Question

prove that root19 is an irrational number​

Answer 1

Prove that √19 is irrational.

This will be a proof by contradiction. We will assume that √19 is rational.

That means that there are two integers, a and b, with no common division where,

a/b = √19

Let us square both sides

(a/b)² = (√19)²

Which equals

a²/b² = 19

Now solve for a²

a² = 19b²

But this last statement means that the right hand side is even because it is a product of integers and one of those integers (at least) is even. So a² must be even, and therefore a must be even.

Since a is even, there must be some integer c that is half of a, or

2c = a

Now let us replace this in the previous conclusion, which was

a² = 19b²

(2c)² = 19b²

4c² = 19b²

4/19c² = b² ---------->(Divide out)

But now we can argue the same thing for b. The left hand side is even, so the right hand side must be even and that means that b is even.

But this is a contradiction to our initial assumption. If a and b are both even then they do have a common division.

Our initial assumption is therefore false and √19 is irrational