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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?

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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). – 2017-02-14
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    What do mean by only possible prime factors of 61 are 2,3,5,7 ? – 2017-02-14
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    @Fawad If 61 is composite, then its prime factors are less than or equal to $\sqrt61$ . – 2017-02-14
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    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. – 2017-02-14
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    "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 $11 > 7$. – 2017-02-14

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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.

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    OP means 61 , and 51 was a typo – 2017-02-14
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    "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. – 2017-02-14
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    Sorry, typo. I mean $a^2$ is multiple of $61^2$. – 2017-02-14