# For what numbers $n$ is $\sqrt{n}$ irrational?

I would say it has something to do with the numbers that can be expressed as a factor of different prime numbers, but when I get to $8$, that can be changed to $2^3$, which goes against this. Is there some kind of pattern I'm missing?

• What kind of numbers $n$ are you considering? Commented Feb 8, 2016 at 8:58
• If you mean $\;n\in\Bbb N;$, then $\;\sqrt n\;$ is either exact (and thus a natural number and this happens iff $\;n\;$ is the square of another natural number), or else irrational. The proof is a reazonably easy generalization of the proof that $\;\sqrt2\;$ isn't rational. Commented Feb 8, 2016 at 8:58
• Positive integers. Commented Feb 8, 2016 at 8:58
• (Natural) numbers which are not perfect squares. Try proving it! (hint: fundamental theorem of arithmetic). Commented Feb 8, 2016 at 9:00

Let $n\in \mathbb N$. Then $$\sqrt n \text{ is rational} \iff \sqrt n \text{ is integer}$$ Proof:
Suppose $\sqrt n = \frac{p}{q}$ is a rational number in lowest terms , so $p,q>0$ and $\operatorname{gcd}(p,q) = 1$.
Then $n = \frac{p^2}{q^2}$, and $\operatorname{gcd}(p^2,q^2) = 1$
And since $q^2$ divises $p^2$, $q^2 = 1$, so $q = 1$
Therefore $\sqrt n=p$ is an integer.

In other words, the only integers with rational square roots are the square numbers: $1, 4,9,16,25,36,...$

There is a pattern yes. And it follows from the fact that prime numbers cannot be factorized further. The basic idea for checking is -

1)prime factorize n.

2)If all prime factors have an even power, then n satisfies the required condition.

Often when we say "pattern", we are either looking for something which is intuitive like an AP or GP. Or else, we are looking for a generative formula which can generate all possible cases of the set. I don't think that's possible here, but it's pretty easy to generate the series by taking only prime numbers and taking products of even powers of these prime numbers. For example - $2^4.3^2.7^4$, when you take a root, you get $2^2.3.7^2$, which is a rational number!

Oh, and if your mind tries to trick you by thinking of examples like $(1.4)^4.(3.1)^2.(7.2)^4$, Remember that each of these decimal point numbers can be converted to fractions of prime numbers. The only difference would be that some prime numbers would have negative powers. Again, as long as they are even, it will work out without any problems!

Hope this helps :)

• Ah, I never saw that, thanks! Would it be fair to say that $\sqrt{n}$ is irrational for any n that consists of prime factorizations that are not all raised to an even power? For example, $\sqrt{12}$ is irrational, but 12 consists of $2^2 * 3$, so it is irrational because there is at least 1 prime factor that is odd. Commented Feb 8, 2016 at 9:19
• Yes that's exactly what I've written above too :D That's the correct reason :) Commented Feb 8, 2016 at 9:22

For any real number $n$: if $n$ is square of a rational, then $\sqrt{n}$ is rational. Otherwise, it's irrational.
Can you apply this to your case ($n$ - positive integer)?