Proving square root of three is an irrational number.

        People often prove that square root of two is an irrational number. But it is also possible to prove that the square root of three is an irrational number on the same lines, rather this thought can be extended to prove square root of 5 or 7 or 6 for that matter is an irrational number. Here's how the proof proceeds.

The proof here is by reductio ad absurdum (proof by contradiction)

Assumption:- Let sqrt(3)=p/q such that gcd(p,q)=1.

Squaring both the sides,

3q^2=p^2.------------(i)

Now since p squared is congruent to 0 mod 3,

p must be congruent to 0 mod 3.

hence we have p = 3a.

substituting this in eq i,

q^2=3p^2.----------(ii)

Thus q must be congruent to 0 mod 3.

Both p and q are multiples of 3 violates gcd(p,q)=1

Hence assumption is wrong and sqrt(3) is an irrational number.

        In proving square root of 2 is irrational, often the even odd terminologies are used but they restrict your thinking to make you feel this proof works only for square root of 2 but actually it does work for 3, 5, 7 etc, as well in the very same manner.