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First, take $x$. $x=x^1$. $1=\frac22$, so $x^1 = x^\frac22 = (\sqrt{x})^2 = |x|$. Therefore $x = |x|$.

I checked my proof and think that all my other steps are valid. I think my mistake has to do with $x^\frac22 = (\sqrt{x})^2$

So, if there is anything wrong with this step, what is it?

J. W. Tanner
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Some Guy
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  • That is the step that's incorrect. If $x = -1$, then $\sqrt{x}$ is undefined. The square root restricts your domain to $x \ge 0$. – Amaan M Jan 08 '21 at 01:10
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    Cf. this question; fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n=a^{m\times n}$ does not always work when $m$ and $n$ are not integers. – J. W. Tanner Jan 08 '21 at 01:10
  • $a = 1$, $b= -1$. Not the same. Done. – ncmathsadist Jan 08 '21 at 01:14
  • Whoever votes to close questions like this in favor of citing it as a duplicate to a completely different question really confuse me. Might as well look at large classes of questions of diophantine equations, and citing one question asking about Catalan's Conjecture, or Fermat's Last Theorem, etc. – Derek Luna Jan 08 '21 at 01:19
  • It answered my question, though, both questions made similar mistakes. – Some Guy Jan 08 '21 at 01:21
  • When we define exponents (on the reals) for non-integer powers, or definition distinctly and specifically require that our base be a positive number. So for any negative $x$ then $x^r$ is not defined if $r$ is not an integer. So $(x^{\frac 12})$ is not defined if $x < 0$ and it is not true that $x^{\frac 22} =(x^{\frac 12})^2$. That is only true if $x > 0$. ( And if $x > 0$ then there is nothing wrong with this proof of the true statement $x=|x|$) – fleablood Jan 08 '21 at 01:26
  • "Whoever votes to close questions like this in favor of citing it as a duplicate to a completely different question really confuse me." Why? They both ask completely equivalent question: Why does an exponent with a negative base raised to a rational power yield conflicting results if we write the ration as a fraction with an even denominator. And they both have the exact same answer. – fleablood Jan 08 '21 at 01:32

1 Answers1

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You must use the definition of $|x|$ and consider the cases $x > 0$ and $x \leq 0$.

What you have done does not work as $\sqrt{x}$ is not defined for $x < 0$ when working over the real numbers. Even then, after you square the square root of $x$ you are left with $|x| = |x|$...

Derek Luna
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