We all know the the square root of 2 is irrational. So, the answer should be no. But how do we prove it?
Let's assume for contradiction that x2=2 and that x is rational. Then x can be expressed as a rational number: x=m/n, where m and n are integers, n is not equal to 0. Moreover let's assume that m,n have no common factor greater than 1.
Why can we assume this?
1. Every rational number can be written as m/n where m,n are rationals and n is not equal to 0. This is the definition of rational numbers.
2. We can always simply a fraction so that m and n have no common factor greater than one.
Now, based on our assumption we end up with: (m/n)2=2. This implies that m2=2n2 and so m2 is an even integer. Then m is also an even integer (refer to my next question) and so m=2p where p is an integer.
Putting everything together, we get that: m2=4p2=2n2, leading to n2=2p2. Thus n2 is even and so n is even. This implies that n=2k for some integer k different from 0.
We got that: m=2p and n=2k. Both m and n are divisible by 2, which contradicts our initial assumtion that they have no common factor greater than 1.
We end up with the fact that x is not rational. So there is no rational number whose square is 2.