Prove that `sqrt(3)` is an a irrational number.

Explanation

Let assume , `sqrt3` be a rational number and simplest form be `a/b` then, `a` and `b` are integers having no common factor other than `1`.
Now, `sqrt3= a/b`
​⟹ `3=(a/b)^2` (On squaring both sides )
or, `3b^2=a^2` .......(i)
⟹ `3` divides `a^2` (∵`3` divides `3b^2`)
⟹ `3` divides `a`
Let `a=3c` for some integer c
Putting `a=3c` in (i), we get
or, `3b^2=9c^2`
⟹ `b^2=3c^2`
⟹ `3` divides `b^2` (∵`3` divides `3c^2`)
⟹ `3` divides `b`
we observe that `a` and `b` have at least `3` as a common factor. so, `a` and `b` are co-prime. This means that our assumption is not correct.
Hence, `sqrt3` is an irrational number.