Is my proof of the irrationality of $\sqrt{3}$ legitimate? I read the proof of the irrationality of $\sqrt{3}$ in my textbook (Richard Hammack's Book of Proof), and I was wondering if my proof is legitimate as well. 
Prop: $\sqrt{3}$ is irrational.
Suppose by way of contradiction that $\sqrt{3}$ is rational. Hence, $\sqrt{3} = \frac{m}{n}, m \in \mathbb{Z}, n \in \mathbb{N}$. Furthermore, suppose both $m,n$ are not even, so the fraction is reduced. Then, $3n^2=m^2$. Suppose $n$ is even, so $n = 2a, a \in \mathbb{Z}$. Therefore, $3 \cdot 4a^2=m^2 \longrightarrow m^2=12a^2 \longrightarrow m^2=2(6a^2)$ and $m^2$ is even, hence $m$ is even. This contradicts our assumption that both $m,n$ were not even, and hence $\sqrt{3}$ must be irrational. $\blacksquare$
Also, I'm wondering if I also need to show the case where $m$ is even?
Edit: Thank you for all the help. I realize that I was essentially trying to make the same argument as the classic proof of the irrationality of $\sqrt{2}$, but it doesn't quite work the same. I've done some research about the prime factorization theorem and I agree that $3n^2=m^2$ being a contradiction is definitely a more elegant proof that a case-by-case approach. 
 A: There are at least two flaws:


*

*you say $m,n$ are both not even, which in fact means neither $m$ nor $n$ are even.

*you say "so the fraction is reduced", but is $\frac{15}{25}$ reduced ?

Allowing only irreducible fractions, $$\sqrt3=\frac pq\iff p^2=3q^2.$$
So $p^2$ is a multiple of $3$, and so must $p$ be. Then $p^2$ is a multiple of $9$ and $q^2$ is a multiple of $3$. And so must $q$ be !
A: It's incomplete. You showed that $n$ cannot be even. But $n$ can still be odd and then your proof does not say that there is a contradiction (which there should be).

Here is a shorter proof:
Assume $3= m^2/n^2$ where $m,n \in \mathbb{Z}$, $n \neq 0$. Then
$$3n^2 = m^2$$
Comparing the unique prime factorizations of both sides we find that the prime $3$ occurs to odd power on the left side, but even power on the right side, contradicting the uniqueness of prime factorizations.
This same proof generalises to $\sqrt{p}$ is irrational for every prime $p$.
A: The $nth$ root of a number is rational precisely when $N$ is the $nth$ power of the rational number $\frac{a}{b}$.
$$\sqrt[n] N=\frac{a}{b} \iff N=\left (\frac{a}{b} \right )^n$$
Suppose that $\sqrt[n] N$ is not. There is a positive number $r$ subtracting $N$ so that their root is rational. We just need to define what $r$ is.
$$\sqrt[n] {N-r}= \frac{a}{b} \iff N=\left (\frac{a}{b} \right )^n+r, r>0$$
If $(\frac{a}{b})^n$ is the first term of a binomial expansion, then because we're adding something to it, $r$ is defined by its following terms, otherwise known for this summation starting at $1$.
$$r=\sum_{k=1}^n {n \choose k} \left (\frac{a}{b} \right )^{n-k} \left (\frac{c}{d} \right )^k$$
And when you replace and expand this in the expression, you end up with the following binomial, whose purpose is to show that N is a $nth$ power.
$$N=\left (\frac{a}{b}+\frac{c}{d} \right )^n$$
However, does it have a rational base, when it's expressed by the sum of two numbers? The only way to find out is to make it a linear combination.
$$N=\left (\frac{a}{b}+\frac{c}{d} \right )^n \iff \sqrt[n] N=\frac{a}{b}+\frac{c}{d} 
\\ \frac{a}{b}=\lambda\sqrt[n] N\land\frac{c}{d}=\left (1-\lambda\right)\sqrt[n] N\implies1\sqrt[n] N=\lambda\sqrt[n] N+\left(1-\lambda\right)\sqrt[n] N$$
Therefore, using our assumption that $\sqrt[n] N$ is irrational, $\frac{a}{b}$ and $\frac{c}{d}$ must be too for some rational $\lambda$. Should the former be rational, the latter two are as well. The number $r$ that we defined is not necessarily irrational due to its products in the expansion.

Let's use the proposition in the opening question: $\sqrt 3$ is irrational.
$$\sqrt {3-r}=\frac{a}{b} \iff 3=\left (\frac{a}{b}\right )^2+r,r>0$$
We consider $r$ to be the binomial expansion using $n=2$.
$$r=2\left (\frac{a}{b}\right )\left (\frac{c}{d}\right )+\left (\frac{c}{d}\right )^2 \implies 3=\left (\frac{a}{b}+\frac{c}{d} \right )^2$$
Then find two numbers for the linear combination to hold.
$$\frac{a}{b}=\frac{\sqrt 3}{2}\land\frac{c}{d}=\frac{\sqrt 3}{2}\implies\sqrt 3=\frac{a}{b}+\frac{c}{d}$$
Therefore, since our assumption was that $\sqrt 3$ was irrational, those two numbers are as well. The number $r$ is rational:
$$r=2\left (\frac{\sqrt 3}{2}\right)\left (\frac{\sqrt 3}{2}\right )+\left (\frac{\sqrt 3}{2}\right )^2=\frac{9}{4}$$

Should this exercise be the square root of $4$, and because we can find two integers whose square is equal to it, the expression would unfold differently and you would find that the numbers in that linear combination are equal to $1$. It means that you can find a lower square by following that process even after the assumption that it has integer roots.
