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This question is about extending a common proof of the irrationality of $\sqrt{2}$ to $\sqrt{p}$ for prime $p$, and then asking which other kinds of number $n$ does the same proof tell us that $\sqrt{n}$ is irrational.


case: $\sqrt{2}$

The following outlines the very common proof that $\sqrt{2}$ is irrational.

  • Intending to prove by contradiction, assume $\sqrt{2}$ is rational.
  • Let $\sqrt{2} = \frac{a}{b}$ is a fraction in the lowest terms, that is, $gcd(a,b)=1$.
  • Then $2 = \frac{a^2}{b^2}$, or $2b^2 = a^2$.
  • From this we observe that $2|a^2$, which implies $2|a$.
  • We can write $a = 2 a_1$, which gives us $b^2=2{a_1}^2$.
  • That is, $2|b^2$, which implies $2|b$.
  • But $2|a$ and $2|b$ contradicts $gcd(a,b)=1$, and so $\sqrt{2}$ can't be written as a ratio of integers.

case: $\sqrt{p}$

Now, I saw a different proof in "Guide to Abstract Algebra, 2nd Edition,Carol Whitehead" which generalises to $\sqrt{p}$ where $p$ is prime.

The proof is structurally the same.

  • Intending to prove by contradiction, assume $\sqrt{p}$ is rational.
  • Let $\sqrt{p} = \frac{a}{b}$ is a fraction in the lowest terms, that is, $gcd(a,b)=1$.
  • Then $p = \frac{a^2}{b^2}$, or $pb^2 = a^2$.
  • From this we observe that $p|a^2$, which implies $p|a$.
  • We can write $a = p a_1$, which gives us $b^2=p{a_1}^2$.
  • That is, $p|b^2$, which implies $p|b$.
  • But $p|a$ and $p|b$ contradicts $gcd(a,b)=1$, and so $\sqrt{p}$ can't be written as a ratio of integers.

case: $\sqrt{n}$

So this led me to ask myself, what kinds of number $\sqrt{n}$ does this proof work for?

I am not an expert in this subject (self-teaching) but my thinking led me to the crucial step being the following.

  • From this we observe that $n|a^2$, which implies $n|a$.

My conclusion is that this can only be true if $n$ is a square-free integer, that is, has no prime factors of power higher than 1.

To illustrate:

  • $2 | (2^2 \cdot 3^2)$ implies $2 | (2 \cdot 3)$
  • $2^2| (2^2 \cdot 3^2)$ does not imply $2^2 |(2 \cdot 3)$

Question: Is it correct that the proof works for $\sqrt{n}$ if $n$ is a square-free integer.


Existing Questions & Answers

The closest other answer that I think applies is this one, but my mathematical abilities aren't strong enough to under it sufficiently: https://math.stackexchange.com/a/1252195/319008

This one also seems too advanced and doesn't appear to directly answer this question: Proof of the irrationality of $\sqrt n$, where $n$ is square free

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    $\begingroup$ Just apply the usual Rational Root Theorem to $x^2-n$. $\endgroup$
    – lulu
    Commented May 29, 2023 at 22:58
  • $\begingroup$ The argument is trivial to extend to any $n$ for which there is a prime $p$ such that $p\mid n$ but $p^2\nmid n$. A bit more argument but still doable to any $n$ for which there is a prime $p$ such that the highest power of $p$ that divides $n$ is odd. $\endgroup$ Commented May 29, 2023 at 23:14

1 Answer 1

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Yes, $n|a^2$ always implying $n|a$ is equivalent (for $n\geq 2$) to $n$ being square free. If $p^2|n$ for some prime $p$ then $n$ divides $(\frac{n}{p})^2=n\cdot\frac{n}{p^2}$, but obviously doesn't divide $\frac{n}{p}$. So this specific proof works for square free natural numbers.

That being said, it can be easily proved from here that $\sqrt{n}$ is irrational whenever some prime divides $n$ with odd multiplicity. Because in this case we can write $n=a^2b$ where $b>1$ is square free. If $\sqrt{n}$ was rational, then so was $\sqrt{b}$, a contradiction.

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