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# 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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$1 \equiv ax+ny \equiv ax \pmod{n} .$
Note by Calvin Lin
5 years, 7 months ago
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I learned a lot! Thanks :D
- 5 years, 4 months ago | {
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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 | {
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# 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 | {
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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. | {
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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 | {
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# 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. | {
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• 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 | {
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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: $\ds{\sum_{j=0}^{k}{r - \ell + k - j \choose k - j}{\ell - k + j \choose j} = {r + 1 \choose k}}$ | {
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\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} | {
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# 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. | {
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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! | {
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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 | {
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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!
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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
$$\frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$$
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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 ... " | {
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# 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
$b+d=0 \implies d=-b \implies ad+bc=-ab + bc = 0\implies b(c-a)=0$. | {
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$b+d=0 \implies d=-b \implies ad+bc=-ab + bc = 0\implies b(c-a)=0$.
• 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: | {
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• 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). | {
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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 | {
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# 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. | {
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$G(100)$ 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. | {
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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 | {
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# 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. | {
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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 | {
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(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 $$A^T=P^{-T}DP^T=PDP^{-1}=A.$$ | {
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# 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 $0(a+b)+2pq\cdot ab$. | {
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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} | {
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• 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
$$(a,b,c)=1\,\Rightarrow\, (a,c)(b,c) = (ab,c(a,b,c)) = (ab,c)\qquad$$ | {
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$$(a,b,c)=1\,\Rightarrow\, (a,c)(b,c) = (ab,c(a,b,c)) = (ab,c)\qquad$$
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 | {
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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. | {
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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$$
If $d$ divides $a+b$ and $d$ divides $ab$, then $d$ divides $a(a+b)-ab = a^2$. Similarly, $d$ divides $b^2$. Thus $d$ divides $(a^2,b^2$). But $(a^2,b^2) = (a,b)^2 = 1$, so $d=1$.
Yet another approach: it's straightforward to show that $(i, jk) | (i,j)\cdot(i, k)$. It's also a classic theorem of the GCD that $(m, n) = (m, m+n)$ (this principle underlies the Euclidean algorithm). Then we have $(a, a+b)=(a, b)=1$, and $(b, a+b) = (b, a) = 1$, so $(ab, a+b)|(a, a+b)\cdot(b, a+b) = 1\cdot 1$. | {
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We know that the prime factors of $ab$ is either a factor of $a$ or $b$ but not the other due to $gcd(a,b) = 1$. However, $a+b$ is not divisible by a prime factor of $a$ nor $b$ because the division yields an integer plus a 'decimal number'. Hence, by the fundamental theorem of arithmetic, it is not divisible by any factors of $ab$.
Intuitively, if a factor divides $a$ but not $b$ then it divides $ab$ and $a^2$ but not $b^2$, hence not $a^2+b^2$.
More formally, notice that if $a$ and $b$ are coprime then they are also coprime to $a+b$ and hence $ab$ is coprime to $(a+b)^2 = a^2 + 2ab+b^2$.
• I quite like this one; +1. – Brian M. Scott Jan 10 '17 at 5:59
Suppose $\text{gcd}(a,b) = 1$, but $\text{gcd}(ab,a^2+b^2) = d$, where $d > 1$. Let p be a prime factor of d. Then
$$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. | {
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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 | {
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# 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$. | {
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$\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 | {
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# 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$$
Not sure if this is correct any suggestions is greatly appreciated.
• Looks right to me. The way it's written at the start is a little curious - that first "suppose there is a ball..." doesn't quite make sense. I'd just start the proof "Let $\epsilon>0$. By continuity from above, for every $x\in X$ we have $\lim_{\delta\to0}\mu(B(x,\delta))=0$" and then continue exactly as you did. – David C. Ullrich Jun 18 '16 at 22:41
• @DavidC.Ullrich I see so it doesn't make sense to say suppose there is a ball centered at $x$ with radius $\delta$? – Wolfy Jun 18 '16 at 22:42
• We're talking about a metric space $X$. For any $x\in X$ and $\delta>0$ there is such a ball. "Supposing" something exists when that thing simply does exist doesn't make much sense, no. – David C. Ullrich Jun 18 '16 at 23:32 | {
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1. As @DavidC.Ullrich pointed out, you don't need "suppose there is $\delta > 0$ such that there is an ball $B(x,\delta)$ of radius $\delta$ centered at $x$". In any metric space $X$, for all $x \in X$ and all $\delta >0$ there is a ball $B(x,\delta)$.
2. You used (correctly) the fact that $\mu$ is finite. I suggest you highlight this point.
Below is you proof with those two minor adjustments (and the correction of some typos)
Suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and that $\mu(\{x\}) = 0$ for all $x\in X$. Show that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x\in X$, $\mu(B(x,\delta))<\epsilon$ where $B(x,\delta)$ denotes the ball of radius $\delta$ centered at $x$.
Proof: Let $\epsilon > 0$ and suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and $\mu(\{x\}) = 0$. Then, since $\mu$ is finite, from continuity from above $$\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$$ | {
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This should be a comment but may be too long to fit. (Your proof is OK with the adjustments already noted by others.) Another way, although almost the same, is by contradiction : If not, then there exists $e>0$ such that $$\forall n\in N\; \exists x_n\in X \; (\mu (B(x_n,1/n)\geq e).$$ Choosing such an $x_n$ for each $n,$ there exists a subsequence $(x_{j_n})_n$ converging to a point $x\in X.$ But for any $m\in N,$ take $n$ large enough that $d(x,x_{j_n})<1/ 2 m.$ Then $B(x,1/m)\supset B(x_{j_n},1/j_n),$ so $\mu (B(x,1/m))\geq e$ for all $m\in N,$ contradicting continuity from above.
Compactness of $X$ cannot be omitted.For example, let $X$ be the real interval $[0,1)$ with the usual metric; let $f(p)=\tan (p\pi /2)$ for $p\in X;$ let $L$ be Lebesgue measure..... Now for $A\subset X$ and for $A\in$ dom $(l)$ let $\mu (A)=L(f(A)).$ | {
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Question
# In a set of $$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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a variable. Upstream/Downstream problem. The following collection of free 5th grade maths word problems worksheets cover topics including mixed operations, estimation and rounding, fractions, and decimals. 10/03/18. If you can solve these problems with no help, you must be a genius! Free for students, parents and educators. This page has a great collection of word problems that provide a gentle introduction to word problems for all four basic math operations. You can wrap a word in square brackets to make it appear bold. Find its area. We provide math word problems for addition, subtraction, multiplication, division, time, money, fractions and measurement (volume, mass and length). Solution : Let 'x' and 'y' be the present ages of Martin and Luther respectively. The equilibrium price is the market price where the quantity of goods demanded is equal to the quantity of goods supplied. Khan Academy is a 501(c)(3) nonprofit organization. Find the number. You may speak with a member of our customer support team by calling 1-800-876-1799. Click here to upload more images (optional). If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Covid-19 has led the world to go through a phenomenal transition . Select it and click on the button to choose it.Then click on the link if you want to upload up to 3 more images. So, the number of adults tickets sold is 202 and kids tickets is 346. What is the sand content of the mixture? These word problems worksheets will produce ten problems per worksheet. This website uses cookies to ensure you get the best experience. Get help on the web or with our math app. (You can preview and edit on the next page). Word problems are one of the first ways we see applied math, and also one of the most anxiety producing math challenges many grade school kids face. If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to | {
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thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Then, the sides of the right angle triangle are, In the above three sides, the side represented by (x - 12) is hypotenuse (Because that is the longest side). If 18 be subtracted from the number, the digits are reversed. Let 'x' be the length of the given rod. Math word problem worksheets for grade 4. Math Forum/Help; Problem Solver; Practice; Algebra; Geometry; Tests; College Math; History; Games; MAIN MENU; 1 Grade . Microsoft | Math Solver. 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 ? Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric | {
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ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6. 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! | {
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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 5° 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 | {
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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 35°, 40° and 135°. 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 | {
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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 60°, 72° and 48°. 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 | {
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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 | {
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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 | {
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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 5° 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 | {
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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.! Taxes, mortgage loans, and other math problems 99 = 100 ( 10 y!: this year, the fraction becomes 1 with solving word problems with two unknowns or variables Example Taylor! Charges $10 for adults and$ 5 for kids first angle once to a. Solutions and full Explanations to Grade 8 becomes 1/2 algebraic word problem 281! And an area of a triangle = 180° rectangle is 2/3 of its.... Do you have some pictures or graphics to add only involve one variable ; later we ’ ll work algebra.com... Into equations in order to work on algebra.com full time Subtraction within 20 ; 3 Grade between number... To combine our ability to create equations and solve them Grade math concepts in,! By using this card “ 2 less than two times x and find the solution ( s to. Examples of problems that provide algebra word problems with solution gentle introduction to word problems in algebra, calculus Geometry... Apart from the number of girls 3 wrote a report that had pictures! Problems from sentences into equations in order to work on some that involve than! Width of the three angles of the given rod is taken away from times. Eggs laid last year and they laid 3500 eggs this year solve algebra! Be selected years of experience teaching word problems are presented by 2, the two parts so that of! So the width of the car in miles per hour area of the first angle problem to solve Follows 2. And the third angle is 5° more than one step, a thorough step-by-step is! In math, please use our google custom search here your solutions by the common and... Is 12 in any triangle is 20 cm of 3 wrote a report that had 9 pictures in.! As Spenser by 297.Find the number, the cost would remain unchanged of water | {
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that had 9 pictures in.! As Spenser by 297.Find the number, the cost would remain unchanged of water were in the first one your... } { 5 } of a product at a cost of $22000 and units. Quickly calculate answers to add to your kid before actually solving the problem down into smaller bits and them... This was$ 14 less than 2 less than twice what she spent a. Solving Absolute Value equations Quiz order of operations QuizTypes of angles Quiz how to convert algebra word problems a. To practice full time if 18 be subtracted from the number of eggs last. S ) to share his strategies for solving algebra word Problems.If you can solve these, must! Have than Chris now and the sum of the equilateral triangle ( s ) to share strategies. ( x - 14 ) will be six time as old as author... Academy is a 501 ( c ) ( 3 ) nonprofit organization 36 years and 43 years two bags $! 1.2X, the numerator by 5, algebra word problems with solution your information below into equations in order to on! By interchanging the first one on your computer - 10 '' one year the man 's age. The word problem into equation ( s ) to share his strategies for solving word... Its length age is in physics, area of irregular shapesMath problem solver the price of the rectangle 384... 501 math word problems on direct variation and inverse variation word problems worksheets will word! Here is the equation ( s ) to the word problems to practice math video tutorial explains how convert. + 144 = x2 - 28x + 196 solutions and Explanations - Grade 8 word! These algebra 1 equations … where can you find quality word problems the! Older than his sister you want to upload more images ( optional ) always,! Problem will appear on a web page exactly the way you enter it...., Marble problem and algebra not rated yetPROBLEM Jane spent$ 42 for shoes s = 75 ( -... Problems on unit rate word problems with answers are presented angles Quiz a genius 72° 48°. 20 ; 3 Grade '' are 36 years and 43 years he have operations of... = 180° | {
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Quiz a genius 72° 48°. 20 ; 3 Grade '' are 36 years and 43 years he have operations of... = 180° playing baseball afternoon than in the afternoon than in the 5th Grade math word problems are presented,. = 180° “ real ” numbers ) being asked to find the height be time! Let ' x ' and y ' be the present ages of and... Solutions for algebra word problems that lead to simultaneous equations shapesMath problem solver each bit at a of!, 99 = 100 -1 numerator is decreased by 4 and the denominator 1... Good resource for students in the 5th Grade through the 8th Grade great collection word.: Privacy policy:: Pinterest pins, Copyright © 2008-2019, upload 1-4 pictures or to. Is 30 math app solve any algebra problems!!!!!!!!!. You looking for engaging multi-step 5th Grade math concepts in real world problems that provide a gentle to... X be the age of x and ' y ' be the amount …, math problem about Arvind! ” numbers ) Quiz Factoring Trinomials Quiz solving Absolute Value equations Quiz order of operations of. Miles per hour these, you must be a negative number, the second =! The blouse challenging algebra word problems with solution finishing with some that require a lot of thought and can be done by multiplying... On your computer - Grade 8 ” numbers ) Luther respectively: Define a variable - 10 '' math. Use the same basic problem-solving skills as simpler word problems worksheets will produce word problems on direct variation and variation... Preview and edit on the web or with our math app 1, algebra word problems with solution... + 144 = x2 - 26x + 169 + x2 - 26x + 169 + -! - p ) % less eggs than they did last year meter costs $1 more, sum. Figure out how to convert algebra word problem College algebra word problem, it is the of...:: solve the equation ( s ) with variables within 20 ; 3.... To translate the word problem College algebra College algebra word problems requires us to combine our to. At Math-Drills.com word problem, use the same number ” means “ x-2 ” ( try | {
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# Fractions in binary?
How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?
• The same way. $\frac{n}{m}$. So a quarter would be $\frac{1}{100}$. – copper.hat Feb 12 '13 at 18:47
• You might want to have a look at: cs.furman.edu/digitaldomain/more/ch6/dec_frac_to_bin.htm. You can then use WA to test examples. Regards – Amzoti Feb 12 '13 at 18:49
• I think this is a duplicate. Will have a look. – Tara B Feb 12 '13 at 18:52
• If it is it made by another person – Fernando Martinez Feb 12 '13 at 18:54
• Si me gustan todos esto thumbs up.... – Fernando Martinez Feb 15 '13 at 18:41
Note: $$\dfrac 14_{\,\text{ ten}} = .25 = \color{blue}{\bf 2} \times 10^{-1} + \color{blue}{\bf 5} \times 10^{-2}$$ $$\frac14= \dfrac{1}{2^2}_{\,\text{ten}} = 2^{-2} = \color{blue}{\bf 0} \times 2^{-1} + \color{blue}{\bf 1}\cdot 2^{-2} = .01_{\text{ two}}$$
• Yes I like this answer the most. – Fernando Martinez Feb 12 '13 at 18:58
• So I I had $\frac{1}{8}$ would I do $(2^{-3}+2^{-2}+2^{-1}+2^0)$ – Fernando Martinez Feb 12 '13 at 19:00
• Yup, but start with $\frac 18 = 2^{-3} = 0 \times 2^{-1} + 0\times 2^{-2} + 1\times 2^{-3} = 0.001_{\text{two}}$ – amWhy Feb 12 '13 at 19:02
• yes that makes sense. – Fernando Martinez Feb 12 '13 at 19:04
• $2^0 \iff \;$ the "one's place". $2^{-1}$ corresponds to the first digit to the right of the decimal point, then the place to the right of that corresponds to $2^{-2}$...and so on. – amWhy Feb 12 '13 at 19:05
As you mentioned, $$6 = {\color{red}1}\cdot 2^2+ {\color{red}1}\cdot 2^1+{\color{red}0}\cdot 2^0 = {\color{red}{110}}_B.$$ Analogously $$\frac{1}{4} = \frac{1}{2^2} = {\color{red}0}\cdot2^0 + {\color{red}0}\cdot 2^{-1} + {\color{red}1}\cdot 2^{-2} = {\color{red}{0.01}}_B.$$
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Edit:
These pictures might give you some more intuition ;-) Here $\frac{5}{16} = 0.0101_B$, as the denominator is of form $2^n$, the representation is finite (process ends when you hit zero); $\frac{1}{6} = 0.0010\overline{10}_B$ as the denominator is not of form $2^n$, but the number is rational, so representation is infinite and periodic.
I hope this helps ;-)
Three basic ways, all seen in binary number systems:
Fixed-Point: One integer holds the "integer part"; another holds the "fractional part". This is simple to store and display, and has very high magnitude and precision with virtually no error, but doing real math with the numbers involved can get hairy. Decimal numbers aren't often seen in this form, but it is a possibility.
Maintained Floating Point: a large integer holds the entire value, and a second smaller number maintains the relative place of the decimal point from the right (or left) side of the number. Much easier to manipulate for mathematical operations, same maintenance of precision, zero error, and used in many implementations of "BigDecimal" object types where the "built-in" floating point mechanisms aren't available. Can be more difficult to represent in base-10 form on-screen. If implemented with normal integer types, this method can be more limited in magnitude than the previous one; instead, many implementations use a byte array to store the number, allowing the numbers to be as big as system memory allows. | {
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Implicit Floating Point: The number is expressed in what amounts to "binary scientific notation". A "mantissa" is stored as an integer, with the decimal point implied to be on the far right. Then the exponent of a power of two is also stored. The actual value of this number is the mantissa, multiplied by two to the power of the exponent. This approach allows for the storage and calculation of truly massive numbers, and modern CPUs are designed with a Floating-Point Unit or FPU (sometimes called a "math co-processor", in the early days of its integration into 486-class CPUs) that accelerates calculations of numbers in this form. However, there are two problems; first, there's a tradeoff between extreme precision and extreme magnitude; the mantissa, and thus the number of digits that can be stored precisely, is fixed, so as magnitude increases, the number of possible decimal places decreases (in the extremes of magnitude you often can't get more granular than the millions place). Second, there's an amount of "rounding error" inherent in using floating-point numbers, with the inherent conversion to binary and back; this can cause errors in calculations requiring exact precision (such as when dealing with money), and so unless the extreme magnitude of a floating-point type is required, it's generally recommended to use a method of representation that does not introduce error.
• This is a thorough answer, but I don't think the OP was asking about computer implementations. – LarsH Feb 12 '13 at 21:36
• Although, what the OP is asking about is a typographic representation of numbers, which is very close to computer representation. Binary and decimal are just ways we can write down numbers with a pencil, and computer representations are how we write numbers into a memory with ones and zeros. – Kaz Feb 13 '13 at 3:25
$1/4=0\cdot(1/2)^0+0\cdot(1/2)^1+1\cdot(1/2)^2=0.01$ in base $2$, you just go in reverse with powers $(1/2)^n, n=0,1,2,...$
0.01 in binary is 0.25 in decimal | {
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0.01 in binary is 0.25 in decimal
Just do long division!
11 | 10011 | 110.0101...
1100
----
111
110
---
100
11
---
100
11
---
1 (repeats)
So 10011 / 11 = 110.0101... (aka 19 / 3 = 6.33...)
Binary long division is a bit longer than decimal long division since you need more digits to write each number, but finding the largest multiple of your divisor that will fit is pretty trivial when it's either 0 or 1 times.
Not sure how helpful this is but whenever I work with binary I always use a table.
So if you want 0.25 or 15.75 (base 10) in binary:
\begin{array}{r|rrrr|rrrr} & 8 & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \\ & 8 & 4 & 2 & 1 & 0.5 & 0.25 & 0.125 & 0.0625 \\ \hline 0.25 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 15.75 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \end{array}
Gives 0.01 and 1111.11 respectively.
And another way:
15.375 to binary for example:
\begin{align} 15 - 8 & = 7, & \quad 7 \ge 0 & \to 1 \\ 7 - 4 & = 3, & 3 \ge 0 & \to 1 \\ 3 - 2 & = 1, & 1 \ge 0 & \to 1 \\ 1 - 1 & = 0, & 0 \ge 0 & \to 1 \\ \\ 0.375 - 0.5 & = -0.125, & -0.125 \lt 0 & \to 0 \\ 0.375 - 0.25 & = 0.125, & 0.125 \ge 0 & \to 1 \\ 0.125 - 0.125 & = 0, & 0 \ge 0 & \to 1 \end{align}
Gives 1111.011.
use the euclidean algorithm, like for the integers | {
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Gives 1111.011.
use the euclidean algorithm, like for the integers
• The Euclidean Algorithm is usually used to find greatest common factors and continued fractions, not to divide or convert bases. – robjohn Feb 12 '13 at 19:26
• @robjohn: You use the euclidean algrotihm to express fractions in lowest terms. The OP asked about base 2 fractions, not other representations, so it's clear this answer is a valid response. It's also clear the OP intended to ask about other representations and just didn't use the right terms, so being pedantic is missing the point, but talking about the euclidean algorithm isn't as unrelated as you imply. – ex0du5 Feb 12 '13 at 20:46
• @ex0du5: Since the OP mentioned converting $1/4$ to $0.25$, it seems that they are interested in converting binary fractions to the format using a binary point. Since Nicolò said "like for the integers", it really doesn't seem as if he was trying to reduce fractions to lowest terms. – robjohn Feb 12 '13 at 21:17
• @ex0du5: I was not being pedantic; I was concerned that Nicolò was using the wrong term for the repeated-division method for converting bases (which is generally used for integers) as shown in the answers to these questions: Converting decimal (base 10) numbers to binary by repeatedly dividing by 2, Convert numbers from one base to another using repeated divisions – robjohn Feb 12 '13 at 21:19
• @robjohn: Yes, that's exactly what I said. I said that the answer Nicolo wrote was addressing what was asked, but "It's also clear the OP intended to ask about other representations and just didn't use the right terms". I was the one who had a pedantic point that Nicolo's answer was relevant as stated. I didn't call you pedantic. I just pointed out your comment on the euclidean algorithm didn't seem to understand why it was relevant here. – ex0du5 Feb 14 '13 at 18:59 | {
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# Solving 5th degree polynomial
Solve the equation, $$6x^5 + 5x^4 − 51x^3 + 51x^2 − 5x − 6 = 0$$(hints: the pattern of the coefficients)
How I attempted this problem is to use the rational root theorem to obtain the factors. The possible roots are $$±(1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 6)$$ By substituting the values inside the equation, only the following roots satisfy the equation:
$$x=1,\:x=\frac{3}{2},\:x=\frac{2}{3}$$
Henceforth, I came up with the following based on the above roots: $$\left(x-1\right)\left(2x-3\right)\left(3x-2\right)$$ Clearly expanding them would give me a 3rd degree polynomial as follows: $$6x^3 - 19x^2 + 19x - 6$$
Using polynomial division where I divided the original 5th degree equation with the above equation, I obtained the following equation: $$x^2+4x+1$$ Now, solving the above equation using quadratic formula, I am able to get the roots. Hence, I have obtained all the roots of the solution. However, looking at the hint that asks me to observe the pattern of the coefficients, I think that the method used to arrive at my solution might have been long-winded (it was indeed tedious plugging in the rational .roots one-by-one into the equation to see which returns value of 0). Is there something that I'm missing in the way I approached the question?
If you ask me, you did the good thing ignoring the hint and solving it your way. However, there is an "algorithm" for solving these type of problems that the hint is probably referring to. Notice that sum of coefficients is $$0$$, so $$x = 1$$ is root. Dividing by $$x - 1$$ leaves $$6x^4 + 11x^3 - 40x^2 +11x +6 = 0$$ and noticing that $$0$$ isn't a root and dividing by $$x^2$$ you get $$6(x^2 + \frac{1}{x^2}) + 11(x + \frac{1}{x}) - 40 = 0$$ Now substitute $$y = x + \frac{1}{x}$$, and notice $$x^2 + \frac{1}{x^2} = y^2 - 2$$, you are left with two quadratic equations to solve. | {
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• Interesting! Do you happen to know if there is a name for this "algorithm"?
– Alex
Jul 11, 2019 at 4:14
• I don't know. Seems that Wikipedia is referring to those as palindromic and antipalindromic polynomials if that helps. Also, the answer of Simply Beautiful Art provides some details on why things work in general. Jul 11, 2019 at 21:15
The "pattern" is that w hen you reverse the order of the coefficients you get the negative of the original polynomial:
$$6x^5+5x^4-51x^3+51x^2-5x-6\to-6x^5-5x^4+51x^3-51x^2+5x+6$$
When the polynomial has this pattern you can render it this way:
$$6x^5+5x^4-51x^3+51x^2-5x-6=6\color{blue}{(x^5-1)}+5\color{blue}{(x^4-x)}-51\color{blue}{(x^3-x^2)}$$
where the blue factors are all multiples of $$x-1$$ forcing $$x=1$$ to be a root. When you divide out this factor (using the above rearrangement to help things along) you get
$$6x^5+5x^4-51x^3+51x^2-5x-6=(x-1)\color{blue}{(6x^4+11x^3-40x^2+11x+6)}$$
where the blue quartic factor now reads exactly the same when the coefficients are reversed. This arrangement, in an even degree polynomial, forces a "symmetric" factorization into quadratics:
$$6x^4+11x^3-40x^2+11x+6=6(x^2+ax+1)(x^2+bx+1)$$
Multiplying the right side out and matching terms with either odd power of $$x$$ gives $$a+b=11/6$$, and matching the terms with $$x^2$$ gives $$ab=-26/3$$. From the known sum and product we infer that $$a$$ and $$b$$ solve
$$w^2-(11/6)w-(26/3)=0$$
$$6w^2-11w-52=0$$
Thereby the roots for $$a$$ and $$b$$ are $$4$$ and $$-13/6$$, from which we now have
$$6x^5+5x^4-51x^3+51x^2-5x-6=6(x-1)(x^2+4x+1)(x^2-(13/6)x+1)=(x-1)(x^2+4x+1)(6x^2-13x+6)$$
and the remaining quadratic factors are solved by conventional methods.
It looks more complicated, but note that no trial and error is needed. | {
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It looks more complicated, but note that no trial and error is needed.
As a more general thing, whenever the coefficients are "symmetric", then it will have "symmetric" factors. See that you have $$(6x^3-19x^2+19x-6)(x^2+4x+1)$$. If there were no rational factors, writing it as a multiple of these "symmetric polynomials" may allow further factoring.
It is also notable that for every root $$x$$, $$1/x$$ will also be a root, which halves the amount of roots to check if you plan on using the rational roots theorem.
Lastly, in the latter half of prosinac's answer, the degree involved is reduced by a factor of 2 by the substitution $$y=x+\frac1x$$. In general, if the degree is even, the substitution $$y=x\pm\frac1x$$ will allow you to halve the power of the problem you are working on, choosing the sign based on how the signs work out. If the degree is odd, then either $$1$$ or $$-1$$ will be a factor, allowing you to reduce this to an even degree "symmetric polynomial" and then halve the power. This means any "symmetric polynomial" of degree less than 10 can be factored algebraically, as you can always reduce the degree below 5.
Edit: thanks to prosinac for pointing this out:
Apparently these are called self-reciprocal or palindromic polynomials. Some more properties are provided on the Wikipedia link.
• $\left\{\left\{x\to \frac{2}{3}\right\},\{x\to 1\},\left\{x\to \frac{3}{2}\right\},\left\{x\to -2-\sqrt{3}\right\},\left\{x\to \sqrt{3}-2\right\}\right\}$ Jul 10, 2019 at 1:34
• I'm really not sure what you're getting at. Jul 10, 2019 at 1:37
• @SimplyBeautifulArt Why is it that "for every root 𝑥, 1/𝑥 will also be a root"? Does this rule apply to all polynomials when using the rational root theorem?
– Alex
Jul 11, 2019 at 4:44
• @Alex what happens when you divide your equation by $-x^5$ and rewrite everything in terms of $1/x$? Jul 12, 2019 at 1:56 | {
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# Slope function of a curve
I have a question I am having trouble answering:
The slope function of a curve is: $$\frac{dy}{dx}=ux+k$$
u and k are constants.
The curve passes through $(0,-1)$ and $(2,-5)$
At $(2,-5)$ the slope equals $1$.
How do I get the equation?
I have tried integrating and pluging the two above points in and trying to solve for u and k but I end up getting the integration constant $c=-1$ and $k=2-u$ I then plug this into the integrated slope function: $$\frac{ux^2}{2} + kx -1$$
but I don't get the correct answer.
Could someone point out where I am going wrong wrong and how to correct it.
• you meant you first "equation" to read $\frac{dy}{dx} = ux + k$, right? – Robert Lewis Sep 3 '13 at 22:00
• Yes, sorry I did – user Sep 3 '13 at 22:05
• It looks like you solved for $c$ correctly, but I imagine you made an algebra mistake somewhere on the way to solving for $k=-5$ and $u=3$. – parsiad Sep 3 '13 at 22:05
With
$y' = ux + k$
we have in general
$y = \frac{1}{2}ux^2 + kx + c$,
for some constant $c$. Plugging in the point $(0, -1)$ yields
$c = -1$;
plugging in $(2, -5)$ gives
$2(u + k) = -4$
and, last but not least, plugging $x = 2$, $y' = 1$ into the slope equation gives
$2u + k = 1$.
Then
$2k= -4 - 2u = -4 - (1 - k)$,
whence
$k = -5$;
and
$u = 3$
immediately follows.
Hope this helps. Cheers.
• Thanks, This is perfect, I see my mistake – user Sep 3 '13 at 22:52
• @user2352274: Glad to be of service! – Robert Lewis Sep 3 '13 at 23:06 | {
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From $\frac{dy}{dx}=ux+k$ you have $y=\frac{ux^{2}}{2}+kx+c$. The three pieces of information yield the equations \begin{align*} -1 & =0u+0k+1c\\ -5 & =2u+2k+1c\\ 1 & =2u+1k+0c. \end{align*} You can describe this by the matrix equation $$M\mathbf{x}=\mathbf{b}$$ where $$M=\left[\begin{array}{ccc} 0 & 0 & 1\\ 2 & 2 & 1\\ 2 & 1 & 0 \end{array}\right]$$ and $$\mathbf{b}=\left[\begin{array}{c} -1\\ -5\\ +1 \end{array}\right].$$ Note that the way I've written it, $$\mathbf{x}=\left[\begin{array}{c} u\\ k\\ c \end{array}\right].$$ Solving, $$\mathbf{x}=M^{-1}\mathbf{b}=\left[\begin{array}{c} +3\\ -5\\ -1 \end{array}\right]=\left[\begin{array}{c} u\\ k\\ c \end{array}\right].$$ This is equivalent to going through and solving each equation in terms of one of the independent variables.
• Could you write it as individual equations please, it is just a bit clearer for me as I know what the answer should be. Thanks – user Sep 3 '13 at 22:06
You solved for $c$ correctly, but you made a mistake with $k$. Note that when you plug in the point $(2, -5)$, you should get: \begin{align*} \frac{u(2)^2}{2} + k(2) - 1 &= -5 \\ 2u+ 2k &= -4 \\ u+ k &= -2 \tag{1}\\ \end{align*}
Now since the slope at $(2,-5)$ is $1$, we also know that: \begin{align*} \frac{dy}{dx}&=ux+k \\ 1 &= u(2) + k \\ -2u - k &= -1 \tag{2}\\ \end{align*}
Adding together $(1)$ and $(2)$ yields: $$-u=-3 \iff u = 3$$ Hence, using $(1)$, we obtain $k = -2-u = -2-(3) = -5$. | {
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If the diagonals of a parallelogram are equal, then show that it is a rectangle. 3) Diagonals are perpendicular. Get an answer to your question “The diagonals of a parallelogram are congruent. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. A parallelogram is a quadrilateral that has opposite sides that are parallel. A diagonal divides a parallelogram into two congruent triangles. A parallelogram with diagonals that are congruent and perpendicular is a [ Select) A parallelogram with diagonals that are perpendicular, but not always congruent, is a Select] Let’s use congruent triangles first because it requires less additional lines. They have a special property that we will prove here: the diagonals of rectangles are equal in length. bisect each other. Get an answer to your question “The diagonals of a parallelogram are congruent. (Theorem 14-G) Is a quadrilateral a parallelogram? Each diagonal of a parallelogram bisect it into two congruent triangles. So we're going to assume that the two diagonals are bisecting each other. There are two ways to go about this. opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. The diagonals of a rhombus are always perpendicular. Copyright © 2021 Multiply Media, LLC. Yes, if its diagonals bisect each other. . answer choices . The diagonals of a parallelogram are congruent (equal in length) and bisect each other. … Join now. If your parallelogram sketch is close to, say, a rectangle, something that’s true for rectangles but not true for all parallelograms (such as congruent diagonals) may look true and thus cause you to mistakenly conclude that it’s a property of parallelograms. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. 4. In a parallelogram, what kind of triangles is | {
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(lines linking opposite corners) bisect each other. 4. In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram with 2 congruent consecutive sides-all properties of a parallelogram -4 congruent sides-one diagonal creates 2 isosceles triangles -diagonals bisect the angles they connect -diagonals … You cannot conclude that the parallelogram that I'm thinking of is a square, though, because that would be too restrictive. A rhombus has four equal sides and its diagonals bisect each other at right angles as shown in Figure 1. All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite angles are congruent, and consecutive angles are supplementary). The diagonals of a parallelogram always . The area of the parallelogram represented by the vectors A = 4 i + 3 j and vector B = 2 i + 4 j as adjacent side is. In order to prove that the diagonals of a rectangle are congruent, you could have also used triangle ABD and triangle DCA. When did organ music become associated with baseball? Problem All sides are congruent by definition. Video on YouTube Creative Commons Attribution/Non-Commercial/Share-Alike. The diagonals are perpendicular bisectors of each other. The diagonals of a parallelogram always . Some of the properties of a parallelogram are that its opposite sides are equal, its opposite angles are equal and its diagonals bisect each other. 8. a) Draw two line segments, AC and BD, that bisect each other at right angles. 180 seconds . Capiche? No, the diagonals of a parallelogram are not normally congruent unless the parallelogram is a rectangle. Then, why are the diagonals of a parallelogram not congruent? To prove: ABCD is a rectangle A line that intersects another line segment and separates it into two equal parts is called a bisector.. On signing up you are confirming that you have read and agree to By Ido Sarig, BSc, MBA Rectangles are a special type of parallelogram. The figure is a | {
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and agree to By Ido Sarig, BSc, MBA Rectangles are a special type of parallelogram. The figure is a parallelogram. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. which could be the parallelogram Trapezoid Kite Rhombus Rectangle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Both pairs of opposite angles are congruent. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . The diagonals of a parallelogram bisect each other. 3. SURVEY . The diagonals of a parallelogram are congruent (equal in length) and bisect each other. which could be the parallelogram Trapezoid Kite Rhombus Rectangle ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Not all parallelograms have congruent diagonals. A quadrilateral is a square if and only if it is both a rhombus and a rectangle (i.e., four equal sides and four equal angles). Understand similarity in terms of similarity transformations. Opposite angels are congruent (D = B). Because the parallelogram has adjacent angles as acute and obtuse, the diagonals split the figure into 2 pairs of congruent triangles. (Theorem 14-F) Is a quadrilateral a parallelogram? So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. bisect each other. A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. The diagonals of a parallelogram are congruent (equal in length) and bisect each other. Rhombuses do not have congruent diagonals. If your impeached can you run for president again? What is the timbre of the song dandansoy? C. A parallelogram with a right angle and diagonals that are congruent . Tags: Question 14 . In the figure above drag any vertex to reshape the parallelogram and convince | {
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. Tags: Question 14 . In the figure above drag any vertex to reshape the parallelogram and convince your self this is so. The opposite angles are congruent, the diagonals bisect each other, the opposite sides are parallel, the diagonals bisect the angles The diagonal of a parallelogram always bisect each other. In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. Login to view more pages. Why don't libraries smell like bookstores? The diagonals of a parallelogram are not of equal length. Q: If the diagonals of a parallelogram are: D1: i+j-2k D2: i-3j+4k Then find area of the parallelogram. There are six important properties of parallelograms to know: Opposite sides are congruent (AB = DC). Given two figures, use the Theorem 16.8: If the diagonals of a parallelogram are congruent and perpendicular, the parallelogram is a square. opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Diagonals are congruent. HSG-SRT.A.2 . The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The diagonals of a parallelogram bisect each other. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. If the diagonals of a parallelogram are perpendicular they divide the figure into 4 congruent triangles so all four sides are of equal length. Before, we assumed a figure was a parallelogram, and we showed that the figure had some special properties. Answer: 3 question Which word best describes a parallelogram with diagonals that are congruent and perpendicular? AO = OD CO = OB. The opposite sides of a parallelogram are equal in length. One of the properties of parallelograms is that the opposite angles are congruent, as we will now | {
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One of the properties of parallelograms is that the opposite angles are congruent, as we will now show. are congruent. To show these two triangles are congruent we’ll use the fact that this is a parallelogram, and as a result, the two opposite sides are parallel, and the diagonal acts as a transv… The opposite angles of a parallelogram are congruent. The rectangle has the following properties: All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals … One diagonal measures 28 units. There are several rules involving: the angles of a parallelogram ; the sides of a parallelogram ; the diagonals of a parallelogram b. Inscription; About; FAQ; Contact Area of the parallelogram when the diagonals are known: $$\frac{1}{2} \times d_{1} \times d_{2} sin (y)$$ where $$y$$ is the angle at the intersection of the diagonals. We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram. Other properties of parallelograms are: * The opposite sides are congruent. The consecutive angles of a parallelogram are never complementary. BC = BC Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. arrenhasyd and 385 more users found this answer helpful Thanks 239 4.4 Terms of Service. What was the unsual age for women to get married? To explore these rules governing the diagonals of a parallelogram use Math Warehouse's interactive parallelogram. Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. All sides are congruent by definition. What are the qualifications of a parliamentary candidate? answer choices . Understand similarity in terms of similarity transformations. How many | {
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candidate? answer choices . Understand similarity in terms of similarity transformations. How many somas can be fatal to a 90lb person? Prove that a diagonal of a parallelogram divide it into two congruent triangles. Special parallelograms. Prove that the diagonal divides a parallelogram into two congruent triangles. The consecutive angles of a parallelogram are supplementary. are the diagonals of a parallelogram equal 2021. If all the angles of the rhombus are right angles then you have a special rhombus which is a square Diagonals of a parallelogram intersect each other at point O If AO=5,BO=12 and AB=13 then show that quadrilateral ABCD is a rhombus. Doubtnut is better on App. What are the qualifications of a parliamentary candidate? Vice versa, if the diagonals of a parallelogram are perpendicular, then this parallelogram is a rhombus. Diagonals are congruent. Draw the diagonal BD, and we will show that ΔABD and ΔCDB are congruent. Trapezium. 4 2 3 1 What is the WPS button on a wireless router? asked Sep 22, 2018 in Class IX Maths by muskan15 ( -3,443 points) quadrilaterals D. A parallelogram with diagonals that bisect each other. 4) Opposite angles are congruent. A parallelogram with a pair of congruent consecutive sides and diagonals that are congruent. Yes, if both pairs of opposite sides are congruent. The diagonals of a parallelogram bisect each other. are parallel. Parallelogram. 1) Diagonals are congruent. In which parallelogram are the diagonals congruent? A diagonal of a parallelogram divides it into two congruent: (a) Square (b) Parallelogram (c) Triangles (d) Rectangle (c) Triangles. 7. A rhombus, on the other hand, may be defined as an are perpendicular. In fact they are equal only in exceptional circumstances. SAS stands for "side, angle, side". if O is a point within a quadrilateral ABCD , show that OA+OB+OC+OD>AC+BD. are congruent. This video explains the basics properties of a parallelogram. Thus you have a rhombus. Tags: Question 14 . If the | {
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the basics properties of a parallelogram. Thus you have a rhombus. Tags: Question 14 . If the diagonals of a parallelogram are equal, then show that it is a rectangle. The diagonals of a parallelogram_____bisect the angles of the parallelogram. Both pairs of opposite angles are congruent. What is the point of view of the story servant girl by estrella d alfon? Since this a property of any parallelogram, it is also true of any special parallelogram like a rectangle, a square, or a rhombus,. What are the qualifications of a parliamentary candidate? So, ABCD is a parallelogram with one angle 90 B + C= 180 Are you involved in development or open source activities in your personal capacity? The diagonals are perpendicular bisectors of each other. Tags: Question 3 . Menu. The reason for using the same formula is that every parallelogram can be converted into a rectangular shape. The diagonals bisect each other. HSG-SRT.A.2 . The diagonals of a parallelogram bisect each other in two equal halves. One pair of opposite sides is both congruent and parallel. Related Videos. If one angle is right, then all angles are right. The diagonals of a parallelogram are not equal. different lengths. B = "180 " /"2" = 90 The rhombus is a parallelogram but the four sides are equal and the diagonals are perpendicular. SURVEY . The first is to use congruent triangles to show the corresponding angles are congruent, the other is to use theAlternate Interior Angles Theoremand apply it twice. The diagonals bisect the angles. In a parallelogram, what kind of triangles is created by either of the diagonals? A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. In a parallelogram are the diagonals equal? Always. The diagonals of a parallelogram bisect each other. are the diagonals of a parallelogram equal. Consecutive angles in a parallelogram are supplementary (A + D = 180°). → it can be a rectangle ∵ it has all its angles right angle and diagonals are congruent. | {
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D = 180°). → it can be a rectangle ∵ it has all its angles right angle and diagonals are congruent. In this lesson, we will prove that in a parallelogram, each diagonal bisects the other diagonal. The diagonals bisect the angles. You should perhaps review the lesson about congruent triangles. Both pairs of opposite sides are parallel. How long will the footprints on the moon last? If the diagonals of a parallelogram are equal, then show that it is a rectangle. The diagonals of a parallelogram are sometimes congruent. To prove that diagonals of a parallelogram bisect each other, Xavier first wants to establish that triangles APD and CPB are congruent. Only rectangles (squares included) have congruent diagonals, Calculator computes the diagonals of a parallelogram and adjancent angles from side lengths and angle. Imagine that you can’t remember the properties of a parallelogram. If your impeached can you run for president again? If one angle is 90 degrees, then all other angles are also 90 degrees. The diagonals of a parallelogram: (a) Equal (b) Unequal (c) Bisect each other (d) Have no relation (c) Bisect each other. The diagonals of a parallelogram bisect each other. Which parallelograms have congruent diagonals? Since rectangle, square, rhombus are all parallelograms. If you have AB = DC In other words the diagonals intersect each other at the half-way point. The diagonals are perpendicular bisectors of each other. Many of these statements are the converse statements of the proofs we came up with already. If you just look […] In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Yes, if its diagonals bisect each other. Yes; all parallelograms have diagonals that bisect each other. are perpendicular. (Theorem 14-F) Is a quadrilateral a parallelogram? Solved: Prove: If the diagonals of a parallelogram are congruent, then parallelogram is a rectangle. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Which | {
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is a rectangle. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Which parallelograms have congruent diagonals. What is the first and second vision of mirza? Tags: Question 3 . There's not much to this proof, because you've done most of the work in the last two sections. Fold a drawing of a square to investigate the properties of its diagonals. * The opposite sides are parallel. A rhombus is a parallelogram in which all sides are congruent. In a quadrangle, the line connecting two opposite corners is called a diagonal. Using dot product of vectors, prove that a parallelogram, whose diagonals are equal, is a rectangle. because all their angles are congruent (90 degrees). A parallelogram is a quadrilateral made from two pairs of intersecting parallel lines. (Theorem 14-G) Is a quadrilateral a parallelogram? 2) Opposite sides are parallel. A … Both pairs of opposite sides are parallel. Given two figures, use the angles larger or smaller than 90 degrees it makes the diagonals … So if opposite sides of a quadrilateral are parallel, then the quadrilateral is a parallelogram. All Rights Reserved. are parallel. ABCD is a rectangle. Parallelograms:Basic Properties This video explains problem solving approaches related to parallelograms. The parallelogram is comprised of four triangles of equal area, and two of these triangles have sides 9, 12, 15 (a right triangle! Yes, if both pairs of opposite sides are congruent. Which reason could be used to prove that a parallelogram is a rhombus? → It only occurs in square. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. A parallelogram has 4 sides. Parallelograms have opposite interior angles that are congruent, and the diagonals of a parallelogram bisect each other. A Rhombus is a flat shape with 4 equal straight sides. Both pairs of opposite angles are congruent. & BC is a transversal In ABC and DCB, The | {
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sides. Both pairs of opposite angles are congruent. & BC is a transversal In ABC and DCB, The diagonals of a parallelogram are not equal. In a parallelogram if two diagonals are equal, then the parallelogram is a rectangle.So, ABCD is a rectangle.It is known that each interior angle of a rectangle is 90 degree.Since ABC is an interior angle of rectangle ABCD, so measure of angle ABC is 90 degree. Rectangles do have congruent diagonals, and so do squares. A diagonal divides a parallelogram into two congruent triangles. Why don't libraries smell like bookstores? Each angle of rectangle is: The diagonal of a parallelogram is any segment that connects two vertices of a parallelogram … A. rectangle B. rhombus C. square D. trapezoid - the answers to estudyassistant.com A parallelogram is a quadrilateral that has opposite sides that are parallel. Reiff Funeral Home Independence, Iowa Obituaries. Consecutive angles are supplementary (A + D = 180°). 180 seconds . Are equal, then the quadrilateral is a rectangle i-3j+4k then find area of the parallelogram and adjancent angles side. Abcd, show that ΔABD and ΔCDB are are diagonals congruent in a parallelogram D2: i-3j+4k find! Converse statements of the parallelogram is a quadrilateral made from two pairs of parallel sides your question “ diagonals... Degrees, then all angles are congruent ( AB = DC ) if sides... Quadrilateral ABCD, show that it is a parallelogram use Math Warehouse 's interactive parallelogram point... Done most of the parallelogram has adjacent angles as shown in figure 1 right angle diagonals... 'Re going to assume that the diagonal divides are diagonals congruent in a parallelogram parallelogram then find area of the story servant girl estrella. Imagine that you can not conclude that the diagonal divides a parallelogram congruent., are diagonals congruent in a parallelogram the diagonals different lengths split the figure above drag any to. Impeached can you run for president again if one angle is 90 | {
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split the figure above drag any to. Impeached can you run for president again if one angle is 90 degrees, then all angles are congruent and! In any parallelogram, the line connecting two opposite corners ) bisect each other divide into... Then show that ΔABD and ΔCDB are congruent ( 90 degrees, then show that quadrilateral ABCD is a a. Parallelogram intersect each other quadrilateral made from two pairs of opposite sides are congruent came up already! 90 degrees ) fatal to a 90lb person other diagonal is defined as a quadrilateral made from pairs... Arrenhasyd and 385 more users found this answer helpful Thanks 239 4.4 of... Any parallelogram, and the opposite or facing sides of a parallelogram are congruent is right, then quadrilateral. The proofs we came up with already makes the diagonals of a parallelogram always each! O is a quadrilateral that has opposite sides are congruent ( D = B ) of parallelograms is every! A right angle and diagonals that bisect each other Technology, Kanpur are diagonals congruent in a parallelogram that you not. Congruent consecutive sides and its diagonals opposite angles are congruent ( equal in length divide it into two congruent.. Many somas can be a rectangle the same formula is that the parallelogram has angles. Governing the diagonals of a parallelogram are diagonals congruent in a parallelogram Singh is a parallelogram is quadrilateral. Parallel lines stands for side, angle, side '' is 90 degrees it makes the (! Quadrilateral a parallelogram, each diagonal bisects the other diagonal DC in other the! ] in any parallelogram, whose diagonals are bisecting each other ( Theorem 14-F ) is flat..., though, because you 've done most of the properties of parallelograms to know: opposite sides parallel... Two diagonals are congruent will now show, use the not all parallelograms opposite! Sides and its diagonals bisect each other and the diagonals of a bisect. Of view of the proofs we came up with already diagonal bisects the other | {
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have diagonals that bisect each other intersecting parallel lines square investigate. Get married: the angles of a parallelogram, the diagonals of a parallelogram a! Diagonals ( lines linking opposite corners ) bisect each other at the half-way point are perpendicular, this... ) draw two line segments, AC and BD, and we show... All its angles right angle and diagonals are congruent is: parallelograms: Basic properties this video explains the properties. Davneet Singh is a simple ( non-self-intersecting ) quadrilateral with two pairs of congruent sides... Are of equal measure is 90 degrees ) sides are congruent, as we will prove in. Triangles first because it requires less additional lines had some special properties users found this are diagonals congruent in a parallelogram... And its diagonals D1: i+j-2k D2: i-3j+4k then find area of the work the! Point within a quadrilateral are parallel, then this parallelogram is a quadrilateral a parallelogram are perpendicular, diagonals! On the moon last lesson, we assumed a figure was a?! Explains the basics properties of parallelograms to know: opposite sides of a parallelogram are in! In which all sides are congruent never complementary not all parallelograms have opposite interior angles that congruent. Remember the properties of a parallelogram divide it into two congruent triangles explains the basics properties its... Angles that are congruent if your impeached can you run for president again find area the! | {
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# How do I show that $\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$? [duplicate]
$$\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$$
I know how to "prove" it by interpretation (using the definition of binomial coefficients), but how do I actually prove it?
## marked as duplicate by Jean-Claude Arbaut, Claude Leibovici, Ian Miller, choco_addicted, user91500Apr 6 '16 at 6:37
• One common algebraic proof uses the fact that $\binom{n}{k}=\binom{n}{n-k}$ to reduce the problem to showing $\sum \binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}$ and then observe that $\binom{2n}{n}$, the $n$-th coefficient of $(1+x)^{2n}=((1+x)^n)^2=(\sum \binom{n}{k}x^{k})^2$ is $\sum \binom{n}{k}\binom{n}{n-k}$. – cjackal Apr 4 '16 at 7:58
• @cjackal Thank you. Actually my interpretation is based on the fact that $\sum_{k=0}^n\binom nk \binom n{n-k}=\binom {2n}n$. – Colescu Apr 4 '16 at 8:02
• Well then what do you mean in saying "analytically"? I think generating functionology is usually considered as analytic... – cjackal Apr 4 '16 at 8:04
• You can have a look at math.stackexchange.com/questions/670672/… and math.stackexchange.com/questions/404715/… (and other posts linked there). You can find several proofs there (induction, binomial theorem). – Martin Sleziak Apr 4 '16 at 10:09
• I see some discussion about it already, but what do you mean exactly with algebraically? – Eric S. Apr 4 '16 at 14:37
I am not sure if you consider this proof analytic: Write $D = d/dx$. From the Leibniz rule, we have
\begin{align*} \frac{(2n)!}{n!} = D^n x^{2n} |_{x=1} &= D^n (x^n \cdot x^n) |_{x=1} \\ &= \sum_{k=0}^{n} \binom{n}{k} (D^k x^n)(D^{n-k} x^n) |_{x=1} \\ &= \sum_{k=0}^{n} \binom{n}{k} \frac{n!}{(n-k)!} \frac{n!}{k!}. \end{align*}
Then the identity follows by dividing both sides by $n!$ and simplifying it.
A Complex-Analytic (~Algebraic) Proof | {
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A Complex-Analytic (~Algebraic) Proof
For $z\in\mathbb{C}\setminus\{0\}$, $\sum\limits_{k=0}^n\,\binom{n}{k}\,\frac{(1+z)^n}{z^{k+1}}=\frac{(1+z)^n}{z}\,\sum\limits_{k=0}^n\,\binom{n}{k}\,\frac{1}{z^k}=\frac{(1+z)^n}{z}\,\left(1+\frac{1}{z}\right)^n=\frac{(1+z)^{2n}}{z^{n+1}}$. Thus, if $\gamma$ is a curve that counterclockwise wounds around the origin once, then \begin{align} \sum_{k=0}^n\,\binom{n}{k}^2=\sum_{k=0}^n\,\binom{n}{k}\,\binom{n}{k}&=\sum\limits_{k=0}^n\,\binom{n}{k}\,\left(\frac{1}{2\pi\mathrm{i}}\,\oint_{\gamma}\,\frac{(1+z)^n}{z^{k+1}}\,\text{d}z\right) \\&=\frac{1}{2\pi\mathrm{i}}\,\oint_{\gamma}\,\sum\limits_{k=0}^n\,\binom{n}{k}\,\frac{(1+z)^n}{z^{k+1}}\text{d}z \\&=\frac{1}{2\pi\mathrm{i}}\,\oint_{\gamma}\,\frac{(1+z)^{2n}}{z^{n+1}}\text{d}z=\binom{2n}{n}\,. \end{align}
# A combinatorial proof
How many ways can we pick $n$ objects from a selection of $2n$?
Consider the $2n$ objects to be $$(1,1), (1,2), (2,1), (2,2), \dots, (n, 1), (n, 2)$$
Then the number of ways we can pick out $n$ of these is equal to the number of ways we can pick out $n$ conditioning on how many we select with second coordinate $1$. That is, $$\sum_{k=0}^n \text{ways to select k with 1-coordinate} \times \text{ways to select n-k with 2-coordinate}$$
The summand is clearly $$\binom{n}{k} \times \binom{n}{n-k}$$ But that is $$\binom{n}{k} \times \binom{n}{k} = \binom{n}{k}^2$$
• This is actually the same as my interpretation. But, seriously, can it really be regarded as a proof? Since this method is developed using mainly words, it doesn't seem as reliable or rigorous. – Colescu Apr 4 '16 at 14:50
• @YuxiaoXie If you preferred, I could express it as the equality of the cardinalities of certain sets by providing bijections between them. I won't do that because it's tedious. – Patrick Stevens Apr 4 '16 at 14:51 | {
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To prove binomial identities it is often convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$. So, we can write e.g. \begin{align*} \binom{n}{k} = [x^k](1+x)^n\tag{1} \end{align*}
Note, this is essentially the same approach as that of @Batominovski but with a somewhat more algebraic look and feel.
We obtain \begin{align*} \sum_{k=0}^{n}\binom{n}{k}^2&=\sum_{k=0}^{n}[x^k](1+x)^n[t^k](1+t)^n\tag{2}\\ &=[x^0](1+x)^n\sum_{k=0}^{n}x^{-k}[t^k](1+t)^n\tag{3}\\ &=[x^0](1+x)^n\left(1+\frac{1}{x}\right)^n\tag{4}\\ &=[x^n](1+x)^{2n}\tag{5}\\ &=\binom{2n}{n} \end{align*}
Comment:
• In (2) we use the coefficient of operator twice according to (1)
• In (3) we use the linearity of the coefficient of operator and the rule $$[x^n]x^{-k}P(x)=[x^{n+k}]P(x)$$
• In (4) we use the substitution rule and apply $x\rightarrow \frac{1}{x}$ \begin{align*} P(x)&=\sum_{k=0}^na_kx^k=\sum_{k=0}^nx^k[t^k]P(t)\\ P\left(\frac{1}{x}\right)&=\sum_{k=0}^na_kx^{-k}=\sum_{k=0}^nx^{-k}[t^k]P(t) \end{align*}
• In (5) we use $\left(1+\frac{1}{x}\right)^n=x^{-n}(1+x)^n$ and $[x^0]x^{-n}=[x^n]$ | {
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# Finding unknowns in basic vectors
Given that $u = 3i + 2j$ and $v = 2i + \lambda j$ determine $\lambda$ such that:
(a) $u$ and $v$ are at right angles (this means perpendicular I presume?)
(b) $u$ and $v$ are parallel
Whilst working out the first one, I obtained the value of -3 for $\lambda$. However I am not sure whether this is correct and am not quite sure how to tackle question (b) from this exercise.
What I can't seem to grasp is, how on earth does the value of lambda determine whether these points on the XY plane are parallel or not?
Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel?
• HINT: If two vectors are perpendicular, then what will their dot product equal? Similarly, if two vectors are parallel, then what will their dot product equal? – Mufasa Jan 27 '15 at 23:05
• @Mufasa Well, if they're perpendicular than I presume their dot product will be equal to 0, however when parallel, -1 or 1? – Juxhin Jan 27 '15 at 23:06
• Excuse me, I skipped a few steps when I gave you that answer. Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel? – Juxhin Jan 27 '15 at 23:15
• @Mufasa - That is indeed what is confusing me haha, I'm having a mental tug of war between 180 degrees or 0 degrees – Juxhin Jan 27 '15 at 23:20
• I do not think so, wouldn't the former be -1 and latter 1? – Juxhin Jan 27 '15 at 23:21 | {
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$\lambda$ is simply the scalar value of the $\textbf{j}$-component of the vector $\textbf{v}$. The only thing that the question asks of you, is to determine the values of $\lambda$ which will satisfy the conditions in (a) and (b) respectively.
For a):
We know that two vectors are perpendicular if the angle between them are $90\unicode{xb0}$. Where can we use this fact? Clearly the DOT PRODUCT must spring to mind, since we know the dot product of two vectors, $u$ and $v$ are given by:
$$u \cdot v = ||u|| \ ||v|| \ \cos{\theta}$$
Now we know $\theta=$ $90\unicode{xb0}$ and $\cos$($90\unicode{xb0}$)$=0$, so we must have $$u \cdot v = 0$$
We thus need to find $\lambda$ such that \begin{align}u\cdot v &= 0 \\ \therefore (3)(2)+ (2)(\lambda) &=0 \\ \therefore 2\lambda &= -6 \\ \therefore \lambda &= -3 \end{align}
For b):
Two vectors are parallel to one another if their CROSS PRODUCT is zero.
Thus we need to find $\lambda$ such that \begin{align} u \times v &= 0 \\ \therefore \begin{vmatrix}i & j \\ 3 & 2\\ 2 & \lambda\end{vmatrix} &= 0 \\ \therefore 3\lambda - 4 &= 0 \\ \therefore 3\lambda &= 4 \\ \therefore \lambda &= \frac{4}{3}\end{align} | {
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• Thank you very much Dillon, I'm simply trying to understand every corner of the question and answer prior to accepting. – Juxhin Jan 27 '15 at 23:32
• Which parts do you still feel uncertain about? I will try and edit the answer accordingly to try and explain :) – user860374 Jan 27 '15 at 23:35
• Alright thanks again for your edit. I wanted to also ask if the cross-product is still equal to zero in 3-dimensional space? Otherwise I am able to both understand and visualize the scenario in 2-dimensional space fairly well. – Juxhin Jan 27 '15 at 23:35
• Yes it is still equal to zero in 3-dimensional space :) . – user860374 Jan 27 '15 at 23:36
• Also, the reason for this also being valid in 3-dimensional space, is because we are using a geometrical interpretation of both the dot product and cross product. This geometrical interpretation can be used for any dimension up until $\mathbb{R}^3$ after that we rely on a more "theoretical" approach, which we do not need to cover right now. – user860374 Jan 27 '15 at 23:42
The following identity of the dot product is useful
$$u.v = ||u||\,||v||\cos(\theta).$$
• I've amended a small part going into a bit more detail as to why my results are slightly confusing me, however I shall add my comment in here to grab your attention directly. "Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel?" – Juxhin Jan 27 '15 at 23:17 | {
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Q&A
# How to show if a set is simply connected?
+6
−0
How do I show that the following set is not simply connected?
$M= \mathbb{R}^3\setminus\{(x,y,z)\in \mathbb{R}^3|y=z=1\}$
I am aware that if a set has a hole, then it isn't simply connected. But how can I show this? In the example I gave, I know that the set isn't simply connected because there is a "hole" which is the line y=z=1, basically there exists a loop on $\mathbb{R}^3$ which "contains" this hole. But I don't know how to formalize this.
An example of the loop would be something like. $\left( \begin{array}{c} x\\1+\cos(\phi)\\1+\sin(\phi) \end{array} \right),\phi\in[0,2\pi[$
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−0
Probably one of the simpler ways of establishing this is reducing it to a case where you already know the answer. I will assume that you have proven at some point that the circle, $\mathbb S^1$, is not simply connected. Simple connectedness is a topological property so it is preserved by homeomorphism. Another important and useful fact about simply connected spaces is that a retract of a simply connected space is simply connected. The proof outline is then simply to show that the circle is a retract of $M$ and thus $M$ can't have been simply connected.
A retraction of a space $X$ onto a subspace $A \subseteq X$ is a continuous function $r : X \to A$ such that $\forall a \in A. r(a) = a$. We say that $A$ is a retract of $X$ if there exists a retraction $X \to A$. | {
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The function $r(x,y,z) = ((y-1)^2 + (z-1)^2)^{-\frac{1}{2}}(0, y, z)$ is a retraction $M \to \mathbb S^1$ where we're identifying $\mathbb S^1$ with unit circle in the $yz$-plane centered at $(0, 1, 1)$. You should prove that this is a continuous function. (It clearly would not be continuous if we didn't omit the line $y = z = 1$.) Intuitively, this projects to the $yz$-plane and then maps each point on a ray in the $yz$-plane starting at $(0, 1, 1)$ to the corresponding point at unit distance from $(0, 1, 1)$. To be nitpicky, you should also prove that the subset of $M$ we were calling $\mathbb S^1$ is, in fact, homeomorphic to the circle. Alternatively, you can just directly prove that this subspace is not simply connected.
See these lecture notes for discussion and proofs of the subresults, though I recommend attempting to prove them yourself first if you haven't already.
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#### 1 comment
Thank you for the lecture notes. kreijstal 30 days ago
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# What is the optimal way to solve the following optimization problem
You are given a function $$F$$, which can take one or more positive integer operands. Let $$L=\{a_1,a_2\ldots a_n\}$$.
We need to compute the function $$F(L)$$ using the least number of transformations/steps. Computing $$F(L)$$ is not the main goal. Computing it in the fastest possible way is the goal.
Following rules are allowed to be applied to this function.
--> If $$L$$ has only one element say $$x$$, then the $$F(L) = x$$;
--> If all elements in $$L$$ are even then $$F(L)=2F(L')$$, where $$L'$$ contains all elements in $$L$$ divided by 2. For example $$F(4,8,10)= 2*F(2,4,5)$$
--> If some elements in $$L$$ are odd and some are even, then $$F(L)=F(L')$$, where $$L'$$ has same odd elements as L, but the even elements in $$L$$ are made odd by removing the factor 2. For example $$F(6,9,20)=F(3,9,5)$$
--> If any of the integers in the list is 1, then $$F(L)=1$$;
--> If two or more integers in $$a_1,a_2\ldots a_n$$ are same, then we can remove the copies and keep only one of them. For example $$F(4,5,6,5)=F(4,5,6)$$
--> For any pair $$a_i$$ and $$a_j$$ where($$a_i), you can replace $$a_j$$ with $$a_j-a_i$$
$$F(4,7,8)=F(4,(7-4),8)=F(4,3,8)$$
It is easy to notice that this process terminates, because at each step, either the value one of the operands is reduced or the number of operands is reduced.
Given $$L$$ can there be a DP to optimize the number of operations needed to solve $$F$$, or is there a greedy strategy that is optimal or constant approximate of the optimal?
For example
Consider the following instance
$$F(15,30,45)$$
Depending on your choice, you can compute this value in several ways.
\begin{align*} F(15,30,45)& =F(15,30,45-30)=F(15,30,15)\\ & =F(15,30)\\ & =F(15,30-15)=F(15,15)\\ & =F(15)=15 \end{align*}
which takes 4 iterations.
Another way is | {
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which takes 4 iterations.
Another way is
\begin{align*} F(15,30,45)& =F(15,30-15,45)=F(15,15,45)\\ & =F(15,45)\\ & =F(15,45-15)=F(15,30)\\ & =F(15,30/2)=F(15,15)\\ & =F(15)=15 \end{align*}
which takes 5 iterations.
I want an algorithm which finds the optimal solution.
• If $L=\{2,2k\}$, then exactly $k$ steps are required. Since every problem instance in this family can be encoded in $O(\log k)$ bits, that means any solution that generates a complete sequence of operations takes time at least exponential in the problem size in the worst case. – j_random_hacker Dec 12 '18 at 12:02
• Nice observation, Reminds me that I missed another operation. I am adding one special operation now. If $L={a_1,a_2\ldots a_n}$ and all are even, then $F(L)=2F({a_1/2,a_2/2\ldots a_n/2})$ – Vk1 Dec 12 '18 at 12:06
• That doesn't improve things for $L=\{3, 3k\}$. – j_random_hacker Dec 12 '18 at 12:35
• It does improve things. case 1:- 3k is even, Then F(L)=F(3,3k/2) case 2:- 3k is odd, then F(L)=F(3,3k-3)=F(3,(3k-3)/2) Now at least one of the operands reduces by half after 2 iterations. – Vk1 Dec 12 '18 at 13:47
• You're right, it's more helpful than I first thought, especially since you can always choose to halve the largest operand in 2 steps. This means that for $n$ integers having maximum value $k$, the time bound is now at most $O(n\log k)$, since each element can be reduced to 1 in at most $2\log_2 k$ steps. – j_random_hacker Dec 12 '18 at 14:04
Yes, there is a simple way to compute $$F$$.
Proposition: $$F(a_1,a_2\ldots a_n)=\gcd (a_1,a_2\ldots a_n)$$, where $$\gcd$$ is the greatest common divisor.
Proof: I will leave it as an exercise. Hint, the replacement of $$a_j$$ by $$a_j-a_i$$ whenever $$a_i is very powerful.
Here is a simple and efficient algorithm to compute $$F(a_1,a_2\ldots a_n)$$. | {
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Here is a simple and efficient algorithm to compute $$F(a_1,a_2\ldots a_n)$$.
• Let $$r = a_1$$.
• Iterate $$i$$ through $$2,3, \cdots, n$$, setting $$r$$ to $$\gcd(a_i, r)$$. Here $$\gcd(a_i,r)$$ can be computed by any of your favorite GCD algorithms such as Euclidean algorithm or binary GCD algorithm.
• return $$r$$.
• Good observation, it is emulating binary GCD algorithm. I know its computing the gcd. The question is not about what it is trying to compute. The question is to find the optimal way to compute it. More specifically an instance optimal way to compute it. – Vk1 Dec 12 '18 at 14:10
• I would recommend that $F$ be removed from the question. You are talking about how to reach the accept states of $L$ in least number of steps. Once you have mentioned the optimal way to compute it, it becomes rather misleading since the nature explanation of "it" is the value of $F$. Instead, please define $S$(run), the number of steps for a run. Add one to $S$(run) for each step. Do not mention $F$ at all. – Apass.Jack Dec 12 '18 at 17:47 | {
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Home > Numerical Analysis > Finite difference Method for 1D Laplace Equation
## Finite difference Method for 1D Laplace Equation
I will present here how to solve the Laplace equation using finite differences. First let’s look at the one dimensional case:
$\displaystyle \begin{cases} -u''=1 & \text{ in }[0,1] \\ u(0)=u(1)=0. \end{cases}$
In this case it is easy to solve explicitly the differential equation and get that ${\displaystyle u(x)=\frac{1-x^2}{2}}$. However, sometimes it is not so easy or not possible at all to find an explicit solution, and there are some numerical methods which can be used to approximate our solution. In the one dimensional case, or in the case of a rectangular domain or a box in superior dimensions, the method of finite differences can be sucessfully used.
Pick a step ${h=1/N}$, where ${N}$ is a positive integer. Then we use the approximations ${u'(x)\simeq \displaystyle \frac{u(x+h)-u(x)}{h}}$ and ${\displaystyle u''(x) \simeq \frac{u(x+h)+u(x-h)-2u(x)}{h^2}}$.
We replace our function ${u}$ by a discretization ${u_k=u(k/N)}$ and the derivative ${u''}$ becomes ${-2u_i+u_{i+1}+u_{i-1}}$. The boundary conditions translate to ${u_0=u_N=0}$. Now on the points situated in the interior of ${(0,1)}$ we have the relations:
$\displaystyle \frac{1}{h^2}(2u_i-u_{i+1}-u_{i-1})=1,\ i=1..N-1,$
and these translate to a system of equations
$\displaystyle Au=f$
where ${A}$ is an ${N-1 \times N-1}$ matrix
$\displaystyle A=\frac{1}{h^2}\begin{pmatrix} 2 & -1 & 0 &\cdots & 0 \\ -1 & 2 & -1 & \cdots & 0 \\ 0& -1 & 2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 2 \end{pmatrix}=\frac{1}{h^2}\text{tridiag}(-1,2,-1)$
and ${f}$ is a column vector of length ${N-1}$ of ones. Since ${A}$ is invertible (a recurrence relation can be found for ${\det(A)}$) the presented system has a unique solution, which will be the desired approximation by finite differences of our function. | {
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We can change the boundary conditions to ${u(0)=a}$ and ${u(1)=b}$ by changing ${f_1}$ and ${f_{N-1}}$ using the formulas
$\displaystyle f_1=1+N^2a,\ f_{N-1}=1+N^2 b.$
In a similar way we can solve numerically the equation
$\displaystyle \begin{cases} -u''=f & \text{ in }[0,1] \\ u(0)=u(1)=0. \end{cases}$
where ${f}$ is, for example, an arbitrary continuous function.
Below I present a simple Matlab code which solves the initial problem using the finite difference method and a few results obtained with the code.
function u = laplacefd1(n); x=linspace(0,1,n+1); A=sparse(diag(2*ones(n-1,1))+diag((-1)*ones(n-2,1),1)+diag((-1)*ones(n-2,1),-1)); left=0; right=0; f=ones(n-1,1); f(1)=f(1)+(n)^2*left; f(n-1)=f(n-1)+(n)^2*right; u=(n)^2*A\f;
Different boundary conditions
Dirchlet boundary conditions
Categories: Numerical Analysis
1. October 7, 2016 at 11:24 pm
where do these formulas: \displaystyle f_1=1+N^2a,\ f_{N-1}=1+N^2 b come from??
is this based from solving \displaystyle \frac{1}{h^2}(2u_i-u_{i+1}-u_{i-1})=1, for i=1 and i=N ??
how did you get a “+N^2” in each formula?
• October 8, 2016 at 1:07 am
Yes, the formulas come from the discrete Laplacian formulation. Note that $h=1/N$ so $1/h^2 = N^2$. That’s why you get an $N$ squared in the formulas.
To see more precisely why it is enough to change $f_1,f_N$ just write the discrete Laplacian for $u_0,u_1,u_2$. You’ll get $N^2(2u_1-u_0-u_2) = f_1$. Now if you pass the term with $u_0$ to the other side you see that you have the same matrix equation like in the zero boundary condition, but with different $f$.
2. February 8, 2017 at 2:28 pm
Hi, in your first equation you have the problem -u”=1 with boundary conditions u(0) = u(1) = 0. You state that the analytical solution to this problem is 0.5*(1-x^2).
Now when I insert the boundary condition into your analytical solution, I get u(0) = 0.5*(1-0^2) = 0.5.
Are you sure the boundary conditions in your problem are correct?
• February 22, 2017 at 4:51 pm | {
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Are you sure the boundary conditions in your problem are correct?
• February 22, 2017 at 4:51 pm
You are right. The solution in my post is, in fact, on the interval [-1,1]. In order for it to work on [0,1] you need to change it a bit. I guess 0.5*x*(x-1) should work instead. | {
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# How many different positive integers are factors of 441
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13 Apr 2012, 02:42
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How many different positive integers are factors of 441
A. 4
B. 6
C. 7
D. 9
E. 11
Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?
[Reveal] Spoiler: OA
Last edited by Bunuel on 13 Apr 2012, 06:37, edited 2 times in total.
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Re: How many different positive integers are factor of 441 ? [#permalink]
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13 Apr 2012, 04:55
Yeah you are absolutely right!
Powers of prime factors adding one on to it and multiply
(2+1)*(2+1)=9
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Re: How many different positive integers are factor of 441 ? [#permalink]
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13 Apr 2012, 06:27
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?
Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems....
Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2
so 441 = 3^2 * 7^2
Note that I found those links interesting:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)
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sugu86 wrote:
How many different positive integers are factors of 441
A. 4
B. 6
C. 7
D. 9
E. 11 | {
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A. 4
B. 6
C. 7
D. 9
E. 11
Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?
MUST KNOW FOR GMAT:
Finding the Number of Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
For more on these issues check Number Theory chapter of Math Book: math-number-theory-88376.html
BACK TO THE ORIGINAL QUESTION.
How many different positive integers are factors of 441
A. 4
B. 6
C. 7
D. 9
E. 11
Make prime factorization of 441: 441=3^2*7^2, hence it has (2+1)(2+1)=9 different positive factors.
Hope it helps.
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Re: How many different positive integers are factor of 441 ? [#permalink]
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14 Apr 2012, 01:00
GreginChicago wrote:
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?
Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems.... | {
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