Prove that the squared root of 2 is an irrational number
Let's start by assuming that the squared root of 2 is a rational number and will prove it is irrational by contradiction.If srqt(2) is rational then it can be represented by an irreducible fraction. Let this fraction be p/q, p and q positive integers. So sqrt(2) = p/q, then we square both sides and get the following: [sqrt(2)]2 = p2/q2 , 2 = p2/q2, 2q2=p2. From this last equation we can conclude that p is an even number because p2 is even. Let p=2m, m being a positive integer. Now we replace the new value of p in the last equation and we get: 2q2=(2m)2, 2q2= 4m2, q2=2m2. From this last equation we can also conclude that q is an even number because q2 is even. If both p and q are even then the fraction is not irreducible, hence contradiction. The squared root of 2 must be irrational.
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