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The question may sound silly but is there a simple logic to counter the paradox.I will be glad to know if there is. Thank You. Edit: x,y $\in R\;\; x,y \neq 0$, m is a integer. Now x = y when m is odd x = y or -y when m is even. and x and y cant be related when m = o.

I didn't mean to say there is a paradox. The question asked above is asked in one of my interview and I was asked to explain mathematically. I gave the above reply but they didn't seem to be satisfied. That is the reason I posted this question to know if there is any mathematical reason. Sorry if anyone thinks such questions shouldn't be posted and thank you everyone for taking their time in commenting.

Rohith Penumala
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  • It will help to define the intervals on which $x,y$ and $m$ are defined on - there are many cases in which your statement "if $x^m = y^m$, then $x=y$ or $x=-y$ when $m$ is even" fail. The "paradox" is in the statement being imprecise. – Irregular User Jun 15 '16 at 19:39
  • Thank you @IrregularUser x,y and m are real. and x, y $\neq$ 0 – Rohith Penumala Jun 15 '16 at 19:42

3 Answers3

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You can reason the same as why $2\neq3$ when $2\cdot0=3\cdot0$. This is because the step you performed is irreversible. It is not allowed to divide by zero to get $2=3$.

In the case $2^0=3^0$, to get a form $2=3$ you would need to take the $0$th root. But remember that $\sqrt[0]{2}=\sqrt[0]{3}$ is equivalent to $2^\frac10=3^\frac10$ with a division by zero, so that is not valid.

ericw31415
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If $m \geq 2$ and $m$ is even, $x$ and $y$ are real, then $x^m=y^m \implies x=y$ or $x=-y$.

$x^0=1 ,\forall x \neq 0, x \in \mathbb{R}$.

There is no contradiction.

Siong Thye Goh
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  • Thank You @SiongthyeGoh for writing my statement in a precise manner. I didnt mean to say there is a paradox. As when m is 0 , x and y cant be related. Since m = 0 is even how can we write that there is no contradiction. – Rohith Penumala Jun 15 '16 at 19:48
  • add more conditinons such as $m$ is even and non-zero? – Siong Thye Goh Jun 15 '16 at 19:50
  • Actually this is a question asked in an interview. And they have asked it in the same manner I asked above. I think it is my mistake to post such question in this forum. But Thank You for your time and patience. – Rohith Penumala Jun 15 '16 at 19:54
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If they were interested in in hearing reasoning around why the claim "$x^m = y^m \implies |x| = |y|$ (if $m$ even)" fails, perhaps you could take a logarithm of each side and note that it is only possible to divide $m$ out of $$m \log(|x|) = m \log(|y|)$$ when $m$ is nonzero (and here we're also assuming that $x$ and $y$ are both nonzero). This is exactly the same reasoning as in other folks' answers, but it's possible that it might be more satisfying to an interviewer because it allows you to go into more detail about why $x^0 = y^0 \not \implies |x| = |y|$.

tyo
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