# Bezout's Lemma

Given integers $x$ and $y$, a linear combination of $x$ and $y$ has the form $ax + by$. Bezout's Lemma relates the question of finding the greatest common divisor of $x$ and $y$ to the question of finding linear combinations of $x$ and $y$.

## Technique

Applying the Euclidean Algorithm backwards gives an algorithm to obtain the integer values of $a$ and $b$. We show how to obtain integers $a$ and $b$ such that $16457a + 1638b = 7$. These numbers are calculated in Euclidean Algorithm.

$\begin{array} {llllllllll} 7 & = & 21 &- & 14 \times 1 \\ & = & 21 \times 1 & - & 14 \times 1\\ & = & 21 &- & (77 - 21 \times 3) \times 1\\ & = & - 77 \times 1 &+& 21 \times 4\\ & = & - 77 \times 1 &+ & (1638 - 77 \times 21) \times 4 \\ &= &1638 \times 4 &-& 77 \times 85\\ & = & 1638 \times 4 &- & (16457 - 1638 \times 10) \times 85 \\ & = &16457 \times (-85) &+& 1638 \times 854\\ \end{array}$

## Worked Examples

### 1. Given integers $x, y$, describe the set of all integers $N$ that can be expressed in the form $N=ax+by$, where $a, b$ are integers.

Solution: Let $k = \gcd(x,y)$. Then any integer of the form $kn$, where $n$ is an integer, can be expressed as $ax+by$. We already know that this condition is a necessary condition, so to show that it is sufficient, Bezout's Lemma tells us that there exists integers $a', b'$ such that $k = a' x + b'y$. Therefore,

$kn = (a'n) x + (b'n) y.$

### 2. [Modulo Arithmetic Property I - Multiplicative inverses] Show that if $a, b$ are integers such that $\gcd(a,n)=1$, then there exists an integer $x$ such that $ax \equiv 1 \pmod{n}$.

Solution: Since $\gcd(a,n)=1$, Bezout's Lemma implies there exists integers $x, y$ such that $ax + n y = \gcd (a,n) = 1$. Then

$1 \equiv ax+ny \equiv ax \pmod{n} .$

Note by Calvin Lin
5 years, 7 months ago

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I learned a lot! Thanks :D

- 5 years, 4 months ago

I did not understand what the 1st worked example mean. Any number of form kn can be represented in the form ax+by but that does not mean that all nos of form ax+by will be of form kn. Actually just explain what the question demands?

- 5 years, 3 months ago

The question asks you to classify all numbers that can be written in the form $ax + by$ for given integers $x$ and $y$. The claim is that these numbers are exactly those of the form $k n$, where$k = \gcd (x,y)$ and $n$ is any integer.

As you realized, there are 2 parts to this.
First, we have to show that any number of the form $ax + by$ can be written as $kn$ for some $n$. This is obvious because we can let $x = k x^*, y = ky^*$ then $ax + by = a k x^* + bky^* = k ( ax^* + by^* )$.
Second, we have to show that any number of the form $kn$ can be written as $ax + by$ for some $a$ and $b$. This follows by applying Bezout's lemma, which tells us that there exists integers which satisfy $k = a' x + b' y$, and hence $kn = ( a'n) x + (b'n) y$.

Staff - 5 years, 3 months ago

thanks....

- 5 years, 3 months ago


# Is f(x) = x smooth?

It may sound too basic to even be a question, but I couldn't find a straight answer in Wolfram Alpha, Wolfram Mathworld or Wikipedia. Several other examples of more complicated functions are given.

In Wolfram Mathworld it is written that

A smooth function is a function that has continuous derivatives up to some desired order over some domain. (...) The number of continuous derivatives necessary for a function to be considered smooth depends on the problem at hand, and may vary from two to infinity.

$f(x) = x$ has derivative 1 of the first order and 0 of second order, so I would say based on this it has at least 2 derivatives. I think it also has an infinite number of derivatives which are also 0.

Another page on Wolfram Mathworld says the following:

A $C^{\infty}$ function is a function that is differentiable for all degrees of differentiation. (...) All polynomials are $C^{\infty}$. (...) $C^{\infty}$ functions are also called "smooth" (...).

Since $f(x) = x$ is a polynomial, I'm concluding that the paragraphs above mean it is also smooth.

• I also think they are smooth Aug 24, 2015 at 18:39
• "Smooth" runs the gamut from continuously differentiable to $C^\infty,$ and even that is probably too narrow a spectrum. There is no one fixed definition. In the literature you'll often see things like "for the purposes of this paper, "smooth" will mean ___," where "____" is a precise definition.
– zhw.
Aug 24, 2015 at 18:55
• @CarstenS The question is in the title: Is the function f(x)=x smooth? Aug 24, 2015 at 23:21
• I made an attempt to answer it based on what I found, and I showed my reasoning to provide something to the question. So yes, the question contains an answer, but I still was not sure if this holds up, and I was still interested in further information and others' input. Aug 24, 2015 at 23:27
• @zhw.: Is there any definition of "smooth" that does not apply to $f(x)=x$? Aug 25, 2015 at 6:32

A function is smooth is it has derivatives of infinite order. $f(x) = x$ is smooth because it has infinitely many derivatives which are all 0, except for the first one. Polynomials are smooth because eventually their derivatives are 0.

• Yes, that's the standard definition Aug 24, 2015 at 18:45
• @MichaelMenke Not so sure that is standard.
– zhw.
Aug 24, 2015 at 18:57
• @IllegalImmigrant yeah technically smooth is $C^\infty$, but many times for all the calculations we only need our function to be $C^1$ or $C^2$ or something.. so we abuse the notation and we call those function "smooth", which basically comes to mean "it's regular enough to employ all the theorems we need with no headaches"
– Ant
Aug 24, 2015 at 21:56
• $x\mapsto x$ satisfies every definition you guys have given. Aug 25, 2015 at 2:36
• What are "derivatives of infinite order"?
– JiK
Aug 25, 2015 at 10:23

Yes, the identity function has derivatives of every finite order, and is therefore smooth. It doesn't matter that most of the derivatives are $0$ everywhere -- being $0$ is a perfectly cromulent way to exist.

• I had to look that word up. Between this and popularizing schadenfreude, The Simpsons has really influenced the language, hasn't it? Aug 25, 2015 at 2:34
• @columbus8myhw yes, it's certainly embiggened our vocabulary Aug 25, 2015 at 3:30

You may be having issues with the difference between existence and triviality.

If $f(x)=x$ then

$f(x)=x$ is continuouss

$f'(x)=1$ is continuous

$f''(x)=0$ is continuous

$f'''(x)=0$ is continuous

etc...

So all its derivatives are continuous. On the other hand, take $g(x)=x\times|x|$

$g(x)=x\times|x|$ is continuous

$g'(x)=\frac{|x|}2$ is continuous

$g''(x) = \frac12$ if $x>0$, $g''(x)=-\frac12$ if $x<0$ $g''(0)$ is undefined

Clearly $g''$ is not continuous because of $g''(0)$ not existing, and so $g$ only has two derivatives.

Intuitively, all of $f$'s derivatives have no breaks in their graph ($y=0$ is simply a nice line), while $g''$ graph has a gaping hole in it at $x=0$.

• Funny definition of a "gaping" hole, that is a nothingth of a unit wide :) Aug 25, 2015 at 15:10
• @thepeer Is a unit more or less than the gap of the grand canyon? Aug 25, 2015 at 17:21
• Does it matter, seeing as we're talking about a nothingth of it? Sep 1, 2015 at 8:01


# Checking an identity involving binomial coefficients

I need some help to check the following identity:

for every $0\leq i\leq l\leq r$ $$\sum_{j=0}^i\binom{r-l+i-j}{i-j}\binom{l-i+j}{j}=\binom{r+1}{i}.$$

Is this true ?

Answering to John, this identity come from a geometric problem. I was computing the top Chern class of a vector bundle of the form $S\otimes Q$, $S$ is a line bundle, and $Q$ a rank $r$ vector bundle whose Chern classes have a particularly easy form. I have a guess for what should be the answer and imposing equality between the two led me to this identity. I am pretty sure my guess is true and this identity indeed holds; it's interesting that it does not depend on $l$. I think it should be possible to prove this with somee induction argument.

• How did you arrive at this answer? – John Jun 3 '16 at 19:12
• Answering to John, this identity come from a geometric problem. I was computing the top Chern class of a vector bundle of the form $S \otimes Q$, $S$ is a line bundle, and $Q$ a rank $r$ vector bundle whose Chern classes have a particularly easy form. I have a guess for what should be the answer and imposing equality between the two led me to this identity. I am pretty sure my guess is true and this identity indeed holds; it's interesting that it does not depend on $l$. I think it should be possible to prove this with some induction argument. – S. S. Jun 3 '16 at 19:56
• I wrote the question as an unregistered user – S. S. Jun 3 '16 at 19:57
• Upper negation (see proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient ) yields $\dbinom{l-i+j}{j} = \left(-1\right)^j \dbinom{j-\left(l-i+j\right)-1}{j} = \left(-1\right)^j \dbinom{i-l-1}{j}$ and $\dbinom{r-l+i-j}{i-j} = \left(-1\right)^{i-j} \dbinom{\left(i-j\right)-\left(r-l+i-j\right)-1}{i-j} = \left(-1\right)^{i-j} \dbinom{l-r-1}{i-j}$ and $\dbinom{r+1}{i} = \left(-1\right)^i \dbinom{i-\left(r+1\right)-1}{i} = \left(-1\right)^i \dbinom{i-r-2}{i}$. Now your identity follows from the Chu-Vandermonde identity, applied to $i-l-1$ and $l-r-1$ and $i$. – darij grinberg Jun 3 '16 at 20:07

Yes, that is true, it is the "double convolution" formula.
To put the formula in more general terms, define the binomial in the extended way: $$\left( \matrix{ x \cr m \cr} \right) = \left\{ \matrix{ {{x^{\,\underline {\,m\,} } } \over {m!}}\quad \left| {\;0 \le {\rm integer}\;m} \right. \hfill \cr 0\quad \quad \left| {\;{\rm otherwise}} \right. \hfill \cr} \right.$$ Then the Upper Negation rule tells that $$\left( \matrix{ x \cr m \cr} \right) = \left( { - 1} \right)^m \left( \matrix{ m - x - 1 \cr m \cr} \right)$$ and therefore, applying it twice \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\,i} \right)} {\left( \matrix{ r - l + i - j \cr i - j \cr} \right)\left( \matrix{ l - i + j \cr j \cr} \right)} = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,\,i} \right)} {\left( { - 1} \right)^i \left( \matrix{ - r + l - 1 \cr i - j \cr} \right)\left( \matrix{ - l + i - 1 \cr j \cr} \right)} = \cr} and by the "simple" convolution $$= \left( { - 1} \right)^i \left( \matrix{ - r + i - 2 \cr i \cr} \right) = \left( \matrix{ r + 1 \cr i \cr} \right)\quad \left| {\;\left\{ \matrix{ {\rm integer}\;i \hfill \cr {\rm real}\;r,l \hfill \cr} \right.} \right.$$

The Question: $\ds{\sum_{j=0}^{k}{r - \ell + k - j \choose k - j}{\ell - k + j \choose j} = {r + 1 \choose k}}$

\begin{align} &\color{#f00}{\sum_{j=0}^{k}{r - \ell + k - j \choose k - j} {\ell - k + j \choose j}} \\[3mm] = &\ \sum_{j=0}^{\infty}{-r + \ell - k + j + k - j - 1\choose k - j}\pars{-1}^{k - j} {-\ell + k - j + j - 1 \choose j}\pars{-1}^{j} \\[3mm] = &\ \pars{-1}^{k}\sum_{j=0}^{\infty}{\ell - r - 1\choose k - j} {k - \ell - 1 \choose j} = \pars{-1}^{k}\sum_{j=0}^{\infty}{\ell - r - 1\choose k - j} {k - \ell - 1 \choose k - \ell -1 - j} \\[3mm] = &\ \pars{-1}^{k}\sum_{j=0}^{\infty}\bracks{\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{\ell - r - 1} \over z^{k - j + 1}}\,{\dd z \over 2\pi\ic}} \bracks{\oint_{\verts{w} = 1^{-}} {\pars{1 + w}^{k - \ell - 1} \over w^{k - \ell - j}} \,{\dd w \over 2\pi\ic}} \\[3mm] = &\ \pars{-1}^{k}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{\ell - r - 1} \over z^{k + 1}} \oint_{\verts{w} = 1^{-}} {\pars{1 + w}^{k - \ell - 1} \over w^{k - \ell}}\sum_{j = 0}^{\infty}\pars{zw}^{j} \,{\dd w \over 2\pi\ic}\,{\dd z \over 2\pi\ic} \\[3mm] = &\ \pars{-1}^{k}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{\ell - r - 1} \over z^{k + 1}} \oint_{\verts{w} = 1^{-}} {\pars{1 + w}^{k - \ell - 1} \over w^{k - \ell}\pars{1 - zw}} \,{\dd w \over 2\pi\ic}\,{\dd z \over 2\pi\ic} \\[3mm] \stackrel{w\ \to\ 1/w}{=}\ &\ \pars{-1}^{k}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{\ell - r - 1} \over z^{k + 1}}\ \overbrace{\oint_{\verts{w} = 1^{+}} {\pars{1 + w}^{k - \ell - 1} \over w - z}\,{\dd w \over 2\pi\ic}} ^{\ds{\pars{1 + z}^{k - \ell - 1}}}\ \,{\dd z \over 2\pi\ic} \\[3mm] = &\ \pars{-1}^{k}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{k - r - 2} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} = \pars{-1}^{k}{k - r - 2 \choose k} \\[3mm] = &\ \pars{-1}^{k}{-k + r + 2 + k - 1 \choose k}\pars{-1}^{k} = \color{#f00}{{r + 1 \choose k}} \end{align}


# Find the sum of α + β.

#### anemone

##### MHB POTW Director
Staff member
Hi members of the forum,

Problem:
The real numbers
$\displaystyle \alpha$, $\displaystyle \beta$ satisfy the equations

$\displaystyle \alpha^3-3\alpha^2+5\alpha-17=0,$

$\displaystyle \beta^3-3\beta^2+5\beta+11=0.$

Find $\displaystyle \alpha + \beta$.

This problem has me stumped. It's not that I didn't try, believe you me, but I kept returning back to square one no matter how hard I tried. It's really annoying and I don't believe the author wants us to use the cubic formula to find each root explicitly (or even a numeric root-finding technique) to get the sum, but at this moment, I just don't see any way to manipulate the given two cubic polynomials to find the desired sum.

Could you please shed some light on this problem for me?

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi members of the forum,

Problem:
The real numbers
$\displaystyle \alpha$, $\displaystyle \beta$ satisfy the equations

$\displaystyle \alpha^3-3\alpha^2+5\alpha-17=0,$

$\displaystyle \beta^3-3\beta^2+5\beta+11=0.$

Find $\displaystyle \alpha + \beta$.

This problem has me stumped. It's not that I didn't try, believe you me, but I kept returning back to square one no matter how hard I tried. It's really annoying and I don't believe the author wants us to use the cubic formula to find each root explicitly (or even a numeric root-finding technique) to get the sum, but at this moment, I just don't see any way to manipulate the given two cubic polynomials to find the desired sum.

Could you please shed some light on this problem for me?

Hi anemone,

Let me outline the method of solving this problem. Firstly you don't know whether each of these equations have any real roots. So use the discriminant of each equation find out how many real roots each equation has (Refer >>this<<). You will see that each equation has only one real root.

Let $$\alpha+\beta=\lambda$$. Add the two equations together and rearrange it to get an cubic equation of $$\lambda$$. You should get,

$\lambda^3-3\alpha\beta(\lambda-2)-3\lambda^2+5\lambda-6=0.$

Try to find a value for $$\lambda$$ that satisfies the above equation (by trial and error). Do you think that the value you obtained for $$\lambda$$ is unique? If so why?

Kind Regards,
Sudharaka.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Another way. Define $f(x)=x^3-3x^2-5x$. We have $f(\alpha)+f(\beta)=6$.We can exoress $f(x)=(x-1)^3+2(x-1)+3$, as a consequence
$$f(\alpha)-3=(\alpha-1)^3+2(\alpha -1)\\ f(\beta)-3=(\beta-1)^3+2(\beta-1)$$
Adding the above equalities we obtain $0=(\alpha-1)^3+(\beta-1)^3+2(\alpha+\beta-2)$. Now, decompose to get $0=(\alpha +\beta-2)g(\alpha,\beta)$ where $g(\alpha,\beta)>0$, so $\alpha+\beta=2$.

#### anemone

##### MHB POTW Director
Staff member
Hi anemone,

$\lambda^3-3\alpha\beta(\lambda-2)-3\lambda^2+5\lambda-6=0.$

Try to find a value for $$\lambda$$ that satisfies the above equation (by trial and error). Do you think that the value you obtained for $$\lambda$$ is unique? If so why?

Kind Regards,
Sudharaka.
Hi Sudharaka, my first instinct is to try the value of $\displaystyle \lambda=2$ and it does work. I think the value of $\displaystyle \lambda$ that I obtained is unique since the other two roots from both of these functions are imaginary roots. Am I correct?

Another way. Define $f(x)=x^3-3x^2-5x$. We have $f(\alpha)+f(\beta)=6$.We can exoress $f(x)=(x-1)^3+2(x-1)+3$, as a consequence
$$f(\alpha)-3=(\alpha-1)^3+2(\alpha -1)\\ f(\beta)-3=(\beta-1)^3+2(\beta-1)$$
Adding the above equalities we obtain $0=(\alpha-1)^3+(\beta-1)^3+2(\alpha+\beta-2)$. Now, decompose to get $0=(\alpha +\beta-2)g(\alpha,\beta)$ where $g(\alpha,\beta)>0$, so $\alpha+\beta=2$.
Hi Fernando, it feels so great to receive two different methods in a day to solve my headache. Thanks enormously!

But another idea hit me when I was having my dinner tonight (I guess there is no rest for the weary.) and I'd really like to hear some opinions about it, if possible.

Let $\displaystyle f(x)=x^3-3x^2+5x-17$ and $\displaystyle g(x)=x^3-3x^2+5x+11$ and if we rewrite the function of g(x) to become $\displaystyle g(x)=(x^3-3x^2+5x-17)+28$, we can see that the function of g(x) is just the function of f(x) that shifted 28 units upward, i.e. g(x) is the transformed function of f(x).

I then use a simple trick by exploiting the fact that the gradient of the function of f(x) and its transformed function, g(x) would be the same at their real root point.

That is, $\displaystyle f'(\alpha)=\displaystyle g'(\beta)$.
$\displaystyle 3\alpha^2-6\alpha+5=3\beta^2-6\beta+5$
$\displaystyle \alpha^2-2\alpha=\beta^2-2\beta$
$\displaystyle \alpha^2-\beta^2=2\alpha-2\beta$
$\displaystyle (\alpha+\beta)(\alpha-\beta)=2(\alpha-\beta)$
$\displaystyle (\alpha-\beta)(\alpha+\beta-2)=0$

Since $\displaystyle (\alpha-\beta)\ne 0$, we can conclude that $\displaystyle (\alpha+\beta-2)=0$, or $\displaystyle \alpha+\beta=2$.

Hmm...how does that sound to you?

#### Fernando Revilla

##### Well-known member
MHB Math Helper
I then use a simple trick by exploiting the fact that the gradient of the function of f(x) and its transformed function, g(x) would be the same at their real root point.
Hmm, why? Consider for instance $f(x)=x^3-1$ and $g(x)=x^3-8$. Here $\alpha=1$ and $\beta=2$, however $f'(\alpha)=3$ and $g'(\beta)=12$ i.e. $f'(\alpha)\neq g'(\beta)$.

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi Sudharaka, my first instinct is to try the value of $\displaystyle \lambda=2$ and it does work. I think the value of $\displaystyle \lambda$ that I obtained is unique since the other two roots from both of these functions are imaginary roots. Am I correct?

Yes that's correct. You are welcome!

#### anemone

##### MHB POTW Director
Staff member
Hmm, why? Consider for instance $f(x)=x^3-1$ and $g(x)=x^3-8$. Here $\alpha=1$ and $\beta=2$, however $f'(\alpha)=3$ and $g'(\beta)=12$ i.e. $f'(\alpha)\neq g'(\beta)$.
I'm so embarrassed to think of this now, as I really should have verified the result by using another pair of simple functions (such as your example) and I understand it now, these two functions (the original and its transformed function) have the same gradient at the same point.

Thanks for commenting on my idea!

Yes that's correct. You are welcome!
Thanks again, Sudharaka!

#### MarkFL

Staff member
I'm so embarrassed to think of this now...
Please don't feel embarrassed, we all make mistakes...I know I have made my share of mistakes, and the potential embarrassment is magnified when it is public, but when we come together in the forums to discuss a problem there will be mistakes, but we have our friends to guide us and hence we learn.

Others reading this topic can benefit from having seen your "mistake" and learn why this does not work. I feel that if one person makes an error, then there are others who will make this same error.

By being willing to take a chance and show your thinking, you have helped others.

#### anemone

##### MHB POTW Director
Staff member
Hey Mark, it's so sweet and nice of you to say that!

#### Fernando Revilla

##### Well-known member
MHB Math Helper
I'm so embarrassed to think of this now, ...
The only way to not make mistakes is to never do anything.

#### anemone

##### MHB POTW Director
Staff member
The only way to not make mistakes is to never do anything.
I'm so touched by the genuine kindness shown by the community at this site, everyone here is so helpful and friendly...

Thanks to all of you, Mark, Fernando and Sudharaka because you all just made my day!

Last edited:


MAT244-2018S > Quiz-2

Q2-T5101

(1/1)

Victor Ivrii:
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3 + x(1 + y^2)y' = 0,\qquad \mu(x, y) = 1/xy^3.$$

Tianjing Ruan:
x2y3/xy3 + x(1 + y2)/xy3y' = 0

x + (y−3 + y−1)y' = 0

∫M dx = 1/2x2 + g(y)

g(y) = ∫y−3 + 1/ydy = -1/2y−2 + ln(y)

The solution is:
1/2x2 − 1/2y2 + ln(y) = C

Junya Zhang:
First, let's show the given DE $x^{2}y^{3} + x(1+y^{2})y' = 0$ is not exact.
Define $M(x,y)=x^{2}y^{3}$, $N(x,y)=x(1+y^{2})$
$$M_y = \frac{\partial}{\partial y}[x^{2}y^{3}] = 3x^{2}y^{2}$$ $$N_x = \frac{\partial}{\partial x}[x(1+y^{2})] = 1+y^{2}$$
Since $3x^{2}y^{2} ≠ 1+y^{2}$, this implies the given DE is not exact.

Now, let's show that the given DE multiplied by the integrating factor $\mu(x,y) = \frac{1}{xy^{3}}$ is exact.
That is to show $$\frac{1}{xy^{3}}x^{2}y^{3} + \frac{1}{xy^{3}}x(1+y^{2})y' = x + (y^{-3}+y^{-1})y'= 0$$ is exact.

Define $M'(x,y) = x$, $N'(x,y) = y^{-3}+y^{-1}$
Since
$$M'_y = \frac{\partial}{\partial y}(x) = 0$$ $$N'_x = \frac{\partial}{\partial x}[y^{-3}+y^{-1}] = 0$$
By theorem in the book, we can conclude that $x + (y^{-3}+y^{-1})y'= 0$ is exact.

Thus, we know there exists a function $\phi(x,y)=C$ which satisfies the given DE.
Also,
$$\frac{\partial \phi}{\partial x} = x$$ $$\frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$$
Integrate $\frac{\partial \phi}{\partial x} = x$ with respect to $x$ we have
$$\phi(x,y) = \frac{1}{2}x^{2} + g(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial\phi}{\partial y} = g'(y)$$
Since we know that $\frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$
Then $g'(y) = y^{-3}+y^{-1}$
Integrate with respect to $y$ we have
$g(y) = -\frac{1}{2}y^{-2} + ln|y| + C$
Altogether, we have $\phi(x,y) = \frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$, which means
$$\frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$$
is the general solution to the given DE.

Besides, notice that the constant function $y(x)=0$ $\forall x$ is also a solution to the given DE.

Victor Ivrii:
ruantian
Please select a proper screen name. Second, your solution is correct but the answer is not due to the lame method to write. Do not use html tags for math, they are inadequate. Use MathJax

Junya
Do not post solutions after correct one was posted. It would be much better if you wrote something like:  "ruantian solved problem correctly, but presented an answer in the confusing way, correct answer is ... "


# Proving complex numbers

Let $z_1,z_2$ be two complex numbers such that $z_1 + z_2$ and $z_1\dot\ z_2$ are each negative real numbers. Prove that $z_1$ and $z_2$ must be real numbers.

My attempt at a solution follows,

Let $z_1 = a+bi$ and $z_2=c+di$. The hypotheses states that $a+c<0$, $b+d = 0$, $ac-bd <0$, and $ad + bc=0$. And I get stuck here since I do not know the logic to proceed in which I can prove that $z_1$ and $z_2$ must be real numbers.

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$d=-b\Rightarrow d(a-c)=0$ in your last equality $\to$ either $d$ or $a-c$ is $0$. So... –  L. F. Jan 17 '13 at 1:21
@LordoftheFlies I am not sure what you are trying to do? So in order to prove that they are real I mist show d or a is 0? –  Q.matin Jan 17 '13 at 1:24
$a-c<0$ so it cannot be $0$. Therefore $d=0$ i.e. $z_2$ is real. Since $d=-b\Rightarrow b=0$ i.e. $z_1$ is real. –  L. F. Jan 17 '13 at 1:26

From $b+d=0$, get $d=-b$. Then from $ad+bc=0$ get $b(c-a)=0$.

So either $b=0$, from which $d=0$ making $z_1,z_2$ real, or else $c=a$ (and $d=-b$ still holds). Now put those into $ac-bd<0$ and it becomes $a^2+b^2<0$, contradiction.

NOTE: Give credit to "Lord of the Flies", whose comment came in while I was writing this.

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Thanks but then what follows directly that says $z_1$ and $z_2$ are real numbers? –  Q.matin Jan 17 '13 at 1:34
The argument above shows that either both $b$ and $d$ are zero, or else a contradiction occurs. Can conclude that is must be that $b=d=0$. The $b,d$ are the imaginary parts of the two complex numbers $z_1,z_2$ using your notation, and saying imaginary part is $0$ is the same as saying a complex number is real. –  coffeemath Jan 17 '13 at 1:36
Ahh, a bulb just turned on in my head when you said, "saying imaginary part is 0 is the same as saying a complex number is real". Thanks a lot coffeemath!! –  Q.matin Jan 17 '13 at 1:41

$b+d=0 \implies d=-b \implies ad+bc=-ab + bc = 0\implies b(c-a)=0$.

• So either $b=0$, from which $d= b = 0$, and thus $z_1,z_2$ are real,
• or else $c-a= a-c = 0$ which contradicts the fact that $a - c < 0.$

Therefore $z_1, z_2 \in \mathbb{R}$.

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Thanks but did you mean $c-a$ instead of $a-c$? –  Q.matin Jan 17 '13 at 1:36
amWhy: Is $a-c<0$ a hypothesis, or is it $a+c<0$? –  coffeemath Jan 17 '13 at 1:38

Alternately, since $z_1+z_2$ is real and $z_1z_2$ is negative real, and $$(z_1-z_2)^2=(z_1+z_2)^2-4z_1z_2,$$ we conclude that $(z_1-z_2)^2$ is positive. Thus $z_1-z_2$ is real. Since $z_1+z_2$ is real, it follows by addition that $2z_1$, and therefore $z_1$, is real.

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Isn't $z_1+z_2$ also negative real? –  Q.matin Jan 17 '13 at 1:51
Yes it is negative real, but it gets squared on the right side so becomes positive in its contribution. –  coffeemath Jan 17 '13 at 1:54
In the problem, it is. But the above proof does not need or use the fact that $z_1+z_2$ is negative. It does use the fact that the product $z_1z_2$ is negative. –  André Nicolas Jan 17 '13 at 1:54

You have chosen to splitting the complex numbers into real and imaginary parts. To that end, you have faithfully translated the givens of your problem into this form.

What you've overlooked is that you haven't translated your goal into this form. Doing this isn't always the simplest way to proceed, but it's almost always a correct and straightforward way to proceed. The translation of your goal is that you seek to prove $b=d=0$.

Finally, even without an idea of how to proceed, you do know something you can do with a system of equations and inequations: you can simply / solve it! Sometimes, the simplified/solved form will suggest what to do next. So without knowing where you're going, we can still do the following:

• Solve $b+d = 0$ to determine $d = -b$
• Simplify $ad+bc = 0$ to $b(c-a) = 0$
• Solve that equation by splitting into cases:
• Case $b=0$:
• Then $d=0$ too.
• The remaining inequalities are $a+c < 0$ and $ac - bd < 0$
• The latter simplifies to $ac < 0$
• We can solve this by cases:
• Case $a < 0$ and $c > 0$:
• The remaining inequation yields $a < -c < 0$
• Case $a > 0$ and $c < 0$:
• The remaining inequation yields $0 < a < -c$
• Case $(c-a)=0$:
• Then $c = a$
• The remaining inequalities are $a+c < 0$ and $ac - bd < 0$
• Thes latter simplifies to $a^2 < -b^2$
• This is a contradiction! (the r.h.s is nonpositive and the l.h.s. is nonnegative)

So the complete solution to the givens is the union of the cases:

• $a < 0$ and $c > 0$ and $|c| < |a|$ and $b=d=0$
• $a > 0$ and $c < 0$ and $|a| < |c|$ and $b=d=0$

And then once you've reached this point, you could return to the problem to see how you could use this information to show $z_1$ and $z_2$ are real numbers.

(Of course, if you translated the goal, you wouldn't have had to do any of the work in the $b=0$ case)

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Thank so much!! THis is very helpful! –  Q.matin Jan 18 '13 at 1:46

$(x\!-\!z_1)(x\!-\!z_2) = x^2\! - (z_1\!+\!z_2) x + z_1 z_2\in \Bbb R[x]$ has discriminant $\,(z_1\!+\!z_2)^2\!-4z_1z_2 > 0,$ since $z_1z_2 < 0,$ hence the roots $z_1,z_2$ are real (by the quadratic formula).

Remark $\$ The proof works more generally for $\ z_1,z_2$ elements of any (integral) domain $\,\Bbb D\supset \Bbb R,$ assuming only that $\,z_1\!+z_2,\, z_1 z_2\,$ are both in $\Bbb R,$ and $\,z_1 z_2 < 0$. The hypothesis that $\Bbb D$ is a domain ensures that the list of roots $z_1,z_2 \in \Bbb R$ obtained by the quadratic formula persists as the unique list of roots in $\Bbb D$ (if the quadratic polynomial had an additional root $z_3\in \Bbb D$ then it would have more roots than its degree, which cannot occur in a domain). The claim may fail in extension rings not domains, e.g. $f(x) = (x+2)(x-1) = x^2+x-2 = (x-w)(x+w+1)\,$ over $\,\Bbb R[w]/(w^2+w-2),\,$ where $f$ has non-real roots $\ w,\, -w-1 \not\in \Bbb R,\,$ contra to the claim.

Notice that said extension ring is not a domain since there $\,(w+2)(w-1) = 0,\,$ but $\,w \ne -2,1$. Informally, one may think of $w$ as an algebraic representation of a "generic" root of $f$, sharing all $(\Bbb R -)$algebraic properties of the roots $-2,1,$ i.e. $w$ satisfies $f(w) = g(w)$ for polynomials $f,g\in\Bbb R[x]\,$ iff $\,-2$ and $1$ also do: $f(-2) = g(-2)$ and $f(1) = g(1).\,$ While we can always construct extension rings containing such generic roots, generally the construction does not preserve the property of being a domain, and the consequent uniqueness of the list/multiset of roots (which, suitably formulated, is a characteristic property of domains among rings). Yet another example that uniqueness theorems provide powerful tools for deducing equalities.

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...from which we also see that the hypotheses can be weakened to "$z_1+z_2 \in \mathbb{R}$ and $z_1 z_2 < 0$". –  Hans Lundmark Jan 17 '13 at 10:13
@Hans Yes, in fact it works much more generally - see my recently added remark. –  Math Gems Jan 17 '13 at 17:19
@MathGems Thanks a lot! –  Q.matin Jan 18 '13 at 1:46


# Tiling a diamond-shaped grid with tetrominoes

You have a grid like this:

(The entire grid isn't shown as it would be too large, but the number of squares in each row are as follows: $2, 4, 6, \ldots, 96, 98, 100, 100, 98, 96, \ldots, 6, 4, 2$.)

We define this grid as $G(100)$, as it is 100 squares across at its widest point and 100 squares high at its tallest point.

You want to tile it with only copies of this tetromino:

You may rotate or flip the tetromino.

1. Is it possible? Why or why not? An explanation in your answer is required.

2. For which even positive integer values of $n$ is tiling the grid $G(n)$ with only the tetromino possible? (Again, you must provide an explanation.)

note: I will post my own solution after either two days have passed or two distinct correct answers (for each question 1 and 2) have been provided.

• Sanity check: $G(100)$ has $4\cdot \binom{50+1}{2}=5100$ squares? – Mike Earnest Apr 25 '15 at 21:28
• @MikeEarnest Correct. (Alternatively, $4 \cdot \sum_{n=1}^{50}n = 5100$.) – Doorknob Apr 25 '15 at 21:34
• In general $G(N)$ has $2⋅N⋅(N+1)$ squares – Ivo Beckers Apr 25 '15 at 22:15
• @IvoBeckers Don't think that's right; it should be $N \cdot (\frac{n}{2}+1)$ (note that there are $5100$ squares for $N = 100$), or equivalently $\frac{N^2+2N}{2}$. – Doorknob Apr 25 '15 at 22:19
• you're right. I was thinking of half the value of $N$ – Ivo Beckers Apr 25 '15 at 22:20

$G(100)$ cannot be tiled.

Divide an infinite checkerboard into $2\times 2$ blocks, then color each block alternately white and black, as shown: $$\begin{array}{ccccc|ccccc} &&\vdots&&\vdots&&\vdots&&\vdots&&\\ \cdots & B & B & W & W & B & B & W & W &\cdots\\ \cdots & B & B & W & W & B & B & W & W &\cdots\\ \hline \cdots & W & W & B & B & W & W & B & B &\cdots\\ \cdots & W & W & B & B & W & W & B & B &\cdots\\ &&\vdots&&\vdots&&\vdots&&\vdots&\\ \end{array}$$ Then, place the $G(100)$ array on top of this so that its axes of symmetry are the two lines (one horizontal, the other vertical) above, and let $G(100)$ be colored to match the the board below it.

No matter how a tile is placed, it will cover either 3 whites and a black, or 3 blacks and a white. This means that whenever an odd number of tiles are placed, the area they cover will be unbalanced between black and white. But covering $G(100)$ entails placing $1275$ tiles, and $G(100)$ itself is white/black balanced, so it cannot be covered. This proof also shows that $G(2n)$ is not tileable whenever $n\equiv 1$ or $2$ (mod $4$), since there are $1+2+\dots+n$ tiles to be placed, which is odd whenever $n=4k+1$ or $4k+2$.

We cannot tile the given grid.

Below is an example coloring of $G(12)$. Notice that a rotation of 180° about the central green dot swaps the colors, thus we have an equal amount of red and white tiles, and we can extend this coloring to any $G(2n)$, in particular, $G(100)$.

Let's consider the S and Z tetrominoes separately. Notice that no matter where an S tetromino is placed, it covers either 4-0, 0-4, or 2-2 of white-red. Thus when we place an S tetromino we do not change the difference between the amounts of red and white tiles modulo 4. Similarly any Z tetromino covers 3-1 or 1-3, thus adding 2 to the difference modulo 4. To end with 0 red and 0 white tiles (ie to cover the entire board) we must have the difference be 0 mod 4, and thus we must have an even number of Z tetrominoes.

This argument is completely symmetrical; there is no real difference between S and Z tetrominoes. If we can tile the board using an odd number of S tetrominoes, simply mirror the board turning all S into Z and vice versa and we have tiled the board using an odd number of Z tetrominoes, which we know is impossible. Thus we have an even number of both S and Z tetrominoes, and an even number of tetrominoes overall. This means that the area of the board must be divisible by 8. But in the case of $G(100)$, it's not.

The size of $G(2N)$ is $4(1 + \dots + N) = 4\frac{N(N+1)}{2} = 2N(N+1)$. If 8 divides this then $8 | 2N(N+1) \iff 4|N(N+1) \iff 4|N$ or $4|N+1$.

• Cool argument! We used different colorings, yet proved that the tiling is impossible for the same values of $n$. I've checked $G(2\cdot 3)$ is not tileable, I guess the next step is to check $G(2\cdot 4)$. – Mike Earnest Apr 26 '15 at 0:06
• @MIkeEarnest, Just case checked. $G(8)$ is impossible as well. – Ben Frankel Apr 26 '15 at 0:17


# Is a real matrix that is both normal and diagonalizable symmetric? If so, is there a proof of this not using the spectral theorem?

Given a quadratic real matrix $$A$$ for which we know it is diagonalizable ($$D = P^{-1}AP$$ for a diagonal matrix $$D$$) and that it is normal ($$AA^T = A^TA$$), is it true that $$A$$ is symmetric? By the spectral theorem over $$\mathbb{C}$$, $$A$$ is orthogonally diagonalizable (say via some unitary matrix $$Q$$). It then seems to me that from this we should get that the eigenspaces are orthogonal and hence there must also exist a real orthonormal basis of eigenvectors. Is this last correct? From this it would then follow by the real version of the spectral theorem that $$A$$ is symmetric.

If it is, is there a way to prove this statement directly, without appealing to the spectral theorem? This would give a nice justification/explanation for the difference in the two spectral theorems over $$\mathbb{C}$$ and $$\mathbb{R}$$ relating it directly to whether the matrix is diagonalizable in the first place.

• I'm not sure I understand the question: since the conjugate of a real number is itself, a real matrix is normal if and only if it is symmetric. (And all normal matrices are automatically diagonalizable, including symmetric real matrices.) Mar 28, 2021 at 17:55
• That’s not true: A real orthogonal matrix is also normal but is in general not symmetric (consider e.g. rotation by 90 degrees). A normal real matrix in general need not diagonalizable over the reals, as the example of orthogonal matrces shows (it is diagonalizable over the complex numbers, of course). Mar 28, 2021 at 17:57
• Ack, you're completely right. I was confusing normal with Hermitian. Mar 28, 2021 at 19:53

Here is a proof avoiding the spectral Theorem.

Let $$\lambda _1, \lambda _2, \ldots , \lambda _k$$ be the distinct eigenvalues of $$A$$ and let $$E_1, E_2, \cdots , E_k$$ be the projections onto the corresponding eigenspaces.

By seeing things from the point of view of a (not necessarily orthonormal) basis of eigenvectors, it is very very easy to prove that (don't let the long expression scare you) $$E_i = \frac{(A-\lambda _1)\ldots \widehat{(A-\lambda _i)}\ldots (A-\lambda _k)}{(\lambda _i-\lambda _1)\ldots \widehat{(\lambda _i-\lambda _i)}\ldots (\lambda _i-\lambda _k)},$$ where the hat means omission.

From this it is clear that each $$E_i$$ is also normal.

Lemma. Any normal, idempotent, real matrix is symmetric.

Proof. Let $$E$$ be such a matrix. We first claim that $$\text{Ker}(E)=\text{Ker}(E^TE)$$. To see this, observe that the inclusion $$\text{Ker}(E)\subseteq \text{Ker}(E^TE)$$ is evident. On the other hand, if $$x\in \text{Ker}(E^TE)$$, then $$\|Ex\|^2 = \langle Ex, Ex\rangle = \langle E^TEx, x\rangle =0,$$ so $$x\in \text{Ker}(E)$$.

We then have that $$\text{Ker}(E)=\text{Ker}(E^TE) =\text{Ker}(EE^T) =\text{Ker}(E^T).$$

Recalling that the range $$R(A^T)$$, of the transpose of a matrix $$A$$, coincides with $$\text{Ker}(A)^\perp$$, we then have that $$R(E^T) = \text{Ker}(E)^\perp = \text{Ker}(E^T)^\perp = R(E).$$ We then see that $$E$$ and $$E^T$$ are projections sharing range and kernel, so necessarily $$E=E^T$$. QED

Back to the question, we then have that $$A=\sum_{i=1}^k \lambda _kP_k,$$ so we conclude that $$A$$ is symmetric.

You can also do this with simultaneous diagonalizability and two classical inequalities: Cauchy-Schwarz and rearrangement inequality.

$$A$$ is real diagonalizable and so is $$A^T$$. Now $$A$$ and $$A^T$$ commute, hence they are simultaneously diagonalizable such that $$A=SD_1S^{-1}$$ and $$A^T = SD_2S^{-1}$$. Since the trace is invariant to conjugation, this implies

(i.)
$$\text{trace}\big(A^2\big)=\text{trace}\big(D_1^2\big)\geq \text{trace}\big(D_2D_1\big)=\text{trace}\big(A^TA\big) =\big\Vert A\Big \Vert_F^2$$
by rearrangement inequality

(ii.)
$$\text{trace}\big(A^2\big)\leq \text{trace}\big(A^TA\big) =\big\Vert A\Big \Vert_F^2$$
by Cauchy-Schwarz

Thus $$\text{trace}\big(A^2\big)= \big\Vert A\Big \Vert_F^2$$ so Cauchy-Schwarz is met with equality$$\implies A = \eta\cdot A^T$$, where $$\eta \in\big\{-1,1\big\}$$ because $$A$$ and its transpose have the same Frobenius norms. Thus $$A$$ is either symmetric or skew symmetric. (And supposing $$A\neq \mathbf 0$$ we can eliminate skew symmetry because $$\mathbf x^T A\mathbf x=0$$ for all $$\mathbf x\in \mathbb R^n$$ by skew symmetry, so $$A$$ could not have any non-zero real eigenvalues, which we know isn't true.)

One can obtain that $$A$$ is symmetric by direct computation. The equality $$A^TA=AA^T$$ is $$\tag1 PDP^{-1}P^{-T}DP^T=P^{-T}DP^TPDP^{-1}.$$ Multiplying by $$P^T$$ on the left and by $$P$$ on the right, $$\tag2 P^TPD(P^TP)^{-1}DP^TP=DP^TPD.$$ Now by $$(P^TP)^{-1}$$ on the right, $$\tag3 \big[P^TPD(P^TP)^{-1}\big]\,D=D\,\big[P^TPD(P^TP)^{-1}\big].$$ We may assume without loss of generality that $$A$$ (and hence $$D$$) is invertible, by adding a suitable scalar multiple of the identity. By writing $$D$$ as a block-diagonal matrix with distinct eigenvalues it is easy to see that the matrices that commute with $$D$$ are block diagonal. So, with $$\tag4X=P^TPD(P^TP)^{-1}$$ we have $$XD=DX$$ and thus $$X$$ is block-diagonal. Knowing that $$X$$ is block-diagonal we rewrite $$(4)$$ as $$XP^TP=P^TPD.$$ The (block) diagonal entries of this equality are $$X_{kk}(P^TP)_{kk}=\lambda_k\,(P^TP)_{kk}.$$ Because $$P^TP$$ is positive definite, its block-diagonal entries are positive definite; in particular, invertible. We conclude that $$X_{kk}=\lambda_{kk}\,I$$, that is $$X=D.$$ We can unravel this as $$D=P^TPD(P^TP)^{-1}=P^TPDP^{-1}P^{-T}.$$ Multiplying by $$P^{-T}$$ on the left and by $$P^T$$ on the right we obtain $$A^T=P^{-T}DP^T=PDP^{-1}=A.$$


# If $(a,b)=1$ then prove $(a+b, ab)=1$.

Let $$a$$ and $$b$$ be two integers such that $$\left(a,b\right) = 1$$. Prove that $$\left(a+b, ab\right) = 1$$.

$$(a,b)=1$$ means $$a$$ and $$b$$ have no prime factors in common

$$ab$$ is simply the product of factors of $$a$$ and factors of $$b$$.

Let's say $$k\mid a+b$$ where $$k$$ is some factor of $$a$$.

Then $$ka=a+b$$ and $$ka-a=b$$ and $$a(k-l)=b$$.

So $$a(k-l)=b, \ a\mid a(k-1)$$ [$$a$$ divides the left hand side] therefore $$a\mid b$$ [the right hand side].

But $$(a,b)=1$$ so $$a$$ cannot divide $$b$$.

We have a similar argument for $$b$$.

So $$a+b$$ is not divisible by any factors of $$ab$$.

Therefore, $$(a+b, ab)=1$$.

Would this be correct? Am I missing anything?

Suppose that the gcd is not $1$. Then there is a prime $p$ that divides $ab$ and $a+b$.

But then $p$ divides one of $a$ or $b$, say $a$. Since $p$ divides $a+b$, it follows that $p$ divides $b$. This contradicts the fact that $a$ and $b$ are relatively prime.

• I like this proof. Short and sweet. – ireallydonknow Dec 20 '13 at 18:59
• This method can be motivated intuitively as a generalization of Euclid's method for generating (co)primes - see this answer here. – Bill Dubuque Dec 16 '19 at 18:55

Your jump from "$k\mid a+b$ with $k\mid a$" to $ka=a+b$ seems to be wrong. Just because $k$ is a factor of $a$ doesn't mean at all that the number of times it divides $a+b$ is exactly $a$. That seems to kill the rest of your argument.

Instead, here is an approach that doesn't mention prime factors at all. It starts from the well-known property that $(a,b)=1$ exactly if there are $p,q\in\mathbb Z$ such that $pa+qb=1$.

Square $pa+qb=1$ to get $$p^2a^2 + q^2b^2 + 2pqab = 1$$ If we can show that each of the terms on the left-hand side of this is an integer combination of $a+b$ and $ab$, then the right-hand side is too.

But $2pqab$ is clearly an integer combination of $a+b$ and $ab$ namely $0(a+b)+2pq\cdot ab$.

And $a^2 = a(a+b)-ab$, so $p^2a^2 = p^2a(a+b)-p^2\cdot ab$.

Similarly $q^2b^2 = q^2b(a+b)-q^2\cdot ab$.

Collecting everything, $(p^2a+q^2b)(a+b)+(2pq-p^2-q^2)ab=1$, so $(a+b,ab)=1$.

• Nice. (+1) I came up with another Bezout Identity approach. The coefficient of $ab$ is the same, so I assume the coefficient of $a+b$ is, too; it is just not as obvious. – robjohn Dec 20 '13 at 20:17
• Worth emphasis - at the heart of this method is a classical (bi-)quadratic composition identity (a generalization of the classical Brahmagupta–Fibonacci identity for composition of squares). See this answer here for that viewpoint. – Bill Dubuque Dec 16 '19 at 19:02

Here is another proof using Bezout's Identity: \begin{align} ma+nb&=1\tag{1}\\ (n-m)b&=1-m(a+b)\tag{2}\\ (m-n)a&=1-n(a+b)\tag{3}\\ -(m-n)^2ab&=1-(m+n)(a+b)+mn(a+b)^2\tag{4}\\ 1&=((m+n)-mn(a+b))\color{#C00000}{(a+b)}-(m-n)^2\color{#C00000}{ab}\tag{5} \end{align} Explanation:
$(1)$: $(a,b)=1$ and Bezout's Identity
$(2)$: subtract $m(a+b)$ from both sides of $(1)$
$(3)$: subtract $n(a+b)$ from both sides of $(1)$
$(4)$: multiply $(2)$ and $(3)$
$(5)$: rearrange and collect terms of $a+b$ and $ab$ from $(4)$

Bezout's Identity and $(5)$ say that $(a+b,ab)=1$.

Henning Makholm and Bill Dubuque also give Bezout Identity based proofs. The coefficient of $ab$ is the same in all of our answers, but the coefficient of $a+b$ in mine looks different from theirs. They are the same however: \begin{align} (m+n)-mn(a+b) &=(m+n)\overbrace{(ma+nb)}^{1}-mn(a+b)\\ &=m^2a+n^2b+mn(a+b)-mn(a+b)\\ &=m^2a+n^2b \end{align}

• Generally $\,(a,c)(b,c) = (ab,c)$ when $\,(a,b,c)=1\,$ as proved in this answer here. The proof in Rob's answer is essentially a Bezout based proof of this in the special case $\,c=a\!+\!b,\,(a,b)=1.\,$ In Rob's $(5)$ if we use his equivalent expression $\,m^2a+n^2b\,$ and replace $1$ on the LHS by $\,(ma+nb)^2\,$ it becomes the first identity in the linked answer, which is a classical composition of squares identity. – Bill Dubuque Dec 16 '19 at 21:09

[This answer was merged from a question concerning $$\rm\color{#0a0}{Bezout}$$-based proofs]

Note $$\ (\color{#c00}{a\!+\!b}) (ai^2\!+\!bj^2)= \color{#0a0}{(ai\!+\!bj)^2}\ \!\!+ \color{#c00}{ab}(i\!-\!j)^2\,$$ by calculation (or by composition - see below)
thus $$\ \ n^{\phantom{|^|}}\!\!\!\mid\color{#c00}{a\!+\!b,ab}\,\Rightarrow\, n\mid\color{#0a0}{(ai\!+\!bj)^2}\!=1^2\$$ by choosing $$\,\color{#0a0}{ai\!+\!bj}=1\,$$ by $$\,(a,b)=1\,$$ & Bezout.

Or: $$\,\ \ (\color{#c00}{a\!+\!b}) (ai\!+\!bj)\,=\, \color{#0a0}{a^2i\!+\!b^2j}\,+\, \color{#c00}{ab}(i\!+\!j)\$$
thus $$\,n^{\phantom{|^|}}\!\!\!\mid\color{#c00}{a\!+\!b,ab}\,\Rightarrow\, n\mid\color{#0a0}{a^2i\!+\!b^2j}\!=1\$$ by choosing $$\,i,j\,$$ to get Bezout for $$\,(a^2,b^2)=1\,$$

Remark $$\$$ The genesis of the proof is ideal-theoretic, as is explained in this post, which presents a handful of proofs of the more general identity $$\, (a\!+\!b,\,{\rm lcm}(a,b)) = (a,b).\,$$ The above is essentially a Bezout form of the the last proof there (which is more general than a proof using primes, since it works in any gcd domain - where primes needn't exist). In fact OP is case $$\, c=a\!+\!b,\,(a,b)\!=\! 1\,$$ of

$$(a,b,c)=1\,\Rightarrow\, (a,c)(b,c) = (ab,c(a,b,c)) = (ab,c)\qquad$$

More classically, the first identity may be viewed conceptually as the special case $$\,x=1=y\,$$ in the biquadratic composition of quadratic forms in the Lemma below (used by Euler) whose proof by norm multiplicativity is an obvious generalization of the classical proof when $$\,a=1=b.$$

Lemma $$\ \ (ax^2+by^2) (ai^2+bj^2)= (aix\!+\!bjy)^2\ \!\!+ ab(iy\!-\!jx)^2$$

$$\!\begin{array}{rll}{\bf Proof}\qquad\qquad\ \ (ax^2+by^2)&\!\!\!\!\!\!(ai^2+bj^2)&\!\! =\ N(\alpha)N(\beta)\\[.2em] \!\!\!\!\!(x\sqrt a\!+\!y\sqrt{-b})(x\sqrt a\!-\!y\sqrt{-b})&\!\!\!\!\!\!(i\sqrt a\!-\!j\sqrt{-b})\ (i\sqrt a\!+\!j\sqrt{-b})&\!\!=\ \ \ \alpha\,\bar\alpha\,\beta\,\bar\beta\\[.2em] (x\sqrt a\!+\!y\sqrt{-b})\ (i\sqrt a\!-\!j\sqrt{-b})&\!\!\!\!\!\!(x\sqrt a\!-\!y\sqrt{-b})(i\sqrt a\!+\!j\sqrt{-b})&\!\!=\ \ \ \alpha\,\beta\,\bar\alpha\,\bar\beta\\[.2em] (aix\!+\!bjy\! +\! (iy\!-\!jx)\sqrt{-ab})&\!\!\!\!\!\!(aix\!+\!bjy\! -\! (iy\!-\!jx)\sqrt{-ab} )&\!\!= \ \ \ \alpha\beta\:\smash[t]{\overline{\alpha\beta}}\\[.2em] (aix\!+\!bjy)^2\ &\!\!\!\!\!\!\!+ ab(iy\!-\!jx)^2 &\!\!= \ \ N(\alpha\beta)\end{array}$$

• Everything looks good. It takes a bit of work to verify that your $c$ and $d$ work, but they do. +1 – robjohn Dec 20 '13 at 19:22
• I just noticed that your solution looks pretty close to Henning's with $p=i$ and $q=j$. – robjohn Dec 20 '13 at 20:14
• @robjohn I hadn't read that. I agree, they do appear to use the same form of the Bezout identity. I derived it as in the linked post. In some sense the Bezout proofs will all be the same, modulo a shift by $\,\pm k\, (a+b, -ab)$ to obtain another solution $\,ab\,x +(a+b)\,y = 1.$ – Bill Dubuque Dec 20 '13 at 20:24
• @robjohn I found some time to elaborate to explain the classical view in terms of composition of quadratic forms. – Bill Dubuque Dec 16 '19 at 6:13

HINT: Suppose that $p$ is a prime that divides $ab$ and $a^2+b^2$. Then $p$ divides both $a^2+2ab+b^2=(a+b)^2$ and $a^2-2ab+b^2=(a-b)^2$. This in turn implies that $p$ divides both $a+b$ and $a-b$. (Why?) Use this to show that $p$ divides both $a$ and $b$.

• If p is even, then both $ab$ and $a^2+b^2$ must be even as well. The second can only be true in a and b are both even or both odd. If they are both even, they are not coprime, and if they are both odd $ab$ is not even. – Miles Johnson Jan 10 '17 at 5:29
• Therefore p is not even. Since $p$ divides $a+b$ and $a-b$, for some integers $k$ and $z$, we may say $(a+b)/p=k, (a-b)/p=z$. Adding the two, we have $2a/p=k+z$, implying, since $p$ is not even, that $p$ divides $a$. Subtracting the second from the first, we have $2b/p=k-z$, implying for the same reason that $p$ divides $b$. But $p$ cannot divide both numbers, so we have found a contradiction. – Miles Johnson Jan 10 '17 at 5:36
• Just in case anyone was wondering how to finish the proof. – Miles Johnson Jan 10 '17 at 5:37

HINT: $a^2 + b^2 +2ab = (a+b)^2.$

You say:

Let’s say $$k\mid a+b$$ where $$k$$ is some factor of $$a$$.

Then $$ka=a+b$$ ...

This step really doesn’t make any sense as written. If $$k\mid a+b$$, there is some integer $$m$$ such that $$km=a+b$$, but there’s certainly no reason to think that $$m=a$$. What you want to do at this point is use the hypothesis that $$k\mid a$$: there is some integer $$n$$ such that $$a=kn$$, and therefore $$km=kn+b$$. Now you can solve for $$b$$ and observe that $$k\mid b$$. But then $$k$$ is a common divisor of $$a$$ and $$b$$, so $$k=\pm1$$, and you’ve shown that $$\gcd(a,a+b)=1$$.

A similar argument shows that $$\gcd(b,a+b)=1$$, and then your last step is fine.

This can be solved intuitively by using a slight twist on Euclid's idea for generating new primes. Euclid employed $$\,1 + p_1\cdots p_n$$ is coprime to $$\,c = p_1\cdots p_n.\,$$ Stieltjes noted the generalization that, furthermore, $$\ \color{#c00}{p_1\cdots p_k} +\, \color{#0a0}{p_{k+1}\cdots p_n}\,$$ is coprime to $$\,c\,$$ too, which motivates the following

Key Idea $$\,$$ Coprimes to $$\,c\,$$ arise by partitioning into $$\rm\color{#c00}{two}\ \color{#0a0}{summands}$$ all prime factors of $$\,c,\,$$ i.e.

Theorem $$\ \ \color{#c00}a+\color{#0a0}b\$$ is coprime to $$\ c\:$$ if every prime factor of $$\,c\,$$ divides $$\,a\,$$ or $$\,b,\,$$ but not both.

Proof $$\$$ If it fails then $$\,a+b\,$$ and $$\,c\,$$ have a common prime factor $$\,p.\,$$ By hypothesis $$\,p\mid a\,$$ or $$\,p\mid b,\,$$ w.l.o.g. say $$\,p\mid b.\,$$ Then $$\,p\mid (a+b)-b = a,\,$$ so $$\,p\,$$ divides both $$\,a,b,\,$$ contra hypothesis.

Corollary $$\ \ (a,b)=1\,\Rightarrow\, (a+b,ab) = 1$$

Remark $$\$$ Note how the solution becomes quite obvious after employing Stieltjes idea. More generally it yields a "coprime" version of Dirichlet's result on primes in arithmetic progression

$$(a,b,c)=1\,\Rightarrow\,(an+b,c) = 1\ \ \ {\rm for\ some}\ n\qquad$$

If $d$ divides $a+b$ and $d$ divides $ab$, then $d$ divides $a(a+b)-ab = a^2$. Similarly, $d$ divides $b^2$. Thus $d$ divides $(a^2,b^2$). But $(a^2,b^2) = (a,b)^2 = 1$, so $d=1$.

Yet another approach: it's straightforward to show that $(i, jk) | (i,j)\cdot(i, k)$. It's also a classic theorem of the GCD that $(m, n) = (m, m+n)$ (this principle underlies the Euclidean algorithm). Then we have $(a, a+b)=(a, b)=1$, and $(b, a+b) = (b, a) = 1$, so $(ab, a+b)|(a, a+b)\cdot(b, a+b) = 1\cdot 1$.

We know that the prime factors of $ab$ is either a factor of $a$ or $b$ but not the other due to $gcd(a,b) = 1$. However, $a+b$ is not divisible by a prime factor of $a$ nor $b$ because the division yields an integer plus a 'decimal number'. Hence, by the fundamental theorem of arithmetic, it is not divisible by any factors of $ab$.

Intuitively, if a factor divides $a$ but not $b$ then it divides $ab$ and $a^2$ but not $b^2$, hence not $a^2+b^2$.

More formally, notice that if $a$ and $b$ are coprime then they are also coprime to $a+b$ and hence $ab$ is coprime to $(a+b)^2 = a^2 + 2ab+b^2$.

• I quite like this one; +1. – Brian M. Scott Jan 10 '17 at 5:59

Suppose $\text{gcd}(a,b) = 1$, but $\text{gcd}(ab,a^2+b^2) = d$, where $d > 1$. Let p be a prime factor of d. Then

$$d|ab \implies p|ab \implies p|a \text{ or } p|b$$

Without loss of generality, assume p|a. Then

$$d|(a^2 + b^2) \implies p|(a^2 + b^2) \implies p|b^2 \implies p|b$$

contrary to $\text{gcd}(a,b)=1$.

By contradiction, let $d>1$ divide both $ab$ and $a^2+b^2$. Then $d$ has a prime divisor $p$, which divides both $ab$ and $a^2+b^2$.

Then, $p|a(a+b) - ab = a^2$ Since $p$ is a prime which divides $a^2$ $\implies$ that $p|a$.

And $p|b(a+b) - ab = b^2$ $\implies$ $p|b$.

Hence $p|a$ and also, $p|b$, so $p$ divides $gcd(a,b) = 1$. However, this forces $p = 1$, which is a contradiction, since 1 isn't a prime.

Thus the only positive common divisor of $ab$ and $a^2+b^2$ is $1$, and hence they are coprime.

A proof in Gaussian integers $$\Bbb Z[i].\,$$ They are Euclidean $$\Rightarrow$$ UFD so enjoy GCDs, Euclid's Lemma, etc.

$$(a,b) = 1\,\Rightarrow\, (a\pm bi,b) = (a,b) = 1.\,$$ Similarly $$\,(a\pm bi,a) = 1.\,$$ So by Euclid $$\,a,b$$ are coprime to $$\,(a-bi)(a+bi) = a^2+b^2\,$$ thus so too is their product $$\,ab.$$

Suppose that gcd(a,b)=1.

Let d=gcd(a+b,ab).

Now d|(a+b) and d|ab.

Suppose that d|a.

Then a=dqa.

Now examining a+b=dq1 we have dqa+b=dq1 b=d(q1−qa).

So then d|b and d=1.

A similar argument shows d|b implies d|a.

So we know that d|a or d|b implies d=1.

Suppose that d∤b and d∤a.

We have a+b=dq1 a=dq1−b.

Here we have that gcd(a,−b)=1=gcd(a,d) (this is from a Lemma used in the proof of the Euclidean Algorithm).

We can also derive that gcd(b,−a)=1=gcd(b,d).

So we know that a and d and b and d are both relatively prime.

Considering d|ab we know by Euclid's Lemma that since a and d are relatively prime implies that d|b.

This contradicts our assumption, so this case is not possible.

Therefore, we may conclude that d=1. □

• I found this question in Elementary Number Theory 7th edition, by Bartle. Section 2.4, Question 6. – fantasticasm89 Jan 22 at 19:24


# Applying Rolle's theorem

This is the first time that I'm doing a proof-problem on my own and I'm not really sure how to check if my answer is correct, so I was wondering if someone could tell me with certainty if my proof is correct.

The problem states:

Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function (on its entire domain) such that $f(0) = f(1) = f(2)$. Prove that there exists some point $x_0 \in (0, 2)$ such that $f''(x_0) = 0$.

So I developed my proof based on Rolle's theorem that states that if a function $f$ is continous on a closed interval $[a, b]$ and differentiable on an open interval $(a, b)$ and if $f(a) = f(b)$ then there exists some point $c \in (a, b)$ such that $f'(c) = 0$.

Proof:

1. $f$ is differentiable $\forall x \in \mathbb{R}$, meaning $f$ must be differentiable on $(0, 1)$ and $(1, 2)$

2. $f$ is differentiable $\forall x\in\mathbb{R}$ meaning $f$ must be continuous $\forall x \in \mathbb{R}$ meaning $f$ must be continuous on $[0, 1]$ and $[1, 2]$.

3. $f(0) = f(1)$ and $f(1) = f(2)$.

From 1., 2. and 3. we conclude that $f$ satisfies the conditions of Rolle's theorem for both of these intervals, meaning: $\exists c_1 \in (0, 1) : f'(c_1) = 0$ and $\exists c_2 \in (1, 2) : f'(c_2) = 0$

Let $F(x) = f'(x)$. We know that the original function is twice differentiable $\forall x \in \mathbb{R}$, therefore $F(x)$ is differentiable on $\mathbb{R}$ meaning it must be differentiable on $(c_1, c_2)$ also meaning it must be continuous on $[c_1, c_2]$.

Now, since $f'(c_1) = f'(c_2) = 0$, the function $F(x)$ satisfies the conditions of Rolle's theorem (on the interval $(c_1, c_2)$) meaning :

$\exists x_0 \in (c_1, c_2) \subset(0, 2): F'(x_0) = f''(x_0) = 0$.

• Hello there. Please do not type all the proof in math mode. This makes the post hard to read. Just type those symbols in mathjax. . Thanks.
– xbh
Sep 16 '18 at 15:31
• Usually, new users have to be encouraged to use mathjax. In your case, I'd say you that you are using mathjax too much. Please, write text without mathjax. Sep 16 '18 at 15:32
• After a cursory reading, i would say the proof seems fine. Although it can be neater.
– xbh
Sep 16 '18 at 15:36
• I've edited your answer, now it's more legible. I would say your proof is right Sep 16 '18 at 15:36
• Your proof is solid. Sep 16 '18 at 15:47


# Finite Borel measure on a compact Metric Space

Suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and that $\mu(\{x\}) = 0$ for all $x\in X$. Show that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x\in X$, $\mu(B(x,\delta))<\epsilon$ where $B(x,\delta)$ denotes the ball of radius $\delta$ centered at $x$.

Attempted proof - Let $\epsilon > 0$ and suppose there is $\delta > 0$ such that there is an ball $B(x,\delta)$ of radius $\delta$ centered at $x$. Further, suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and $\mu(\{x\}) = 0$. Then from continuity from above $$\lim_{\delta\rightarrow 0}B(x,\delta) = 0$$ Thus there exists a $\delta_x$ such that $B(x,\delta_x)<\epsilon$. Now, since $X$ is a compact metric space, we can choose points $x_1,\ldots, x_n$ such that $X\subset \bigcup_{1}^{n}B(x_j,\delta_{x_j/2})$.

Let $\delta = \min\{\delta_{x_1/2},\ldots,\delta_{x_n/2}\}$ and suppose $x\in X$. Then for some $j$, $x\in B(x_j,\delta_{x_j/2})$ thus $B(x,\delta)\subset B(x_j,\delta_{x_j})$. Applying monotonicity we have $$\mu(B(x,\delta))\leq \mu(B(x_j,\delta_{x_j}) < \epsilon$$

Not sure if this is correct any suggestions is greatly appreciated.

• Looks right to me. The way it's written at the start is a little curious - that first "suppose there is a ball..." doesn't quite make sense. I'd just start the proof "Let $\epsilon>0$. By continuity from above, for every $x\in X$ we have $\lim_{\delta\to0}\mu(B(x,\delta))=0$" and then continue exactly as you did. – David C. Ullrich Jun 18 '16 at 22:41
• @DavidC.Ullrich I see so it doesn't make sense to say suppose there is a ball centered at $x$ with radius $\delta$? – Wolfy Jun 18 '16 at 22:42
• We're talking about a metric space $X$. For any $x\in X$ and $\delta>0$ there is such a ball. "Supposing" something exists when that thing simply does exist doesn't make much sense, no. – David C. Ullrich Jun 18 '16 at 23:32

1. As @DavidC.Ullrich pointed out, you don't need "suppose there is $\delta > 0$ such that there is an ball $B(x,\delta)$ of radius $\delta$ centered at $x$". In any metric space $X$, for all $x \in X$ and all $\delta >0$ there is a ball $B(x,\delta)$.

2. You used (correctly) the fact that $\mu$ is finite. I suggest you highlight this point.

Below is you proof with those two minor adjustments (and the correction of some typos)

Suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and that $\mu(\{x\}) = 0$ for all $x\in X$. Show that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x\in X$, $\mu(B(x,\delta))<\epsilon$ where $B(x,\delta)$ denotes the ball of radius $\delta$ centered at $x$.

Proof: Let $\epsilon > 0$ and suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and $\mu(\{x\}) = 0$. Then, since $\mu$ is finite, from continuity from above $$\lim_{\delta\rightarrow 0}\mu(B(x,\delta))=\mu(\{x\}) = 0$$ Thus there exists a $\delta_x$ such that $\mu(B(x,\delta_x))<\epsilon$. Now, since $X$ is a compact metric space, we can choose points $x_1,\ldots, x_n$ such that $X\subset \bigcup_{1}^{n}B(x_j,\delta_{x_j}/2)$.

Let $\delta = \min\{\delta_{x_1}/2,\ldots,\delta_{x_n}/2\}$ and suppose $x\in X$. Then for some $j$, $x\in B(x_j,\delta_{x_j}/2)$ thus $B(x,\delta)\subset B(x_j,\delta_{x_j})$. Applying monotonicity we have $$\mu(B(x,\delta))\leq \mu(B(x_j,\delta_{x_j}) < \epsilon$$

This should be a comment but may be too long to fit. (Your proof is OK with the adjustments already noted by others.) Another way, although almost the same, is by contradiction : If not, then there exists $e>0$ such that $$\forall n\in N\; \exists x_n\in X \; (\mu (B(x_n,1/n)\geq e).$$ Choosing such an $x_n$ for each $n,$ there exists a subsequence $(x_{j_n})_n$ converging to a point $x\in X.$ But for any $m\in N,$ take $n$ large enough that $d(x,x_{j_n})<1/ 2 m.$ Then $B(x,1/m)\supset B(x_{j_n},1/j_n),$ so $\mu (B(x,1/m))\geq e$ for all $m\in N,$ contradicting continuity from above.

Compactness of $X$ cannot be omitted.For example, let $X$ be the real interval $[0,1)$ with the usual metric; let $f(p)=\tan (p\pi /2)$ for $p\in X;$ let $L$ be Lebesgue measure..... Now for $A\subset X$ and for $A\in$ dom $(l)$ let $\mu (A)=L(f(A)).$


Question

# In a set of $$10$$ coins, $$2$$ coins are with heads on both the sides. A coin is selected at random from this set and tossed five times. If all the five times, the result was heads, find the probability that the selected coin had heads on both the sides.

A
16
B
89
C
23
D
210

Solution

## The correct option is B $$\dfrac89$$Let $$E_1, E_2, A$$ be the events defined as follows :$$E_1$$= selecting a coin having a head on both the sides.$$E_2$$= selecting a coin not having a head on both the sides.A = Getting all heads when a coin is tossed five times.To find:The probability that the selected coin had heads on both the sides: $$P\left(\dfrac {E_1}A\right)$$There are 2 coins having heads on both the sides $$P(E_1)=\dfrac {^2C_1}{^{10}C_1}=\dfrac 2{10}$$There are 8 coins not having heads on both the sides $$P(E_2)=\dfrac {^8C_1}{^{10}C_1}=\dfrac 8{10}$$$$P\left(\dfrac A{E_1}\right)=1^5=1\\\implies P\left(\dfrac A{E_2}\right)=\left(\dfrac 12\right)^5$$By Bayee's Theorem, we have$$P\left(\dfrac {E_1}A\right)=\dfrac {P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}=\dfrac {\dfrac 2{10}\times 1}{\dfrac 2{10}\times 1+\dfrac 8{10}\times \left(\dfrac 12\right)^5}\\\implies P\left(\dfrac {E_1}A\right)=\dfrac 2{2+\dfrac 8{32}}=\dfrac 89$$Mathematics

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First, we need to translate the word problem into equation (s) with variables. The problem states that "This" -- that is, $42 -- was$14 less than two times x. So, area of the rectangle is 384 square cm. If the rectangle's diagonal measurement is 32 inches, find the height. If they had laid 3500 eggs this year, how many eggs did they lay last year ? 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Education; Math; Pre-Algebra; The 4 Steps to Solving Word Problems; The 4 Steps to Solving Word Problems. 18 is taken away from 8 times of a number is 30. You can … He prepared this workbook (with full solutions to every problem) to share his strategies for solving algebra word problems. Let 'x' and "x + 4" be the lengths of other two sides. The Algebra Class E-course provides a lot of practice with solving word problems for every unit! Given : 15 years back X's age was 3/4  of Y's age. Type a math problem. Given : Thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find below a wide variety of hard word problems in algebra. If 100 of the marbles in the first bag are placed in the second bag, the number in …, Survey and intersection of three sets Not rated yetIn a survey of 400 students of a college it was found that 240 study mathematics, 180 study Statistics and 140 study Accounts, 80 study mathematics and …, A combination problem  Not rated yetThere are 13 qualified applicants for 4 trainee positions in a fast food management program. 4.2 5 customer reviews. And the third angle is three times of the first angle. In a triangle, the second angle is 5Â° more than the first angle. check your answers against the answer explanation section at the end of each chap-ter. Find the three angles of the triangle. Always let x represent the unknown number. Question 1: There are 47 boys in the class. Find the length of the sides. Click the button and find the first one on your computer. For problems that require more than one step, a thorough step-by-step explanation is provided. Given : If the rod was 2 meter shorter and each meter costs $1 more. So, the two parts of the 25 are 10 and 15. He prepared this workbook (with full solutions to every problem) to share his strategies for solving algebra word problems. Get step-by-step solutions to your math problems. Distance, Rate, and Time Word Problems These Algebra 1 Equations Worksheets will produce distance, rate, and time word problems with ten problems per worksheet. Given : The total cost of 125 units of the product is$28750. Find out below how you can print these problems. \displaystyle \frac {4} {5} of a number is less than 2 less than the same number. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Now that you can do these difficult algebra problems, you can trick your friends by doing some fancy word problems; these are a lot of fun. Question 2 : The denominator of a fraction exceeds the numerator by 5. ... College Algebra Word Problem College Algebra Algebra Word Problem. Section 1: Examples. Area and perimeter word problems. So, the three angles of the triangle are 35Â°, 40Â° and 135Â°. Divide 25 in two parts so that sum of their reciprocals is 1/6. The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. : Step 2:: Solve the equation created in the first step. Identifying Divisibility by Looking at the Final Digits. Author: Created by dh2119. In a group of 120 people, 90 have an age of more 30 years, and the others have an age of less than 20 years. Example 1. Area = length × width …, Math problem about age Arvind is eight year older than his sister. Ticket problem. Get help on the web or with our math app. Search. Below are more complicated algebra word problems Example #6: The ratio of two numbers is 5 to 1. Click below to see contributions from other visitors to this page... Bessy and Bob algebra word problem Bessy has 6 times as much money as Bob, but when each earn $6. Algebra Word Problems. Ratio and proportion word problems. Math Word Problems with Solutions and Explanations - Grade 8. Always let x represent the unknown number . Phone support is available Monday-Friday, 9:00AM-10:00PM ET. 10 â (2x) + 1 â x - 18 = 10 â x + 1 â (2x). Over 600 Algebra Word Problems, Algebra Number Puzzles, Exponents, Polynomials, Radical Numbers, Factoring, Complex Numbers. Just type!...Your algebra word problem will appear on a Web page exactly the way you enter it here. Algebraic word problems | Worked example Our mission is to provide a free, world-class education to anyone, anywhere. There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time. Note that Using Systems to Solve Algebra Word Problems can be found here in the Systems of Linear Equations and Word Problems section. This is three more than four times the number of girls. Therefore, (100 – 1 ) 2 = 100 2 + 1 2 – 2 x 100 x 1 [By identity: (a -b) 2 = a 2 + b 2 – 2ab = 10000 + 1 – 200 = 9801. These Word Problems worksheets will produce word problems involving all basic operations. Many students find solving algebra word problems difficult. Counting Zeros and Writing Exponents. What was the average speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60Â°, 72Â° and 48Â°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx  =  12. x  =  9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let 'x0y' be the three digit number. How much was the blouse? Long distance cost .11 a minute using this card. Here is the equation: 2x − 14 = 42. Expert Answers • 1 Algebra Word Problem College Algebra Word Problem. 17 SEEKING SOLUTIONS JC Burns 18 101 PROBLEMS IN ALGEBRA T Andreescu Et Z Feng. Solution A translation of 2 units up will increase the y coordinate by 2 units and a translation by 5 units to the left will decrease the x coordinate by 5. Solution Then, assign variables to the unknown quantities. If 3 be added to both, the fraction becomes 3/4. About me :: Privacy policy :: Disclaimer :: Awards :: DonateFacebook page :: Pinterest pins, Copyright © 2008-2019. In many problems, what you are asked to find is presented in the last sentence. If the rectangle's diagonal measurement is 32 inches, find the height. Given : If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1. I quit my day job, in order to work on algebra.com full time. of adult tickets and "y' be the no. Percent of a number word problems. Solution: Let x represent Spenser's age Therefore, Taylor's age is 5x x + 5x = 18 6x = … Then: 0.10(10 – y) + 0.30y = 1.5 1 – 0.10y + 0.30y = 1.5 Find the total value of the substance …, Mixture word problem with sand and soil Not rated yet2 m 3 of soil containing 35% sand was mixed into 6 m 3 of soil containing 15% sand. Most tricky and tough algebra word problems are covered here. …, Top-notch introduction to physics. Inequality Word Problem/Math. Solution: We can write, 99 = 100 -1. QuickMath allows students to get instant solutions to all kinds of math problems, from algebra and equation solving right through to calculus and matrices. ” …, A Little Tricky Algebra word problem Not rated yetNavin spent 25% of it and gave 2/5 of the remainder to his brother. 2x = 42 + 14 = 56. x = 56 2 = 28. How much does each have before and after …, Fencing a rectangular garden word problem A farmer has 260 meters of fencing and wants to enclose a rectangular area of 4200 square meters. Also solutions and explanations are included. Great! What was the average speed of the car in miles per hour? how old was they are now? Every problem in 501 Math Word Problems has a complete answer explanation. Given : Cost of each adult ticket is$10 and kid ticket is $5 and tickets were sold for a total of$3750. Let 'x' be the present age of X and 'y' be the present age of Y. Grade 8 math word problems with answers are presented. Word problems are the most difficult type of problem to solve in math. PREFACE This book contains one hundred highly rated problems used in the train-ing and testing of the USA International Mathematical Olympiad (IMO) team. Word problems in algebra. Do you have some pictures or graphics to add? The questions are increasingly challenging, finishing with some that require a lot of thought and can be investigated further. The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment. Worksheets > Math > Grade 4 > Word problems. Online math solver with free step by step solutions to algebra, calculus, and other math problems. Grade 6 math word problem worksheets with answers - Estimation word problems for 6th Grade are made of the following Math skills for kids: estimate to solve word problems, multi steps word problems, identifying word problems with extra or missing information, distance direction to starting point word problems, using logical reasoning to find the order, guest and check word problems. The author, Chris McMullen, Ph.D., has over twenty years of experience teaching word problems and math skills to physics students. Let 'x' be the number of eggs laid last year. Load more. The following are some examples and solutions for algebra word problems that you will commonly encounter in grade 9. A the end of five hours, 16,000 gallons remained. Teachers divided students into groups of 3. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. Problems requires us to combine our ability to create equations and solve them are examples. Every problem ) to find is more than four times the number of adults tickets sold is and. To inform you about new math lessons the button and find the first and third digits is more the! Long distance call times of a fraction is increased by 2, the digits are reversed angle is 5Â° than. With our math app the height are 36 years and 43 years place! Be added to both, the digits are reversed the lengths of other two sides of the is. Loans, and other math problems not be a genius ( 2x ) calculators step-by-step, but they use same. Answer explanation taken away from 8 times of a fraction exceeds the sixth part algebra word problems with solution.... The answers into your online assignment math problems can intimidate sixth-graders but it should.! 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