)) + (1-p)*(2/(x+2))

So

P(A l B) = (P(B l A) * P(A)) / P(B) = (3p/(x+3)) / (p*(3/(x+3)) + (1-p)*(2/(x+2)))

I end up with 3/2 p. I know I did something wrong, but I'm not sure what.

4. Hello, Penguins!

I agree with Mr. F

A sock contains 2 red marbles and an unknown number of blue marbles.
Tom places a new marble in the sock.
Jason puts his hand in the sock and pulls out a red marble.

What is the probability that the marble Tom put in the sock was red?
We want: . $P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})}$ .[1]

The sock contains: 2 Red and $b$ Blue marbles.

There are two cases to consider:

[1] A Red marble is added.
. . .The sock contains: 3 Red and $b$ Blue.
. . . . Then: . $P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3}$ .[2]

[2] A Blue marble is added.
. . .The sock contains: 2 Red and $b+1$ Blue.
. . . . Then: . $P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}$

Hence: . $P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3}$ .[3]

Substitute [2] and [3] into [1]: . $P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}$

5. Originally Posted by Soroban
Hello, Penguins!

I agree with Mr. F

We want: . $P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})}$ .[1]

The sock contains: 2 Red and $b$ Blue marbles.

There are two cases to consider:

[1] A Red marble is added.
. . .The sock contains: 3 Red and $b$ Blue.
. . . . Then: . $P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3}$ .[2]

[2] A Blue marble is added.
. . .The sock contains: 2 Red and $b+1$ Blue.
. . . . Then: . $P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}$

Hence: . $P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3}$ .[3]

Substitute [2] and [3] into [1]: . $P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}$

I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.

I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?

Thanks for the replies!

6. Originally Posted by Penguins
I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.

I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?

Thanks for the replies!
If you're expected to get a numerical answer then you have to make an assumption regarding a priori knowledge. The best one (for various reasons which I won't get into here) is a uniform prior distribution.


# 1 Biased Coin and 1 Fair Coin, probability of 2 Heads?

You have 1 fair coin and 1 coin with 2 heads. Given that the first flip was a heads what is the probability of getting another heads?

My Answer: P(2H|F=H) = P(2H|F=H, Biased Coin)*P(Biased Coin) + P(2H|F=H, Unbiased Coin)*P(Unbiased Coin) = 0.5 + 0.25 = 0.75. In my equation, F refers to the First Throw. But the answer is supposed to be 5/6 and I can't seem to understand how.

Edit: From Arthurs comment I get the following, however, I dont know if this is the correct method, despite getting the correct answer:

P(Biased|F=H) = 2/3.

P(2H|F=H) = P(2H|(Biased|F=H))*P(Biased|F=H) + P(2H|(Unbiased|F=H))*P(Unbiased|F=H) = (1*2/3) + (1/2 * 2/3) = 5/6.

Thank You

• Is it random (and unknown) which coin was flipped first? – Arthur Oct 15 '15 at 12:41
• @Arthur Yes. There is no information about that. – Jojo Oct 15 '15 at 12:42
• I think this is the biggest thing you've missed: Given that the first toss was a head, what is the proability that the first coin was the biased one? – Arthur Oct 15 '15 at 12:46
• @Arthur Could you please check the edit to my question. – Jojo Oct 15 '15 at 13:08
• @Jojo, yes, although that should be $P(2H\mid\text{ Biased} \cap F=H)$ and so forth. – Graham Kemp Oct 15 '15 at 13:11

The first flip was a head.

The $3$ heads have equal probabilities to be the head that appeared at the first flip.

$2$ of the $3$ heads have another head as other side.

$1$ of the $3$ heads has tail as other side.

So there is a chance of $\frac23.1+\frac13.\frac12=\frac56$ of throwing a second head with that coin.

• Very intuitive answer. Thanks. – Jojo Oct 15 '15 at 13:22
• Glad to help, and indeed cherish your intuition in maths. – drhab Oct 15 '15 at 13:24
• Does this assume that the same coin is flipped twice, or that one coin is flipped and then the other? I'm getting a different answer when I assume that latter, but the question is ambiguous. – Bill the Lizard Oct 15 '15 at 14:56
• It assumes that the same coin was flipped twice. – drhab Oct 15 '15 at 15:00

You missed a conditioning in your formula. Let the event of choosing biased coin be B and fair be F. Getting heads on second toss be 2H and on first toss 1H. You want to calculate the probability of getting heads on second throw given that you it landed heads on first throw, which is

P(2H/1H) = P(2H/1H,F)P(F/1H) + P(2H/1H,B)P(B/1H)

To evaluate P(F/1H) and P(B/1H) use bayes rue.

P(F/1H) = P(1H/F)*P(F)/P(1H) = 0.5*0.5/(0.5*0.5 + 1*0.5) = 1/3

P(B/1H) = 2/3

P(1H) = P(1H/F)P(F) + P(1H/B)P(B)

P(2H/1H,F) = 0.5 and P(2H/1H,B) = 1 Therefore, P(2H/1H) = 0.5*(1/3) + 1*(2/3) = 5/6


$f$ has a zero integral on every measurable set. Prove $f$ is zero almost everywhere

I am trying to solve the following exercise:

Let $f$ be integrable. Assume that $\int_A f d\mu = 0$ for every measurable set $A$. Prove that $f = 0$ a.e. [$\mu$].

I have the following proof but it seems to me too simple to be true. What is wrong with it?

For any $a>0$, define $W_a:=\{w|f(w)\geq a\}$. Now we have: $$\int_{W_a} fd\mu = 0 \geq \int_{W_a} ad\mu=a\mu(W_a)$$ and therefore $\mu(W_a)=0$. Same holds for the negative $a$'s. This completes my proof.

• You might want to add why it suffices to show $\mu(W_a) = 0$ for $a>0$ and $\mu(f \leq a)=0$ for $a<0$. – saz Feb 7 '16 at 19:58

Indeed, your proof shows that $\{f > 0\}$ has measure zero. Similarly, you show $\{f < 0\}$ has measure zero, and hence $f=0$ a.s. Yes it really is this easy.
It remains to show that the set $\{w\mid f(w)\ne0\}$ has measure $0$. Consider the nonzero rational $a$'s to complete the proof.


# Surprising Summation (1) $\sum_{i=1}^n(n-i+1)(2i-1)=\sum_{i=1}^n i^2$

Show that

$$\sum_{i=1}^n(n-i+1)(2i-1)=\sum_{i=1}^n i^2$$

without expanding the summation to its closed-form solution, i.e. $\dfrac 16n(n+1)(2n+1)$ or equivalent.

E.g., if $n=5$, then

$$5(1) + 4(3)+ 3(5)+2(7)+1(9)=1^2+2^2+3^2+4^2+5^2$$

Background as requested:
The summands for both equations are not the same but the results are the same. The challenge here is to transform LHS into RHS without solving the summation. It seems like an interesting challenge.

Further edit
Thanks to those who voted to reopen the question!

• Hmm I don't see what is so slow about expanding and collecting the terms by powers of $r$. Summation by parts works too, but would take about the same amount of computation. – user21820 Aug 15 '15 at 3:03
• I think this is a duplicate. Looks a lot like this: math.stackexchange.com/questions/1408182/… – marty cohen Apr 4 '16 at 3:21
• @martycohen - It is a duplicate only if the question asks to show that it is equivalent to the sum of squares without expanding to the closed-form solution. – hypergeometric Apr 4 '16 at 3:25
• @martycohen - The summations may be related but it is definitely not a duplicate - the summands are different and the other question is a double summation. In any case this question was posted first so it cannot be a duplicate! – hypergeometric Apr 4 '16 at 3:50

Drawing pictures for this kind of problem helps. If you draw

There are

• $n$ sets of $1$ black squares,
• $n-1$ sets of $3$ blue squares,
• $n-2$ sets of $5$ purple squares,
• $n-3$ setes of $7$ red squares,
• $\vdots$
• $n - r + 1$ sets of $2r-1$ squares of any given color

but the shapes together form a sum of squares pyramid. Algebraically that is:

\begin{align} % \sum_{r=1}^{n} (n - r + 1)(2r - 1) % &= \sum_{r=1}^{n} \left(\sum_{s=r}^n 1\right)(2r - 1) % \\ &= \sum_{r=1}^{n} \sum_{s=r}^n (2r - 1) % \\ &= \sum_{s=1}^{n} \sum_{r=1}^s (2r - 1) % \\ &= \sum_{s=1}^{n} s^2 % \end{align}

where the last step uses the fact that the sum of odd numbers up to $2k-1$ is $k^2$.

• Yes very nice solution! (+1) That was how I started, hoping to find a shorter solution to the sum of squares by recognising the each square is the sum of odd numbers. – hypergeometric Aug 15 '15 at 6:39
• @hypergeometric I felt like the logic was backwards when I was writing it. Anyway, if you want a short solution to sum of squares, I suggest looking at $\sum (k^3 - (k - 1)^3) = \sum (3k^2 - 3k + 1)$ , use telescoping on the left and the known closed form for $\sum k$ on the right. – DanielV Aug 15 '15 at 7:32

$$(n-r+1)(2r-1)=(n+1)(-1)+r[2(n+1)+1]-2r^2$$

$$\implies\sum_{r=1}^n(n-r+1)(2r-1)=-(n+1)\sum_{r=1}^n1+\{2(n+1)+1\}\sum_{r=1}^nr-2\sum_{r=1}^nr^2$$

$$=-(n+1)\cdot n+(2n+3)\cdot\dfrac{n(n+1)}2-2\cdot\dfrac{n(n+1)(2n+1)}6$$

$\begin{array}\\ \sum_{r=1}^n(n-r+1)(2r-1) &=\sum_{r=1}^n(n+1)(2r-1)-\sum_{r=1}^nr(2r-1)\\ &=(n+1)n^2-2\sum_{r=1}^nr^2+\sum_{r=1}^nr\\ &=(n+1)n^2-2\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\\ &=\frac{6(n+1)n^2-2n(n+1)(2n+1)+3n(n+1)}{6}\\ &=n(n+1)\frac{6n-2(2n+1)+3}{6}\\ &=n(n+1)\frac{2n+1}{6}\\ &=\frac{n(n+1)(2n+1)}{6}\\ \end{array}$

Now that's quite a surprise!

I'll stop here a try to find a magic trick later.

• Well spotted! (+1) – hypergeometric Aug 15 '15 at 6:38

Make the substitution $i = n - r + 1$ (like change of variable in integration), then \begin{align} & \sum_{r = 1}^n (n - r + 1)(2r - 1) = \sum_{i = 1}^n i(2n - 2i + 1) \\ = & (2n + 1) \sum_{i = 1}^n i - 2\sum_{i = 1}^n i^2 \end{align} I think I didn't expand the term "brutally", but the last step is still slight expansion.

Here's the "magic" solution in more general form, followed by an continuous analog.

Let $f(n) =\sum_{i=1}^n (n+1-i)g(i)$. Note that $f(1) = g(1)$ and $f(2) =2g(1)+g(2) =g(1)+(g(1)+g(2))$. Then

$\begin{array}\\ f(n+1) &=\sum_{i=1}^{n+1} (n+2-i)g(i)\\ &=\sum_{i=1}^{n+1} (n+1+1-i)g(i)\\ &=\sum_{i=1}^{n+1} (n+1-i)g(i)+\sum_{i=1}^{n+1} g(i)\\ &=\sum_{i=1}^{n} (n+1-i)g(i)+\sum_{i=1}^{n+1} g(i) \quad\text{(since }n+1-i = 0 \text{ for } i=n+1)\\ &=f(n)+\sum_{i=1}^{n+1} g(i)\\ so\\ f(n+1)-f(n)&=\sum_{i=1}^{n+1} g(i)\\ so\ that\\ f(n)&=\sum_{j=1}^n\sum_{i=1}^{j} g(i)\\ \end{array}$

Setting $g(n) = 2n-1$, since $\sum_{i=1}^{n} g(i) =n^2$, $f(n) =\sum_{i=1}^{n} i^2 =\dfrac{n(n+1)(2n+1)}{6}$.

The continuous analog:

Let $f(x) =\int_0^x (x-y)g(y)dy$. Then

$\begin{array}\\ f(x) &=\int_0^x (x-y)g(y)dy\\ &=\int_0^x xg(y)dy-\int_0^x yg(y)dy\\ &=x\int_0^x g(y)dy-\int_0^x yg(y)dy\\ \text{so that}\\ f'(x)&=xg(x)+\int_0^x g(y)dy- xg(x)\\ &=\int_0^x g(y)dy\\ \text{Integrating,}\\ f(x)&=\int_0^x \int_0^y g(z)dz dy\\ \end{array}$

If $g(y) = 2y$, then $\int_0^x g(y)dy =x^2$ so $\int_0^x (x-y)g(y) dy =\int_0^x y^2 dy =\dfrac{x^3}{3}$

\begin{align} \sum_{i=1}^n(n-i+1)(2i-1) &=\sum_{i=1}^n\sum_{j=i}^n(2i-1)\\ &=\sum_{j=1}^n\sum_{i=1}^j(2i-1) &&(1\le i\le j \le n)\\ &=\sum_{j=1}^n\sum_{i=1}^ji^2-(i-1)^2\\ &=\sum_{j=1}^n j^2&&\text{(by telescoping)}\\ &=\sum_{i=1}^n i^2\qquad\blacksquare \end{align}

Posting another solution which has just been suggested by a friend:

\begin{align} \sum_{i=1}^n i^2 =&\sum_{i=1}^n \color{blue}{-}(\color{blue}{n-i})i^2+\sum_{i=1}^n(\color{blue}{n-i}+1)i^2\\ =&\sum_{i=2}^{n+1} -(n-i+1)(i-1)^2+\sum_{i=1}^n(n-i+1)i^2\\ =&\sum_{i=1}^{n} -(n-i+1)(i-1)^2+\sum_{i=1}^n(n-i+1)i^2\\ =&\sum_{i=1}^{n} (n-i+1)\big[i^2-(i-1)^2\big]\\ =&\sum_{i=1}^{n} (n-i+1)(2i-1)\qquad\blacksquare\\\ \end{align}


## Sunday, May 15, 2005

### Coprime Numbers: More Results

In this blog, I will review some basic results regarding coprime numbers.

Lemma 1: p,q coprime -> pq, p2 - q2 are coprime.

(1) Assume gcd(pq,p2 - q2) = d and d is greater than 1

(2) So, there is some prime P which divides d [Fundamental Theorem of Artihmetic]

(3) Since d divides pq. P either divides p or q. [Euclid's Lemma]

(4) Let's assume P divides p and the same argument will apply if it divides q. So that, there exists R such that p = PR.

(5) Since d divides p2 - q2, there exists a value Q such that p2 - q2 = PQ

(6) This means that q2 = p2 - PQ = (PR)2 - PQ = P(PR2 - Q).

(7) But this means that P divides q by Euclid's Lemma which is a contradiction since P also divides p. [Because gcd(p,q)=1]

(8) Therefore, we reject our assumption.

QED

Lemma 2: p,q are relatively prime and are of different parity (one is odd, one is even), then p + q, p - q are relatively prime.

(1) Assume that gcd(p+q,p-q) = d and d is greater than 1.

(2) Then there exists a prime x which divides d. [Fundamental Theorem of Arithmetic]

(3) We know that this x is odd since p+q and p-q are odd. [Since they are of different parity]

(4) Now x divideds 2p since 2p = (p + q) + (p - q) which means x divides p. [By Euclid's Lemma Since x is odd]

(5) Likewise x divides 2q since 2q = (p + q) - (p - q) = p + q - p + q which means x divides q. [Same reason as above]

(6) But this is a contradiction since p,q are relatively prime.

QED

Lemma 3: if S2 = P2 + Q2 and P2 = T2 + Q2, then Q,S,T are relatively prime.

(1) We can assume that S,P,Q are relatively prime and P,T,Q are relatively prime. [See Lemma 1, here for details]

(2) We can also assume that S,P are odd which means that Q is even and that T is odd. [See details above]

(3) From the two equations, we can derive that:
S2 = T2 + 2Q2

(4) Now we need only prove that any factor that divides 2 values, will divide the third.

Case I: f divides S,Q

(a) There exists s,q such that S = fs, Q = fq
(b) T2 = S2 - 2Q2 = f2 [ s2 - 2q2 ]
(c) Which means that f divides T [See here.]

Case II: f divides T,Q

(a) We can use the same argument as Case I.

Case III: f divides S,T

(a) There exists s,t such that S = fs, T = ft
(b) 2Q2 = f2[ s2 - t2 ]
(c) Since S,T are odd, f must be odd and s,t must be odd.
(d) Then, s2 - t2 must be even.
(e) Then, there exists u such that 2u = s2 - t2
(f) So, we have:
2Q2 = (2)u(f2)
(g) And canceling out the 2s gives us:
Q2 = f2u
QED

Lemma 4: S,T relatively prime, both odd, 2u = S - T, 2v = S + T, then u,v are relatively prime.

(1) Assume f divides u,v such that u = fU, v = fV.
(2) The f divides S
S - T + S + T = 2S = 2u + 2v = 2f(U + V)
(3) And f divides T
S + T - (S - T) = 2T = 2v - 2u = 2f(V - U)l.
(4) This contradicts our S,T beings coprime so we reject our assumption.

QED

Oscar Rojas said...

Hi Larry,

I think that in Lemma 3, the assumption 1 (about some of the numbers being relative primes) should be part of the hypothesis.

In fact the Lemma does not hold if that assumption is not satisfied.

The proper assumption must be made in the part of fermat's one proof that is using the Lemma 3.

Since Lemma 3 is only used in that proof, then one could say that it is irrelevant where do you place the assumption... but, for pure formalism...

Larry Freeman said...

Hi Oscar,

The assumption is justified by the proof in Lemma 1 in the link.

Since we can reduce any solution to x^n + y^n = z^n to a form where x,y,z are coprime (see Lemma 1, here), we can assume that S,P,Q and P,T,Q are relatively prime because if they weren't, then we could still find a form that was.

-Larry

Oscar Rojas said...

Thanks Larry.

I am sorry; I failed to state more clear my comment. I will try to do it better:

I agree that the assumption is justified... but it is justified in the context outside the lemma. If you take the lemma out of that context, it is not true as written.

The only fact in the hypothesis is that the numbers satisfy the equations, and that simple fact is not sufficient to derive the conclusion. If you take another set of numbers that are multiples of the originals, the new set still satisfy the unrestricted hypothesis but obviously they do not satisfy the conclusion.

So, I am not against the assumption, I am against using it as part of the proof of the lemma.
To save the consistency, the hypothesis must state that the original numbers are relative primes, precisely to bring the lemma in the restricted context.

As I said: pure formalism!


# Conditional probability

#### Romanka

##### New member
Please, help me to solve the problem
Details at a factory are tested randomly to check if they are faulty. It is known from previous experience that the probability of a
faulty detail is known to be 0.03. If a faulty detail is tested the probability of it testing faulty is 0.82. If a non-faulty detail is
tested the probability of it testing faulty is 0.06. Given that the detail was tested as not been faulty, calculate the
probability that it was faulty.

I understand that it's about conditional probability, but can't get it.
Thanks!

#### tkhunny

##### Well-known member
MHB Math Helper
Have you considered drawing a 2x2 grid and filling in the boxes?

#### Plato

##### Well-known member
MHB Math Helper
Details at a factory are tested randomly to check if they are faulty. It is known from previous experience that the probability of a faulty detail is known to be 0.03. If a faulty detail is tested the probability of it testing faulty is 0.82. If a non-faulty detail is tested the probability of it testing faulty is 0.06. Given that the detail was tested as not been faulty, calculate the probability that it was faulty.
Use $F$ for faulty and $T$ for a positive test.
From the given: $\mathcal{P}(F)=0.03,~\mathcal{P}(T|F)=0.82,~\&~ \mathcal{P}(T|F^c)=0.06$

Now you want $\mathcal{P}{(F|T^c)}$$=\dfrac{\mathcal{P}(F\cap T^c)}{\mathcal{P}(T^c)}. #### Romanka ##### New member thanks, I'll try ---------- Post added at 03:27 PM ---------- Previous post was at 03:26 PM ---------- Use F for faulty and T for a positive test. From the given: \mathcal{P}(F)=0.03,~\mathcal{P}(T|F)=0.82,~\&~ \mathcal{P}(T|F^c)=0.06 Now you want \mathcal{P}{(F|T^c)}$$=\dfrac{\mathcal{P}(F\cap T^c)}{\mathcal{P}(T^c)}$.
that's exactly what I need! thank you so much!

#### HallsofIvy

##### Well-known member
MHB Math Helper
Here's how I like to do problems like this: Imagine there are 10000 details (chosen to avoid fractions).

"It is known from previous experience that the probability of a faulty detail is known to be 0.03."
Okay, so our 10000 sample includes (0.03)(10000)= 300 faulty details and 10000- 300= 9700 that are not faulty.

"If a faulty detail is tested the probability of it testing faulty is 0.82."
Of the 300 faulty details, (0.82)(300)= 246 will test faulty, the other 54 will test not-faulty.

"If a non-faulty detail is tested the probability of it testing faulty is 0.06."
Of the 9700 non-faulty details, (.06)(9700)= 582 will test faulty. 9700- 582= 9118 will test not-faulty.

"Given that the detail was tested as not been faulty, calculate the probability that it was faulty."
There are a total of 9118+ 54= 9172 details that test non-faulty of which 54 are faulty.

thanks!

#### HallsofIvy

##### Well-known member
MHB Math Helper
By the way, "detail" doesn't seem like quite the word you want. "Detail" means a small part of something larger. I suspect this was translated from another language and you really wanted "item".

#### Romanka

##### New member
Yes, maybe. But I got the problem about "details"

#### Plato

##### Well-known member
MHB Math Helper
Yes, maybe. But I got the problem about "details"
YES. But what was language of the question?
Did you use a translation program to post it here?


# Uncountability of increasing functions on N

I believe I have made a reasonable attempt to answer the following question. I would like a confirmation of my proof to be correct, or help as to why it is incorrect.

Question: Let $f : \mathbb{N} \to \mathbb{N}$ be increasing if $f(n+1) \ge f(n)$ for all $n$. Is the set $A$ of functions $f$ countable or uncountable?

For each $f$ in $A$, there is a function $g: \mathbb{N}\cup \{0\} \to \mathbb{N} \cup\{0\}$ defined by $g(0)=f(1)$ and $g(n)=f(n+1)-f(n)$ for $n>0$. Conversely, for each $g: \mathbb{N}\cup \{0\} \to \mathbb{N}\cup\{0\}$ with $g(0)>0$, there is an increasing function $f$ defined recursively by $f(1)=g(0)$ and $f(n+1)=f(n)+g(n)$ for $n>0$. Consequently, there is a bijection between $A$ and the set $B$, of $g: \mathbb{N}\cup\{0\} \to \mathbb{N}\cup\{0\}$, $g(0)>0$. If $B$ is uncountable, then $A$ must be uncountable.

The set $C$ of $h: \mathbb{N}\cup\{0\} \to \{0,1,2,3,4,5,6,7,8,9\}$ with $h(0)=1$is a subset of $B$. For each real $x \in [1,2)$, there is a unique decimal expansion with 1 before the decimal point, ending NOT in trailing 9's. By defining for $n>0$, $h(n)=$ (n'th digit of x after decimal point), we have an injection from $[1,2)$ to the set $C$ (different $x$ => different decimal expansions => x maps to different h). Since the set of reals on $[1,2]$ is uncountable, the set C is uncountable, => the set B is uncountable, => the set A is uncountable.

This proof works but seems needlessly verbose. It is easier to prove that strictly increasing functions form an uncountable set (which implies that the monotone increasing are uncountable as well). This is because there is an obvious 1-1 correspondence between strictly increasing functions and infinite subsets of N (each such function is uniquely determined by its range). It is a standard result that there are uncountably many subsets of N, of which only countably many are finite. Hence, there are uncountably many infinite subsets.

It is rather easy to apply the diagonalization argument directly. Let $\langle f_n\rangle$ be a sequence of increasing functions from the naturals to itself. Define a function $f$ recursively by $f(1)=1$ and $$f(n+1)=f(n)+f_n(n+1)-f_n(n)+1.$$ The resulting function $f$ is clearly different from every function in the sequence $\langle f_n\rangle$.

• Oh I really like this. +1 – Pete Caradonna Jul 15 '16 at 11:30
• The resulting function $f$ need not be increasing, so doesn't belong to the set being considered. – hunter Jul 15 '16 at 13:21
• (More precisely, $f(n+1) - f(n) = (f_n(n+1) - f_n(n)) - (f_{n-1}(n) - f_{n-1}(n-1))$; since there's no relation between the rates of growth of $f_n$ and $f_{n-1}$ you can't determine if this difference is positive or negative. – hunter Jul 15 '16 at 13:23
• @hunter Thank you, you are right, what I wrote was different from what I actually intended. I have fixed it now. That is exactly where the recursive definition comes in. – Michael Greinecker Jul 15 '16 at 13:38
• Note that this is exactly the diagonalisation argument for general functions composed with the OP's transformation. – filipos Jul 15 '16 at 20:49

Your proof looks OK to me. It's a little longer than needed though.

Right after you define your set $A$, what if we just restrict ourselves to a subset of $A$ given by the subset of monotone functions $t: \mathbb{N} \to \mathbb{N}$ such that $t(n+1)-t(n) \in \{0,1\}$ and $t(0)=k$.

Then in your spirit, we can identify each $t$ with the infinite binary sequence of its step-wise differences. Moreover, clearly for any infinite binary sequence we can construct the unique $t$ that gets related to it. Then by Cantor's diagonalization argument we know $\{0,1\}^\mathbb{N}$ is uncountable and we're done.

Another way to show that $A$ is uncountable : Show that if $B=\{f_n: n\in N\}\subset A$ then $B\ne A$ as follows:

Let $g(n)=\max \{1+f_j(n): j\leq n\}.$

We have $g(n)> f_n(n)$ for every $n.$ So $g\ne f_n$ for every $n.$ So $g \not \in B.$

For each $n$ we have $g(n)=1+f_{j_n}(n)$ for some $j_n\leq n .$ So $g(n+1)=\max \{1+f_j(n+1):j\leq n+1\}\geq 1+f_{j_n}(n+1)\geq 1+f_{j_n}(n)=g(n).$ So $g\in A.$


# Probability involving exponentially distributed random variabl

Two buses A and B arrive independently at a bus station at random rate of 5/hour and 4/hour respectively. A passenger comes to the bus station at 10 am. What is the probability that it takes at most 5 minutes before the first bus arrives at the station?

Attempt

Let X and Y be random variables indicating time of arrivals of Bus A and B respectively. Then, X and Y are exponentially distributed random variables with rate 5 and 4 respectively. So the probability density functions of X and Y are 5*exp(-5*t) and 4*exp(-4*t).

Now, let Z = min(X,Y) be a new random variable. Then, I need to calculate P(Z<5/60).

Question: My solution sheet goes as follows: P(Z<5/60) = 1-P(Z>5/60) = 1-P(X>5/60, Y>5/60) and goes on like this. But why cannot I just calculate P(Z<5/60) = P(X<5/60, Y<5/60) directly? These two give me different answers, meaning my attempt might be wrong. Why do we need to calculate the complementary probability?

• It does not matter if the second bus arrives before 5 minutes if the first one does arrive in 5 minutes. You only care about the complement of the event that no busses arrive within 5 minutes. Apr 24, 2018 at 18:22
• I did not get it. @TimDikland Apr 24, 2018 at 18:24
• There are four possible interesting events. P(X<5, Y<5), P(X<5, Y>5), P(X>5, Y<5), P(X>5, Y<5). Now ask yourself which of those events are desirable and which are not. (Here X<5 means bus X arrives within 5 minutes) Apr 24, 2018 at 18:29
• I am still confused. So, if the question was: What is the probability that she waits at least 5 mins before the first bus arrives? Would I directly calculate P(Z>5/60) = P(X>5/60,Y>5/60)? Apr 24, 2018 at 18:34
• Exactly. Your mistake in your first calculation is that $P(Z < 5/60) \neq P(X<5/60,Y<5/60)$. If you would do it correctly then you would find that $P(Z<5/60) = P(X<5/60, Y<5/60) + P(X<5/60, Y>5/60) + P(X>5/60, Y<5/60)$, which is obviously more work to calculate Apr 24, 2018 at 18:38

Here's another interesting way to approach this problem. I'll start out with a more general version, and then we'll look at the case you need.

Let $X_1, X_2, \cdots X_n$ be independent and let $X_{(1)}$ be the minimum observation. Let's find the CDF of $X_{(1)}$.

\begin{align*} F_1(x) &= P(X_{(1)} \leq x) & \text{(definition)} \\ &= 1 - P(X_{(1)} > x) & \text{(complement)} \\ &= 1 - P(X_1 > x, \ X_2 > x, \ \cdots, \ X_n > x) & \text{(min > x $\Leftrightarrow$ all > x)} \\ &= 1 - P(X_1 > x)P(X_2 > x)\cdots P(X_n > x) & \text{(independence)} \\ \end{align*}

## Back to the problem

Let $$X \sim Exp(4) \quad Y \sim Exp(5) \quad Z = min(X, Y)$$ $$P(X > x) = e^{-4x}$$ $$P(Y > x) = e^{-5x}$$ Hence, by the above result we can write the CDF of the minimum as: $$P(Z \leq x) = 1 - e^{-4x}e^{-5x} = 1 - e^{-9x}$$

Now you can calculate the desired probability easily.

Note that this is equivalent mathematically to the approach that Davis gives, but with this approach we see that the the minimum of independent exponential random variables is itself Exponentially distributed! Specifically, $Z \sim Exp(4+5)$.

$Z < \frac{5}{60}$ might occur without $X < \frac{5}{60}$ occurring, specifically, if $Y < \frac{5}{60}$.

Multiplying probabilities of independent events gives you the probability of their intersection (both events happen), but what you want is the probability of their union. Call $E$ the event that bus A arrives before the given time, and $F$ the event that bus B arrives before the given time. Then, the event you are interested in is:

$E \cup F = (E^c \cap F^c)^c$

The equality is by DeMorgan's law. Now you can figure out the probability by multiplying the probabilities of $E^c$ and $F^c$.


An included angleis an angle formed by two given sides. For example, both of these triangles are isosceles, since they have two equal sides and angles. (iii) If two figures have equal areas, they are congruent. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. Congruence is our first way of knowing that magnitudes of the same kind are equal. Triangles are congruent when all corresponding sides & interior angles are congruent. = False (v) If two sides and one angle of a triangle are equal to the corresponding two sides and angle of another triangle, the triangles are congruent. Hence all squares are not congruent. You have one triangle. TRUE. find the price at which it was sold. Triangle Congruence Theorems (SSS, SAS, ASA), Congruency of Right Triangles (LA & LL Theorems), Perpendicular Bisector (Definition & Construction), How to Find the Area of a Regular Polygon, Do not worry if some texts call them postulates and some mathematicians call the theorems. asked by priyanka. Geometricians prefer more elegant ways to prove congruence. Since b/e = 1, we have a/d = 1. = False (iv) If two triangles are equal in area, they are congruent. - the answers to estudyassistant.com Asked on January 17, 2020 by Premlata Pandagre In a squared sheet, draw two triangles of equal areas such that (i) the triangles are congruent. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. -If two squares have equal perimeters, then they have equal ...” 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. prove that they are congruent. Think: Two congruent triangles have the same area. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Yes, they are similar. Triangles can be similar or congruent. one could look upside-down compared to the other but if they have the same dimensions (i.e. (if you don't know the a False. Two triangles are congruent if they have the same shape and size, but their position or orientation may vary. they have to be congruent to have same perimeters AND area con. FALSE. All three triangle congruence statements are generally regarded in the mathematics world as postulates, but some authorities identify them as theorems (able to be proved). After you look over this lesson, read the instructions, and take in the video, you will be able to: Get better grades with tutoring from top-rated private tutors. You can think you are clever and switch two sides around, but then all you have is a reflection (a mirror image) of the original. Congruent triangles are triangles which are identical, aside from orientation. An included side is the side between two angles. It is congruent to ∠WSA because they are alternate interior angles of the parallel line segments SW and NA (because of the Alternate Interior Angles Theorem). it the shopkeeper sells it at a loss of 10% . Converse of the above statement : If the areas of the two triangles are equal, then the triangles are congruent. If the areas of two similar triangles are equal, prove that they are congruent. But just to be overly careful, let's compute a/d. So once you realize that three lengths can only make one triangle, you can see that two triangles with their three sides corresponding to each other are identical, or congruent. To be congruent two triangles must be the same shape and size. For example, these triangles are similar because their angles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. A, B and C are three sets, n(A) = 14, n(C) = 13, n(A U BUC) = 28, Move to the next side (in whichever direction you want to move), which will sweep up an included angle. Congruent Triangles. b=e. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. If two triangles have the same area, they must be congruent. The given statement - "If two triangles are congruent, then their areas are equal." (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. We use the symbol ≅ to show congruence. Log in. In the sketch below, we have △CAT and △BUG. (iv) If two triangles are equal in area, they are congruent. Their interior angles and sides will be congruent. In the simple case below, the two triangles PQR and LMN are congruent because every corresponding side has the same length, and every corresponding angle has the same measure. 3) They are equal areas. Local and online. …, how much loss and profit computed on his original capital. Answer: 3 question What would prove that the two triangles are congruent? Now you have three sides of a triangle. Similar triangles will have congruent angles but sides of different lengths. Why should two congruent squares have the same area? Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. The SAS Postulate says that triangles are congruent if any pair of corresponding sides and their included angle are congruent. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. triangles have same shape/size but just placed/viewed differently cidyah for that to … Click here to get an answer to your question  if the area of two triangles are equal. the triangle which are equal in area also congruent, Here ur answer ....pls mark it brainlist..pls, This site is using cookies under cookie policy. It is equal in length to the included side between ∠B and ∠U on △BUG. Congruent triangles will have completely matching angles and sides. Join now. You could cut up your textbook with scissors to check two triangles. So let's just start That triangle BCD is congruent Let’s use congruent triangles first because it requires less additional lines. This forces the remaining angle on our △CAT to be: This is because interior angles of triangles add to 180°. Testing to see if triangles are congruent involves three postulates. Theorem 1 : Hypotenuse-Leg (HL) Theorem. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. Get help fast. You will see that all the angles and all the sides are congruent in the two triangles, no matter which ones you pick to compare. Congruence is denoted by the symbol ≅. AAS is equivalent to an ASA condition, by the fact that if any two angles are … So now you have a side SA, an included angle ∠WSA, and a side SW of △SWA. Similar triangles. You can't do it. 0 votes . We could also think this angle right over here. 1. (ii) If two squares have equal areas, they are congruent. A postulate is a statement presented mathematically that is assumed to be true. By applying the Side Angle Side Postulate (SAS), you can also be sure your two triangles are congruent. Triangles can be considered congruent if following conditions are satisfied. https://www.mathsisfun.com › geometry › triangles-congruent.html ... maths. Answer In two similar triangles, the ratio of their areas is the square of the ratio of their sides. Compare them to the corresponding angles on △BUG. 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. 1. For two triangles to be similar, it is sufficient if two angles of one triangle are equal to two angles of the other triangle. The legs of each of these isosceles triangles could have any lengths as long as they are equal, but the legs of these two triangles need not be the same. So the fact that two triangles have the same area does not necessarily tell us that they are congruent (they could be congruent but that is often not the case) since they can have different measures on their sides or a different size and still have the same area. monojmahanta4 monojmahanta4 09.04.2020 Math Secondary School If the area of two triangles are equal. Conditional Statements and Their Converse. You can check polygons like parallelograms, squares and rectangles using these postulates. There are exactly two isosceles triangles with given perimeter and area. https://www.mathsisfun.com/geometry/triangles-congruent.html Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. Thus, a=d. are both 90 degrees. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. If triangle RST is congruent to triangle WXY and the area of triangle WXY is 20 square inches, then the area of triangle RST is 20 in.². e.g. 2) Their corresponding angles are equal. Two triangles, ABC and A′B′C′, are similar if and only if corresponding angles have the same measure: this implies that they are similar if and only if the lengths of corresponding sides are proportional. And for sensible cases. If you have two similar triangles, and one pair of corresponding sides are equal, then your two triangles are congruent. (g) If two triangles are equal in area, they are congruent. Write the following statements in words, farhan sarted his bussiness by investing 75000 in the first year he made a profit of 20% he invested the capital in new business and made loss of 12% The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Now you will be able to easily solve problems on congruent triangles definition, congruent triangles symbol, congruent triangles Class 8, congruent triangles geometry, congruent t The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. A rectangle and a triangle with the same base and height have the same area. = False (vi) If two angles and any side of a triangle are equal to the corresponding angles of another triangles then the triangles are congruent. So go ahead; look at either ∠C and ∠T or ∠A and ∠T on △CAT. "If two triangles are congruent, then their areas are equal." You can compare those three triangle parts to the corresponding parts of △SAN: After working your way through this lesson and giving it some thought, you now are able to recall and apply three triangle congruence postulates, the Side Angle Side Congruence Postulate, Angle Side Angle Congruence Postulate, and the Side Side Side Congruence Postulate. Therefore, triangles are congruent. If the area of two similar triangles are equal, prove that they are congruent. Corresponding sides and angles mean that the side on one triangle and the side on the other triangle, in the same position, match. The two are not the same. You now have two triangles, △SAN and △SWA. Suppose you have parallelogram SWAN and add diagonal SA. Using any postulate, you will find that the two created triangles are always congruent. Theorem 1 : Hypotenuse-Leg (HL) Theorem. In most systems of axioms, the three criteria – SAS, SSS and ASA – are established as theorems. I could have one triangle with sides measuring 7 in + 5 in + 8 in (perimeter = 20 in) and another triangle with sides of 6 in + 4 in + 10 in (perimeter = 20 in). Congruent triangles sharing a common side. …, nswer then don't give) give me with full steps.don't spam ❌❌❌❌❌. This is the only postulate that does not deal with angles. That is not very helpful, and it ruins your textbook. You can now determine if any two triangles are congruent! For the two triangles to be congruent, those three parts -- a side, included angle, and adjacent side -- must be congruent to the same three parts -- the corresponding side, angle and side -- on the other triangle, △YAK. Ask your question. All the sides of a square are of equal length. Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. You can only make one triangle (or its reflection) with given sides and angles. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. Now shuffle the sides around and try to put them together in a different way, to make a different triangle. Definition: Triangles are congruent when all corresponding sides and interior angles are congruent.The triangles will have the same shape and size, but one may be a mirror image of the other. (iv) If two triangles are equal in area, they are congruent. So I guess the answer is not always. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. An included angleis an angle formed by two given sides. This triangle has two angles with equal measures of 75^@. However, they do not have to be congruent in order to be similar. ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. Both the areas would be 9, but the figures would not be congruent. Illustration of SAS rule: Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. Light1729 Light1729 If two triangles are congruent then they must overlap each other completely, so, they have exactly equal area. Furthermore, your question is about congruent triangles and not similar triangles.l. prove that they are congruent. Notice we are not forcing you to pick a particular side, because we know this works no matter where you start. Here, instead of picking two angles, we pick a side and its corresponding side on two triangles. , the radic of two circles are 8 cm and 6 cm representively find the radius of the circle having area equal to the sum of the areas of the two circles, bnRlinP) = 16 and n(Q).10, what are the greatest and the least values of n(PU QI?8. The congruence of two objects is often represented using the symbol "≅". Only if the two triangles are congruent will they have equal areas. Comparing one triangle with another for congruence, they use three postulates. Isosceles triangles are triangles with two equal sides, and thus two equal angle measures. In this case, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle. SSSstands for "side, side, side" and means that we have two triangles with all three sides equal. Since two triangles are similar therefore the ratio of the area is equal to the square of the ratio of its corresponding side a r e a ∆ A B C a r e a ∆ D E F = B C E F 2 = A B D E 2 = A C D F 2 B C E F 2 = A B D E 2 = A C D F 2 = 1 N o w , t a k i n g a n y o n e c a s e 1 = B C E F 2 1 = B C E F E F = B C However they can share a side, and as long as they are otherwise identical, the triangles are still congruent. …, n(A) = n(B), nn Byn(B, C) = 4 and n(An C) = 3. However, triangles … Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. The ASA Postulate was contributed by Thales of Miletus (Greek). Testing to see if triangles are congruent involves three postulates, abbreviated SAS, ASA, and SSS. Pairs - The classic pairs game with simple congruent shapes. You already know line SA, used in both triangles, is congruent to itself. For two triangles to be congruent, the corresponding angles and the sides are to be equal. The four triangles are congruent with each other regardless whether they are rotated or flipped. Divide a square sheet in two triangles of equal area so that they are congruent. The Angle Side Angle Postulate (ASA) says triangles are congruent if any two angles and their included side are equal in the triangles. (f) Two circles having same circumference are congruent. Congruent triangles. But, it is not necessary that triangles having equal area will be congruent to one another they may or may not be congruent. Contrapositive of the given statement : If the areas of two traingles are not equal then the triangles are not congruent. So also, sides d and e must be equal because DEF is isosceles. They have the same area and the same perimeter. (ii) If two squares have equal areas, they are congruent. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. $\endgroup$ – Moti Feb 7 '15 at 16:42 add a comment | Your Answer In the figure below, the triangle LQR is congruent to PQR even though they share the side QR. Congruent triangles are triangles having corresponding sides and angles to be equal. If the areas of two similar triangles are equal, prove that they are congruent. But that does not mean that they have to be congruent. If the corresponding angles of two triangles are equal, then they are always congruent. Guess what? If the areas of two similar triangles are equal, prove that they are congruent. If 2 sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent. Cut a tiny bit off one, so it is not quite as long as it started out. So, for equal area, all sides are equal. FALSE. For example: (See Solving SSS Trianglesto find out more) Put them together. Cut the other length into two distinctly unequal parts. It is true that all congruent triangles have equal area. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. Congruence in triangles is important in the study of plane geometry. Yes, if two triangles have two congruent angles and two congruent sides then the triangles are guaranteed to be congruent. What is the solution set of y=3x+16 and 2y=x+72? Join now. Only if the two triangles are congruent will they have equal areas. And why does a $1 \times 1$ square have an area of $1$ unit?) the cost price of a flower vase is rs 120 . Remember that the included angle must be formed by the given two sides for the triangles to be congruent. This will happen when the area is half the multiplication of the sides, or maximum area for such two sides. 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# Integrating for approximation of a sum

1. May 15, 2014

### goraemon

1. The problem statement, all variables and given/known data

Find an N so that $∑^{\infty}_{n=1}\frac{log(n)}{n^2}$ is between $∑^{N}_{n=1}\frac{log (n)}{n^2}$ and $∑^{N}_{n=1}\frac{log(n)}{n^2}+0.005.$

2. Relevant equations
Definite integration

3. The attempt at a solution
I began by taking a definite integral: $\int^{\infty}_{N}\frac{log(n)}{n^2}dn$ and, using integration by parts, arrived at the following answer: $\frac{log(N)+1}{N}$. (Is this right? If not, I could post the steps I used to try to see where I made an error)

Next we need $\frac{log(N)+1}{N}$ to be within 0.005 as given by the problem, so:

$\frac{log(N)+1}{N}=0.005=\frac{1}{200}$

But I'm having trouble how to solve for N algebraically. Would appreciate any help.

2. May 15, 2014

### BiGyElLoWhAt

I actually got -1 times what you have. remember: $\int \frac{1}{n^2} = -\frac{1}{n}$

3. May 15, 2014

### goraemon

Right but since we're taking a definite integral on the interval from N to ∞, and since $\frac{-log(N)-1}{N}$ as N approaches infinity equals zero, shouldn't the definite integral work out to:
$0-\frac{-log(N)-1}{N}=\frac{log(N)+1}{N}$?

4. May 15, 2014

### pasmith

The graph of $x^{-2}\log(x)$ has a maximum in $[1,2]$, so it's difficult to say whether the integral is an over-estimate or under-estimate of the sum. To be safe I'd require that $$\frac{\log(N) + 1}{N} < \frac{1}{400}$$.

Equations of that type have to be solved numerically. You're looking for zeroes in $x > 0$ of $$f(x) = \frac{x}{200} - 1 - \log x$$. The derivative of this function is $$f'(x) = \frac{1}{200} - \frac{1}{x}$$ so the second derivative ($x^{-2}$) is everywhere positive. There is a minimum at $x = 200$ where $f(200) < 0$. Since $f(x) \to +\infty$ as $x \to 0$ or $x \to \infty$ there are exactly two solutions. Logic dictates you want the larger, since if $N$ works then any larger $M$ should also work.

5. May 15, 2014

### goraemon

Thanks, I tried solving it numerically and came up with N = 1687, which the textbook confirms is correct.


# Combinations with exclusions

Given five objects $A, B, C , D , E$. I'd like to calculate all possible sets of these objects such that

1. Two pairs of objects, $B$ and $C$, and $D$ and $E$, cannot coexist. For example the set $\{ABD\}$ would be valid but the set $\{CDE\}$ would not.
2. Each object appears no more than once in a set, e.g. $\{ACD\}$ is valid but $\{ADD\}$ is not.
3. There can be no empty set.
4. Apart from the above, there is no other limit on the cardinality of a set, e.g. $\{D\}$, $\{AB\}$, $\{ACE\}$. I know there can be no set of cardinality greater than 3, but it would be nice to show this.
5. The objects can be listed in any order, i.e. $\{AB\} = \{BA\}$

Edit My intention is to generalize the process of generating these combinations given any number of elements and restrictions.

• "Sets of these objects" is somewhat incompatible with the ambiguous "objects can be listed in any order." So is AB different from BA? Jul 3 '13 at 4:13
• I apologize for the confusion. {A,B} = {B,A} Jul 3 '13 at 9:44

Since your options are $A$, one of $\{B, C\}$, and one of $\{D, E\}$, it's easy to see why the cardinality can only be at most 3.

I'm going to tackle this case by case:

# No objects

There is only one possibilitiy for the empty set.

# One object

We can choose any of the 5 objects, so there are 5 possibilities.

# Two objects

If we pick A first, we can choose any of the other four objects to go with it. If we pick any of the other objects first, then there are only three objects that can follow, due to the first restriction (e.g. $BA, BD, BE$ for $B$)

Thus there are $4 + 3 \times 4 = 16$ permutations for two objects.

# Three objects

Pick one item from each of the sets $\{A\}, \{B, C\}, \{D, E\}$. There are $1 \times 2 \times 2 = 4$ possible selections that can be made.

For each selection, there are $3! = 6$ ways to permute the objects, giving a total of $4 \times 6 = 24$ permutations for three objects.

Altogether, this is a grand total of $1 + 5 + 16 + 24 = 46$ permutations.

From a response to a question, it appears that order does not matter.

It is not clear whether the empty set ("none") is to be considered a valid choice. We will assume that it is allowed. If it is not, subtract $1$ from the answer we will obtain.

With such a small number of objects, a carefully drawn up list may be the best approach.

Another good option is the approach taken by Sp3000. There, permutations were being counted, but the ideas can be readily modified to disregard order. For choices of $0, 1, 2, 3$ objects, the answers become $1$, $5$, $8$, and $4$ for a total of $18$ (or $17$ if we don't allow the empty set).

But you may like the following idea. Line up the $5$ objects in front of us, with $B$ and $C$ close to each other, and also $D$ and $E$, like this: $$A\qquad BC \qquad DE.$$

Stand in front of $A$ and decide whether to include her in the set we are building. There are $2$ choices, yes or no.

Now walk over to the $BC$ group. For every choice we made about $A$, we now have $3$ choices, $B$, $C$, or neither. So far, we have $2\times 3$ distinct possibilities.

Finally, walk over and stand in front of the $DE$ group. For every choice we made earlier, there are again $3$ choices, this time $D$, $E$, or neither.

Thus the total number of possible choices is $2\times 3\times 3$, that is, $18$. Again, if the empty set is not allowed, we have $17$ possibilities only.

• You hit the nail on the head. My intention is to generalize the process of generating these combinations, for any number of elements and restrictions. I've updated my updated question accordingly. Jul 3 '13 at 11:17
• For setups precisely of this form (the people belong to disjoint groups of $1$ or more, possibly of different sizes, and at most one member of a group can be selected) the above idea generalizes immediately, and gives implicitly a mechanical way to generate and list. Jul 3 '13 at 12:28

Your question asks for the number of all subsets $S$ of $\{A,B,C,D,E\}$ which are not a superset of $\{B,C\}$ or $\{D,E\}$.

This is a typical application of the principle of inclusion-exclusion.

The total number of subsets of $\{A,B,C,D,E\}$ is $2^5 = 32$. The number of supersets of $\{B,C\}$ in $\{A,B,C,D,E\}$ is $2^3 = 8$ (each of the elements $A,B,C$ may be selected or not). In the same way, the number of supersets of $\{D,E\}$ in $\{A,B,C,D,E\}$ is $8$. The number of common supersets of $\{B,C\}$ and $\{D,E\}$ is 2.

Now the principle of inclusion-exclusion yields the result $$32 - 8 - 8 + 2 = 18.$$