I need to prove that $p^{m/n}$, $m$ and $n$ naturals, $n > 1$, with $p$ prime and $\gcd(m, n)=1$, is irrational. It's suggested that this proof should be by contradiction or contraposition.
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Do you know how to show that $\sqrt{2}$ is irrational? Try to generalise that. – stochasticboy321 Jul 30 '16 at 22:51
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Try a proof by contradiction, assume there is a $p,m,n$ such that $p^{\frac{m}{n}}=\frac{k}{l};k,l \in \Bbb Z$ – marshal craft Jul 30 '16 at 22:53
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1It looks like the hypotheses got garbled, Presumably you want $,n>1, m>0,$ to exclude obvious counterexamples. I presume that in my answer. – Bill Dubuque Jul 30 '16 at 23:30
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Suppose $\,p^{m/n}$ is rational. It's a root of $\,x^n\! = p^m$ so by the Rational Root Test $\,x\,$ is an integer $\,a.$ Therefore, $ $ comparing powers of $\,p\,$ in $\,a^n = p^m$ $\Rightarrow$ $\,n\alpha = m\,$ so $\,n\mid m\,$ contra $\,n,m\,$ coprime.
Remark $ $ Without RRT: $\ $ if $\ p^{m/n}\! = a/b\ $ is rational then $\, p^m b^n = a^n\,$ hence comparing powers of $\,p\,$ on both sides of the prior equation yields $\, m+n\beta = n\alpha\, $ so $\, n\mid m\, $ contra $\,n,m\,$ coprime
Bill Dubuque
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Since you want to prove that $p^{m/n}$ is irrational, you have to assume that it is rational and derive a contradiction.
So, assume that $p^{m/n} = \frac{a}{b} $ where $(a, b) = 1$.
Then $p^{m} = \frac{a^n}{b^n} $ or $b^np^{m} = a^n $.
$p$ must divide $a$, so let $a = p^kq$ where $p \not\mid q$. Since $(a, b) = 1$, $p \not\mid b$. Then $b^np^{m} = a^n = (p^k q)^n = p^{kn} q^n $.
Since $p$ does not divide either of $b$ and $q$, $m = kn$, so that $n \mid m$. Since $(m, n) = 1$, this implies that $n = 1$, which is the only way that $p^{m/n}$ is rational.
And we are done.
marty cohen
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First a comment: the problem should have the hypothesis $m\ne0$ and $n>1$, otherwise the statement is false.
We can assume $m<n$; indeed, with the division algorithm, write $m=nq+r$, with $0\le r<n$. Then $$ p^{m/n}=p^q\,p^{r/n} $$ and this is rational if and only if $p^{r/n}$ is rational. It can't be $r=0$, because otherwise $\gcd(m,n)=n>1$.
So we can assume $0<m<n$.
Suppose $$ p^{m/n}=\frac{a}{b} $$ for $0<m<n$. Then $$ b^np^m=a^n $$ and so $p\mid a^n$, which implies $p\mid a$. Therefore $a=pc$ and $$ b^np^m=p^nc^n $$ Since $0<m<n$ we can write this as $$ b^n=p^{n-m}c^n $$ and so $p\mid b^n$. Hence $p\mid b$: a contradiction.
egreg
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$n>1$ suffices - then $m$ can't be zero if the g.c.d. is 1. I edited the original post to add $n>1$. – Jul 30 '16 at 23:39
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I couldn't figure out why can you assume n > m. I guess I understood the steps of the argument, but I can't see why does it entail the conclusion. Were m > n, everything said would still be true, wouldn't it? Also, it is fundamental to the proof this assumption? – izzorts Jul 31 '16 at 01:02
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@izzorts The assumption just simplifies the proof. If you write $m=nq+r$, then $p^{m/n}=p^qp{r/n}$. Since $p^q}$ is rational, we just need to show $p^{r/n}$ is irrational. – egreg Jul 31 '16 at 06:19
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I tried to do it without supposing $n$ > $m$. Suppose $p^{m/n}$ rational. So $p^{m/n}$ = $a/b$ ($gdc(a, b) = 1$). Also suppose $n$ > 1, otherwise it's trivial that $p^{m/n}$ is an integer or undefined. Now take $p^{m}$ = $a^{n}/b^{n}$. Then $a^{n}$ = $b^{n}p^{m}$. Therefore $p$ divides $a^{n}$ and consequently $a$. So $a$ = $pk$ for a $k$ in N. So $p^{m}$ = $p^{n}k^{n}/b^{n}$, and it follows that $b^{n} = p^{n-m}k^{n}$. Now, if $n$ > $m$, it implies that $p$ also divides $b$, contradicting $gdc(a, b) = 1$. Else, being $m$ > $n$, we have $b^{n} = k^{n}/p^{m-n}$. – izzorts Aug 01 '16 at 00:33
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Then $b=k/p^{{(m−n)}/n}$ = $k/p^{m/n}p^{−1}$ = $pk/p^{m/n}$ Ergo $b$ is divisible by $p$ in this case also, implying the same contradiction. So assuming $p^{m/n}$ rational implies contradiction in all cases possible. – izzorts Aug 01 '16 at 00:55
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@izzorts You are using $p^{m/n}$ in the last part, which is only supposed to be rational, not integer. – egreg Aug 01 '16 at 09:17
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I don't see where I assumed it to be an integer, or when it were necessary to the argument, could you show me specifically, please? – izzorts Aug 01 '16 at 21:25
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@izzorts You use it when from $b=pk/p^{m/n}$ you deduce $p$ divides $b$. – egreg Aug 01 '16 at 21:39
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Alright, thanks, now I see it. Anyway, any tips on how would I prove this last part without assuming $n > m$, then? – izzorts Aug 01 '16 at 23:52
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Assume the negation of the statement and set out for a contradiction. That is $$\frac{k}{l}^{\frac{n}{m}}=p$$
But the greatest common denominator is one so
$$\frac{k^{\frac{n}{m}}}{l^{\frac{n}{m}}}$$
Which must be a multiple of $l^{\frac{n}{m}}$. So $p=j^{\frac{n}{m}}$.
$p=({g^{\frac{1}{m}}})^n$ which contradicts the fact that $p$ was prime.
marshal craft
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