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i have some problems to get a contradiction:

problem: Does not exist $n > 0$ such that $a = p^{n}$. prove if $\log_{p}(a)$ is irrational (where $a \in \mathbb Z$ and $p$ prime)

i try this:

suppose $\log_{p} (a)$ is rational

then $\exists s,t$ such

$\log_{p} (a) =\frac{s}{t}$

$\implies p^{\frac{s}{t}}$ = a

$\implies \sqrt[t]{p^{s}}$ = a

$\implies p^{s} = a^{t}$ ( i think this is a contradiction)

if not how can i get a contradiction, can someone give me a hint

thanks for reading me

Robert Lewis
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Erick VH
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  • It is not clear what you want to prove. It looks like the sentence has two pieces that are spliced together in the wrong order. If your objective is "If $\log_p a$ is irrational then..." you can't start by assuming it is rational. You need to assume the negation of the conclusion to argue by contradiction. – Ross Millikan Mar 18 '20 at 17:47
  • Uh..... if $a=p^n$ then $\log_p a = n$ so if $\log_p a$ is irrational then $\log_p a$ is not an integer. – fleablood Mar 18 '20 at 18:18

3 Answers3

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When you arrive at $p^s = a^t$, use the fundamental theorem of arithmetic to argue that the only prime factor of $a$ is $p$.

Hence $a$ must be an integer power of $p$, contradicting the non-existence of $n$ such that $a = p^n$.

WhatsUp
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Suppose ${\rm log}_p(a) = \frac{n}k,\ \gcd(n,k)=1.\,$ Then $\, a = p^{\large n/k} \Rightarrow\, a^{\large\color{#c00} k} = \color{#c00}p^{\large n},\,$ hence this theorem implies that $\,\color{#c00}p\,$ is a $\color{#c00}k$'th power, $ $ so $\:k=1,\:$ so we conclude that $\,a = a^{\large k} = p^{\large n}.$

Bill Dubuque
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Um...... $a = p^n \iff \log_p a = n$. That's a definition.

So if $\log_a p$ is irrational then $\log_a p\ne n$ for any integer and $a\ne p^n$ for any integer.

(Of course, we have to assume (prove) that the definition of "$\log_p a$" makes any sense. That is that for $a>0, a\ne 1, p > 0$ there is a unique $x\in \mathbb R$ so that $a^x =p$. And for that we have to define/figure out what $a^x$ means. .... But none of that is in the realm of number theorem, I think.)

fleablood
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