Can any improvements be made to my proof that "$\sqrt{3} $ is irrational"?

Question

Suppose, for contradiction that $\sqrt{3}$ is rational. Then there exists $a,b \in \mathbb{Z}$ such that $$\frac{a}{b}= \sqrt{3},$$

where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is even, which is a contradiction since $a/b$ is therefore not in its simplest form.

Now, consider $b$ to be odd, then a is odd. Then for $m,n \in \mathbb{Z}$, we have $$(2m+1)^2=3(2n+1)^2\\ 4m^2+4m+1=12n^2+12n+3\\ 2(2m^2+2m)=2(6n^2+6n+1)\\2(m^2+m)=2(3n^2+3n)+1.$$

The LHS is even since $m^2+m \in \mathbb{Z}$ and the RHS is odd since $ 3n^2+3n\in \mathbb{Z}$. This is a contradiction, and we therefore conclude that $\sqrt{3}$ is irrational.

Answer 1

Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 \equiv 1 \mod 4,$$ and \begin{align*} b^2 &\equiv 1\mod 4\\ 3b^2 &\equiv 3\mod 4, \end{align*} contradiction.

Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.

Answer 2

Other proof

we know that $$1<\sqrt{3}<2$$

assume $a=b\sqrt{3}\in \Bbb N$

with $b>1$.

Let $$A=\{c\in \Bbb N, c>1 : c\sqrt{3}\in \Bbb N\}$$

$$A\neq \emptyset $$ let $m=\min A$.

then

$$\alpha=m(\sqrt{3}-1)\in A \text{ and } \alpha<\min A$$ which is a contradiction, thus $\sqrt{3}\notin \Bbb Q$.