Math HW

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