Question

Prove that root3 is an irrational number ​

Answer 1

3–√3 is irrational because it cannot be expressed as a fraction of integers.

Let’s say it was rational. Then it could be expressed as a fully simplified fraction between two integers p and q.

3–√=pq3=pq

3=p2q23=p2q2

3q2=p23q2=p2

So now we know: p2p2 is a multiple of 3.

But since p2p2 is a perfect square, and it’s multiple of 3, then it must be a multiple of 9.

If we look at our equation, that means q2q2 is also a multiple of 3.

Why?

p2p2 is a multiple of 9, so it could equal 9n9n for some integer n.

3q2=9n3q2=9n

q2=3nq2=3n

Anyway, q2q2 is a multiple of 3. This means that p and q will always have a common factor of 3, which is a contradiction to the beginning. This also means that the fraction can never be fully simplified. This also means that this fraction does not exist.

Thus, 3–√3 cannot be expressed as a fraction, and it is irrational.

Hope this helped!