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Let $a$ be an irrational number where $a^2$ is a rational number. Prove by mathematical induction or generalized mathematical induction that $a^n$ is an irrational number for all odd integers $n ≥ 1$.

So I know that a (rational) * (irrational) -> (irrational) but I'm not sure how to go about this with induction.

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    $\begingroup$ I think you may find it easier to prove that $a^{2m+1}$ is irrational for all non-negative integers $m$, by induction on $m$. $\endgroup$ – André Nicolas Nov 6 '15 at 0:15
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As succinctly as I can put it...

Basis: $a^1 = a$ is irrational.

Induction: Suppose that $a^m$, $m$ odd, is irrational. Then $a^{m+2} = a^m \cdot a^2$ is an irrational number times a rational number, and therefore irrational.

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If $n$ is odd, then $$ a^n = \underbrace{a \cdots\cdots a}_{ \large \text{$n$ factors}} = \underbrace{(aa)(aa)(aa) \cdots (aa) a}_{\begin{smallmatrix} \large \text{One left over when} \\[4pt] \large \text{they're paired; that's} \\[4pt] \large \text{what “odd'' means.} \end{smallmatrix}}. $$ Each $aa$ is $a^2$, which you said is rational. So you're multiplying a bunch of rational numbers, getting a rational number, and the one left-over $a$ is irrational.

To make this a proof by mathematical induction, you can say $$ a^\text{next odd number after $n$} = a^{n+2} = a^2 a^n $$ and $a^2$ is rational and the induction hypothesis says $a^n$ is irrational, so you're multiplying a rational number by an irrational number.

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  • $\begingroup$ Now how would I word that with induction? $\endgroup$ – RiGid Nov 6 '15 at 0:23
  • $\begingroup$ @RiGid : I've added a new paragraph to my answer. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 6 '15 at 0:26
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If $n$ is odd, then $n+2$ is odd and so $a^{n+2}=a^n a^2$ is irrational, being the product of an irrational and a rational.

If you want to do induction in steps of $1$ instead of $2$, write $n=2k+1$ and use induction on $k$.

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  • $\begingroup$ So I can do an I.H. with $a ^ (2k+1)$ then an I.S. with $a ^ (2k+2)$ ? $\endgroup$ – RiGid Nov 6 '15 at 0:20
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    $\begingroup$ @RiGid : "Induction on $k$" means proving that if it's true in case $k$ then it's true in case $k+1$, so you put $k+1$ where $k$ was, and that gives you $2(k+1)+1$ where $2k+1$ was. That is not $2k+2$; it is $2k+3$. Then next odd number after $2k+1$ is $2k+3$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 6 '15 at 0:28

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