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Suppose we want to prove that some number $x$ is transcendental. Can we do so by proving that $$x^n\ne\frac{a}{b}$$ where $a, b, n \in \Bbb{Z}$ where $n \ne 0$? The reasoning behind the statement is as follows:

Consider a polynomial equation: $0 = x^n + f(x)$ where $f(x)$ is a polynomial function where $f(x) \in \Bbb Q$ if $x \in \Bbb Q$. We can rearrange the polynomial as $-f(x)=x^n$. The LHS is rational and therefore the RHS is rational for rational values of x. We can solve the equation by taking the $nth$ root of both sides: $$\sqrt[n]{-f(x)}=x$$

The LHS and RHS now may be irrational, but remain algebraic. Therefore, if $x$ is transcendental, then $x^n\ne\frac{a}{b}$

Badr B
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    There are algebraic numbers which are not $n^{th}-$roots of rational numbers. $1+\sqrt 2$, for example. – lulu Jan 21 '18 at 18:55
  • @lulu That's true, didn't think of that. Thank you for the comment! – Badr B Jan 21 '18 at 19:01
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    Your "if" is going the wrong way. In order to prove $x$ is transcendental, you'd need something of the form "if ... then $x$ is transcendental", not "if $x$ is transcendental then ...". – Robert Israel Jan 21 '18 at 20:20
  • @RobertIsrael Thank you. I will keep that in mind! – Badr B Jan 21 '18 at 22:22

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