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This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)

Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.

Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.

Case 1. $x \geq 0$ and $y \geq 0$. Now suppose, for the sake of contradiction, $x \neq y$. Then $x > y$ or $x < y$. If $x > y$, then $0 \leq y < x$, so $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < y$ leads to a contradiction. Thus $x = y$.

Case 2. $x \geq 0$ and $y < 0$. Now suppose, for the sake of contradiction, $x \neq -y$. Then $x > -y$ or $x < -y$. If $x > -y$, then $0 < -y < x$, so $(-y)^n < x^n$. Since $n$ is even, it follows that $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < -y$ leads to a contradiction. Thus $x = -y$.

Case 3. $x < 0$ and $y \geq 0$. Applying case 2 with $x$ and $y$ interchanged, we get $y = -x$. Therefore $x = -y$.

Case 4. $x <0$ and $y < 0$. Then $-x > 0$ and $-y > 0$. Applying case 1 to $-x$ and $-y$, we get $-x = -y$. Therefore $x = y$.

Is my proof correct? Is there a shorter way to prove this?

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4  
For Case 2, you can apply Case 1 to $x$ and $-y$. To avoid four cases entirely, note that if $n$ is even, then $x^n=|x|^n$, etc. –  user86418 Aug 20 '13 at 11:37
    
@Andrew: Good observation. Make it an answer? –  Cameron Buie Aug 20 '13 at 11:42
    
@AndrewD.Hwang: Nice observation. Thank you! –  Stavros Mekesis Aug 20 '13 at 11:43
    
The other question would be (since this is Chapter I) whether all the things you use have been previously established. –  GEdgar Aug 20 '13 at 11:49
    
@GEdgar: I've proved by induction that for all $n \in N$, if $0 \leq y < x$ then $y^n < x^n$. I've also proved that if $n$ is even then $(-y)^n = y^n$. I think everything has been established. –  Stavros Mekesis Aug 20 '13 at 12:00

5 Answers 5

up vote 6 down vote accepted

The argument for Case 1 handles all four cases if you note that $x^n = |x|^n$ (etc.) when $n$ is even. (Just answerifying an earlier comment.)

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In general

$$x^n=y^n\iff 0=x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)$$

If $\,x\neq y \;$ then we must have that the second factor to the right above equals zero.

But if $\,x,y >0\,$ ($\;x,y<0\;$), then that second factor is always positive (negative), so we can already focus on the case with different signs, and thus we can assume WLOG $\,x>0>y\;$ and the second factor's zero, thus:

$$x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}=P+N\;,\;\;\text{with}$$

$$P:=x^{n-1}+x^{n-3}y^2+\ldots+xy^{n-2}\;,\;\;N:=x^{n-2}y+x^{n-4}y^3+\ldots y^{n-1}$$

Observe that $\,P>0\;,\;N<0\;$ (why? Remember $\,n\;$ is even...) , so we can put

$$x^{n-1}+x^{n-3}y^2+\ldots+xy^{n-2}=-\left(x^{n-2}y+x^{n-4}y^3+\ldots y^{n-1}\right)\iff$$

$$x\color{red}{\left(x^{n-2}+x^{n-4}y^2+\ldots+y^{n-2}\right)}=-y\color{red}{\left(x^{n-2}+x^{n-4}y^2+\ldots y^{n-2}\right)}$$

End now the proof that it must be $\,x=-y\;$ ...

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The case $\ x=0,\ y=0\ $ is of the form $\ x=y$.

Now without loss of generality set $\ y=c\cdot x$, which implies $\ c^n=1$, which implies $\ y=\pm\ x$.

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If $x$ and $y$ are complex (it is not stated that they are real), this is false, since $i^4 = 1^4 = 1$.

Just goes to show that in math, as in aerospace contracting, it's always a requirements problem.

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I would do it this way:

We have $x^n=y^n$ and $n=2k$ so we have $(y^2)^k=(x^2)^k$, taking $k$-th root of the both sides we obtain $y^2=x^2$ (beacuse $k$-th root function is an injection) so it follows $(x-y)(x+y)=0$ from which it follows that either $x=y$ or $x=-y$, short enough?

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The $k$-th power function is not an injection in general but it is an injection for non-negative arguments, which is the case. –  lhf Aug 20 '13 at 13:40
    
@lhf It is always an injection. If not, then for even $k$ it would be multivalued real function and that is not the case and because of that it is said that the $k$-th root of $x$ when $k$ is even is a positive number and because of that is an injection. –  user90628 Aug 20 '13 at 13:44
    
$y^2$ and $x^2$ are non-negative so my proof applies, right? –  user90628 Aug 20 '13 at 13:45
1  
@lhf The issue is not injectivity, but rather the fact $k$-th root is not necessarily well-defined. The argument works, it's just that in general you have to be careful when working with this function. –  M Turgeon Aug 20 '13 at 14:46

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