Some questions about proofs of irrational numbers I have some questions about some things I want to clarify in regard to basic questions that ask to show that roots are irrational, for example $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{6}$. To me, I think there are a few little lemmas/thereoms I am using without explicility understanding, particularly in my assumptions about division. I will write some examples to clarify what I am wondering. I am looking for any advice, critiques, tips etc.
I am wanting to show that the following are irrational;
$\mathbf{\sqrt{3}}:$
The hint given for this one that I want to use is , to note that all integers can be written in one of the following ways, $3n$, $3n+1$ or $3n+2$.
Suppose that $\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ where $p,q \in \mathbb{Z}$ and $q \neq 0.$ We also can note that p and q are relatively prime, that is, there exists no common denominator. ( because if there was we could just divide by it and continue).
so we then have, $$\frac{p^2}{q^2}=3$$
giving $$p^2=3q^{2}$$
Now, how can I use that property of all numbers at this point? Or do I even need it? I am just not sure what to do that is valid, does this make any sense? from the property of all numbers we have that q is either of the form $3n$, $3n+1$ or $3n+2$ and hence $q^2$ must be of the form of one of $9n^2$, $9n^2+6n+2$ or $9n^2+12n+4$ but I'm not sure if I can make anything with this.
$\mathbf{\sqrt5}:$
With similar initial assumptions, we could write $p^{2}=5q^{2}$, which implies $p^2$ is divisible by $5$, and so does this imply p is divisible by 5 as well?
I think this is what I am looking for, something to do with if the square is divisible by some prime, what can we say about it?
Im sorry if what I am asking is confusing, I have more but I will leave it as is for now and update it soon. I hope it makes some sense , thanks!
 A: All you have to do is to make a contradiction. So you can just begin with a discussion on whether $p$ and $q$ are multiples of $3$ (or $5$).
A: I think the hint is asking you to consider the equation:
$p^2 = 3q^2$ (mod $3$).
Since the right hand side is $0$ (mod $3$), so is the left, so this means $3|p^2 \implies 3|p$ (since $3$ is prime).
So say $p = 3t$. Then we have:
$9t^2 = 3q^2 \implies 3t^2 = q^2$. Taking this mod $3$ as well, we have:
$0 = q^2$ (mod $3$), so that $3|q^2 \implies 3|q$. Since we assumed $\gcd(p,q) = 1$, this is a contradiction.
A: It's a theorem that if a prime number $p$ divides $ab$, (integers), then $p$ divides $a$ or $p$ divides $b$. Now if $p$ divides $a^{2}$, then $p$ divides $a$ or $p$ divides $a$ so $p$ divides $a$. Edited:Now you can apply this as $5$ is prime, further, if a squarefree integer divides $a^{2}$, then it also divides $a$.
A: You can also argue this... For any integer, $n$, and for any prime number, $p$, $n^2$ will contain an even number of $p's$ as factors. For example, $12^2 = (2^2\cdot3)^2 = 2^4 \cdot 3^2$. So there are four $2's$, two $3's$, zero $5's$, zero $7's$, and so on. So if $p^2 = 3q^2$, there there will be an even number of $3's$ on the left side and an odd number of $3's$ on the right side. This can't be true because then $p^2$ would factor into primes two distinct ways.
