Respuesta :
This proof is similar to the one for [tex] \sqrt{2} [/tex], and it's done by contradiction.
Suppose that [tex] \sqrt{3}[/tex] is rational. So, there must exist a fraction such that
[tex] \sqrt{3}=\dfrac{a}{b},\quad a,b\in\mathbb{Z},b\neq 0 [/tex]
Suppose that the fraction is already reduced, i.e. a and b are coprime.
Squaring both members give
[tex] 3 = \dfrac{a^2}{b^2} \implies a^2 = 3b^2 [/tex]
So, [tex] a^2 [/tex] is divisible by 3, and thus so is [tex] a [/tex], which means that we can write [tex] a=3k [/tex] for some [tex] k \in \mathbb{Z} [/tex]
So, the expression becomes
[tex] 9k^2 = 3b^2 \iff b^2 = 3k^2 [/tex]
So, [tex] b^2 [/tex] is also divisible by 3, and thus so is [tex] b [/tex].
Here is the contradiction, becuase we assumed that a and b were coprime, but it turned out that they are both divisible by 3.
