# Square root of a prime is irrational [duplicate]

How can one prove that the square root of a prime number (e.g. $\sqrt 61$) is irrational.

First we need to prove that $61$ is prime. This can be done by simply showing that $$49 < 61 < 64$$ and so $$7 < \sqrt 61 < 8$$ then the only possible prime factors of $61$ are $2 , 3, 5, 7$. Then by contradictions and DIC we can show that $61$ is prime.

Now how can we show that $\sqrt 61$ is irrational?

## marked as duplicate by dxiv, Gerry Myerson, Martin Sleziak, Claude Leibovici, NamasteFeb 14 '17 at 12:08

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• If you want to do this for a prime number, do "the standard thing". Let $p$ be prime, assume that $\sqrt{p} = \frac{a}{b}$ where $a,b$ are coprime. Then, rewrite it as $a^2 = pb^2$, and derive a contradiction (an easy one - write out the prime factorization of both sides and count how many times $p$ occurs in both sides - the right side should have an odd number of $p$'s, the left side should have an even number). – Mark Feb 14 '17 at 6:09
• What do mean by only possible prime factors of 61 are 2,3,5,7 ? – Fawad Feb 14 '17 at 6:11
• @Fawad If 61 is composite, then its prime factors are less than or equal to $\sqrt61$ . – u123435 Feb 14 '17 at 6:15
• You don't mean that 2,3,5 and 7 are the only possible prime factors. You mean one o 2,3,5 or 7 must be a prime factor. If $p*q = x$ and $p, q$ are prime the one of them must be $\le \sqrt{x}$ and the other must be $\ge \sqrt{x}$. And if we find the smaller one, then we can find the larger one by dividing be the smaller one. So we only have to check to the square root. But that's not to find all prime factors; just to find one. 51 has 3 as a prime factor. AND it has 51/3 = 17. We only had to check to 7 but in doing so we found 17. – fleablood Feb 14 '17 at 6:22
• "If 61 is composite, then its prime factors are less than or equal to 6–√1" Incorrect. Then some of it's prime factors must be less than or equal to $\sqrt{61}$. AND some of its factors (maybe prime) must be GREATER than or equal to $\sqrt{61}$. Example $55$ has a prime fact $5 < 7$. AND it has another prime factor $11 > 7$. – fleablood Feb 14 '17 at 6:26

## 1 Answer

Suppose $61=a^2/b^2$, where $a$ and $b$ are both in lowest terms. Then $$61 \cdot b^2 =a^2$$ since $a^2$ is multiple of $61^2$, or not at all, this simply cannot happen in lhs --- it has odd number of factor of 61, if any.

• OP means 61 , and 51 was a typo – Fawad Feb 14 '17 at 6:12
• "since a is multiple of $61^2$, or not at all, this simply cannot happen." What does "or not at all" mean? And why can that "simply not happen? Why can't $a = 61^k$? Your proof is not done yet. – fleablood Feb 14 '17 at 6:30
• Sorry, typo. I mean $a^2$ is multiple of $61^2$. – Aminopterin Feb 14 '17 at 6:31