Irrationality of ${5^{1/7}}$

Question

I am struggling with elementary proofs, and would appreciate any feedback as to the logic and structure of my work.

Show that ${5^{1/7}}$ does not represent a rational number.

Suppose ${5^{1/7}}$ is rational. Then there exists some $p,q$ in $Z$ for which $q$ is not equal to zero and $(p,q)=1$, and ${\dfrac{p}{q}}$ = ${5^{1/7}}$. Then, for some $k$ in $Z$, $p$ = $5k$ . So, ${5^{7}}{k^{7}}$ = $5{q^{7}}$. Then, $q^{7}$ is divisible by $5^{6}$ so $q$ must be divisible by $5$. But by the premise that $(p,q) = 1$, and $p$ is divisible by $5$, a contradiction has been reached.

I followed the structure of similar proofs that I understand how to do, but I know my logic is flawed because I don't really understand why if $q^{7}$ is divisible by $5^{6}$ then $q$ must be divisible by $5$.

Thanks.

Answer 1

You have a typo, I hope, when you say "Suppose $5^{1/7}$ is irrational". You mean "rational".

I don't know how you've got $p^7 = 5k$ and then $5^7 k^7 = 5 q^7$. The lines which actually follow are:

Then $p = 5^{1/7} q$, so $p^7 = 5 q^7$. Therefore $p$ is divisible by 5, so $q^7$ is divisible by $5^6$.

Since $5^6 \vert q^7$, we have that $5 \vert q^7$, and so (since $p \vert a b \Rightarrow p \vert a$ or $p \vert b$) have $5 \vert q$, by taking $a = q, b = q^6$.

There's an easier way to do this from the line $p^7 = 5 q^7$, using the fundamental theorem of arithmetic. Simply count the power of 5 on each side of the equality sign. It's 1 mod 7 on the RHS, and 0 mod 7 on the LHS - immediate contradiction.