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Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$

I can't seem to wrap my head around that.

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If you have one bread and you take away half of it, then you remain with...? –  azimut May 2 '13 at 23:02
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This probably needs more than five answers. One more effort! –  1015 May 2 '13 at 23:20
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@azimut I find your comment confusing, given that most breads are not square. –  1015 May 2 '13 at 23:33
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I don't believe the result unless I see it proven via $\mathbb{N}$-induction using $\text{Nats}$ within Homotopy Type Theory. –  NikolajK May 3 '13 at 7:49
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By golly! That looks exactly like the Grothendieck-Riemann-Roch Theorem!!! –  Alexei Averchenko May 3 '13 at 13:02

15 Answers 15

up vote 21 down vote accepted

$$n^2 = \dfrac{2n^2}{2}$$ $$\dfrac{2 n^2}{2} - \dfrac{n^2}{2} = \dfrac{2n^2 - n^2}{2}= \dfrac{n^2}{2}$$


In short, for any $X$: $\;(X = x; X = n^2; \text{ or}\;X = n^{917})\;\text{etc.}:\;$ $${\bf two}\text{ halves of X}\;- \;{\bf one} \text{ half of X}\; = \;{\bf one} \text{ half of X}$$

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This reduces to proving that $1-\frac 12 = \frac 12$. In order to do this, expand $1$ into the geometric series $$1=\sum_{i=1}^{\infty}\frac 1{2^i},$$ and divide both sides by $2$ to get $$\frac 12=\sum_{i=1}^{\infty}\frac 1{2^{i+1}}=\sum_{i=2}^{\infty}\frac 1{2^i}.$$ It follows that $$1-\frac 12 = \sum_{i=1}^{\infty}\frac 1{2^i}-\frac 1{2^1}=\sum_{i=2}^{\infty}\frac 1{2^{i}}=\frac 12.$$

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+1 for mathematical Rube Goldburg way. –  tia May 3 '13 at 6:02
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Sorry but this is also an attention-seeking answer. –  1015 May 3 '13 at 12:34
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@julien: Exactly, my comment on the other answer was sarcasm. ;) –  Karl Kronenfeld May 3 '13 at 12:36
    
Ha, ha! I was not totally sure. –  1015 May 3 '13 at 12:37

$$\begin{array}{cccccccc}\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\ \end{array}\hspace{15pt}-\hspace{15pt}\begin{array}{cccc}\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}&\color{blue}{\star}\\\end{array}\hspace{15pt}=\hspace{15pt}\begin{array}{cccc}\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\color{red}{\star}&\color{red}{\star}&\color{red}{\star}&\color{red}{\star}\\\end{array}$$

$$ $$

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45  
This only proves the special case $n=8$. How might we proceed for other $n$? –  Karl Kronenfeld May 3 '13 at 4:00
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The generalization to even $n$ is somewhat straightforward, but to odd $n$ is nontrivial. :-) –  ShreevatsaR May 3 '13 at 5:00
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These are the results of the 2000 US presidential election. Off-topic. –  1015 May 4 '13 at 3:31

Did you know that $$x-\frac{x}{2}=\frac{x}{2}$$ whatever $x$ you take?

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We proceed by induction on $n$. The base case for $n = 0$ is $0^2 - \frac{0^2}{2} = 0 =\frac{0^2}{2}$, which follows from the fact that $0$ is the additive identity in $\mathbb{Q}$. Suppose the claim holds for some $n = k$. Now \begin{aligned} (k+1)^2 - \frac{(k+1)^2}{2} &= k^2 + 2k + 1 - \frac{k^2}{2} - \frac{2k}{2} - \frac{1}{2} \\ &= \color{blue}{\left ( k^2 - \frac{k^2}{2} \right )} + k + \frac{1}{2} \\ &= \color{blue}{\frac{k^2}{2}} + k + \frac{1}{2} \quad \color{blue}{\text{(hypothesis)}} \\ &= \frac{k^2 + 2k + 1}{2} \\ &= \frac{(k+1)^2}{2}. \end{aligned} This concludes the inductive step, so our claim holds for all $n \in \mathbb{N}$.

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Maybe an ancient greek-style of answer can be fun to read.

Let $A$ be the square with lenght of side $n$. Thererefore the area of $A$ is $n \cdot n = n^2$. Now draw one of the two diagonals. Note that the sum of the areas of the two right triangles you create (call it $B,C$) is equal to the area of $A$, and they have the same area, because they have two equal legs and the right angle between. Consider, for example $B$: the area of it is $\frac{n \cdot n}{2}$. So if you take from $A$ of area $n^2$, $B$ of area $\frac{n^2}{2}$ there is still the other, $C$ of area $\frac{n^2}{2}$.

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$$x-\frac x2=\frac x2$$ Take from something its half and you're left with half of it

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That equation is saying $$n - \text{ half of } n =\text{ half of } n$$

If you substitute n for integers it's quite easy to see why it works.

$$1 - 1/2 = 1/2$$ $$2 - 2/2 = 2/2$$

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This would be a better answer if you used some variable other than $n$. –  Micah May 2 '13 at 23:49

Since $\left(1-\dfrac{1}{2}\right)=\dfrac{1}{2}$, then $x\cdot\dfrac{1}{2}=\left(x-\dfrac{x}{2}\right)=\dfrac{x}{2}$.

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I can't say this is very enlightening. Could I also conclude that "If $(1 - \frac{1}{2}) = \frac{1}{2}$ then $(1 - \frac{1}{x}) = \frac{1}{x}$."? –  TMM May 2 '13 at 23:18
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@TMM Do you know distributive laws? –  Gastón Burrull May 2 '13 at 23:20
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@GastónBurrull I changed a word from your answer. I hope you don't mind. Maybe an intermediate step might be useful, to show the application of the distributive law. –  Pedro Tamaroff May 2 '13 at 23:26
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If you added the step $x\left(1-\frac{1}{2}\right)$, this would be the most understandable proof of this fact. From my perspective at least. –  1015 May 3 '13 at 3:46
    
@Gaston: Of course I know what you mean. But if the user who asked the question does not know why $x - x/2 = x/2$, he also may now know what you did to go from $1 - \frac{1}{2} = \frac{1}{2}$ to $x - \frac{x}{2} = \frac{x}{2}$. –  TMM May 3 '13 at 9:11

Simple arithmetic: \begin{equation} n^2 - \dfrac{n^2}{2} = n^2(1 - \dfrac{1}{2}) = n^2(\dfrac{1}{2}) = \dfrac{1}{2}n^2 = \dfrac{n^2}{2} \end{equation}

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Subtracting half of something yields half of that thing.

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Let $n= \sin x$, so that $n^2 = \sin^{2} x = 1- \cos^{2} x$. Thus $$ n^{2}- \frac{n^{2}}{2} = 1-\cos^{2} x - \frac{1-\cos^{2}x}{2}. $$

Multiplying both sides by $2$ yields

$$ 2n^{2} - n^{2} = 2-2\cos^{2} x - (1-\cos^{2}x) = 2-2\cos^{2} x - 1 +\cos^{2}x = 1 - \cos^{2} x = \sin^{2} x = n^{2}. $$

Now dividing both sides by $2$ gives

$$ n^{2} - \frac{n^{2}}{2} = \frac{n^{2}}{2}. $$

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$n=\sin x$ has not sense if $n>1$. –  Gastón Burrull May 3 '13 at 0:02
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Do you know that $|\sin x|\leq 1$ ? –  Gastón Burrull May 3 '13 at 0:07
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@GastónBurrull I know that the government wants you to think that. –  Quinn Culver May 3 '13 at 0:07
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@GastónBurrull It's only a joke answer. –  A.P. May 3 '13 at 0:43
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@GastónBurrull How is using $-2\cos^2x+\cos^2x=-\cos^2x$ more circular than using $x-x/2=x/2$ or $2n^2-n^2=n^2$? The only important thing here is $1-1/2=1/2$. Every approach uses an equivalent form of that. –  1015 May 3 '13 at 3:38

We want to prove that $n^2 - \frac{n^2}{2} = \frac{n^2}{2}$. In order to better understand what we want to do, let us simplify this equation a little bit by adding $\frac{n^2}{2}$ to both sides. We get $n^2 = \frac{n^2}{2} + \frac{n^2}{2}$. So in order to solve the problem, we just need to show that $\frac{n^2}{2} + \frac{n^2}{2} = n^2$. In order to show that these two things are equal, let's start with the left hand side and show that it equal to the right hand side. Since we know how to add fractions with the same denominator, we get $$\frac{n^2}{2} + \frac{n^2}{2} = \frac{2n^2}{2} = \frac{2}{2}n^2= n^2$$ Now we're done!

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We know that $n^2=n^2$.

Hence $n^2=\frac{2n^2}{2}$ $$\implies n^2=\frac{n^2}{2}+\frac{n^2}2$$ $$ \implies n^2- \frac{n^2}2=\frac{n^2}2$$ as we wanted.

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All you've shown is that if $n^{2} - \frac{n^{2}}{2} = \frac{n^{2}}{2}$ holds, then something 'obviously true' holds. –  Quinn Culver May 2 '13 at 23:59
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yeah you're right edit will be made –  John Marty May 3 '13 at 0:04

Actually, if you consider an integer division (so called Euclidian division) the proposition becomes: $$ n^2 - \frac{n^2}{2} = \frac{n^2}{2} + \text{mod}{(n^2,2)} $$

Indeed, by definition of the Euclidian division of the integer $n^2$ by the integer $2$, one has: $$ n^2 = 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} $$ where $\text{mod}{(a,b)}$ denotes the remainder of the euclidian division of an integer $a$ by an integer $b$.

Now, by (cleverly enough) remarking that the term $2\times \frac{n^2}{2}$ equals to $\frac{n^2}{2} + \frac{n^2}{2}$ one can write: $$ \begin{aligned} n^2 &= 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} \\ n^2 &= \frac{n^2}{2} + \frac{n^2}{2} + \text{mod}{(n^2,2)} \\ \Leftrightarrow n^2 - \frac{n^2}{2}&= \frac{n^2}{2} + \text{mod}{(n^2,2)} \\ \end{aligned} $$ that ends the proof ;)

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protected by Alexander Gruber Jun 2 '13 at 0:49

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