Prove that the square root of 3 is irrational [duplicate]

Solution 1:

Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.

So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.

What is the contradiction?

Solution 2:

suppose $\sqrt{3}$ is rational, then $\sqrt{3}=\frac{a}{b} $ for some $(a,b) $ suppose we have $a/b$ in simplest form.
\begin{align} \sqrt{3}&=\frac{a}{b}\\ a^2&=3b^2 \end{align} if $b$ is even, then a is also even in which case $a/b$ is not in simplest form.
if $b$ is odd then $a$ is also odd. Therefore:
\begin{align} a&=2n+1\\ b&=2m+1\\ (2n+1)^2&=3(2m+1)^2\\ 4n^2+4n+1&=12m^2+12m+3\\ 4n^2+4n&=12m^2+12m+2\\ 2n^2+2n&=6m^2+6m+1\\ 2(n^2+n)&=2(3m^2+3m)+1 \end{align} Since $(n^2+n)$ is an integer, the left hand side is even. Since $(3m^2+3m)$ is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

Solution 3:

A supposed equation $m^2=3n^2$ is a direct contradiction to the Fundamental Theorem of Arithmetic, because when the left-hand side is expressed as the product of primes, there are evenly many $3$’s there, while there are oddly many on the right.