Question

prove that√2 is irrational​

Answer 1

Answer:

Step-by-step explanation:

Let's suppose √2 is a rational number. Then we can write it √2  = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality √2  = a/b it follows that 2 = a²/b²,  or  a² = 2 · b².  So the square of a is an even number since it is two times something.

From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a² would be odd too. Odd number times odd number is always odd.

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a²/b², this is what we get:

2 = (2k)²/b²

2 = 4k²/b²

2*b² = 4k2²

b² = 2k²

This means that b² is even, from which follows again that b itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.