Video states that a corollary of the Rational Root Theorem is that $\left(\frac{a}{b}\right)^n != 2$ for integers $a,b,n$, where $n \gt 1$.

I'm simply looking for a proof by contradiction of this.

I started of by assuming $\left(\frac{a}{b}\right)^n = 2$, but couldn't arrive at a contradiction. (I'm probably very close).

Any ideas?

I got to: $n(\log a - \log b) = \log 2 = 0.30102..$

  • $\begingroup$ It can be shown that if $\frac ab$ is in lowest terms, then $\frac{a^{\large n}}{b^{\large n}}$ is in lowest terms. $\endgroup$ Oct 14, 2015 at 1:30

1 Answer 1


Suppose that we have $(\frac{a}{b})^n = 2$. WLOG $a$ and $b$ are in lowest terms (we can do this because we can effectively "cancel" out any common factors and obtain $c$ and $d$ in lowest terms with $\frac{a}{b} = \frac{c}{d}$).

Then we have $a^n = 2b^n$. So $a^n$ is even and hence $a$ is even. Write $a = 2k$. Then $2^n k^n = 2b^n$ and hence $b^n = 2^{n-1}k^n$ (note that $n \ge 2$ by hypothesis). Thus $b^n$ is even and hence $b$ is even.

This is a contradiction because we assumed $a$ and $b$ are in lowest terms. That completes the proof.

If you require proof that $p$ prime, $p | x^n \implies p | x$ then the easiest way is to look up the FTA and work with prime power factorizations.

  • $\begingroup$ You mean that $a $ is even, early in the proof. $\endgroup$
    – Ian
    Oct 14, 2015 at 1:17
  • $\begingroup$ @Ian Corrected, thanks. $\endgroup$
    – MathMajor
    Oct 14, 2015 at 1:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .