First....if a number is irrational, then any multiple of it is also irrational....therefore, let us prove - by contradiction - that the sqrt(2) is irrational
Suppose there exists a/b reduced to lowest terms such that a/b = sqrt(2)....square both sides
a2/b2 = 2 mutiply both sides by b2
a2 = 2b2
The right side is even....thus, a is even
Thus, we can write a as 2n...... and a2 as 4n2
So we have
4n2 = 2b2 divide both sides by 2
2n2 = b2 and since the left side is even, then the right side is even....which means that b is also even
But, if a,b are even, then a/b isn't in lowest terms which we assumed was true....thus....we have a contradiction ......then, the sqrt(2) is irrational....and so is any multiple of it
