How you can prove that root 2 is an irrational number?

Answer 1

Let's start out with the basic inequality 1 < 2 < 4.

Now, we'll take the square root of this inequality:

1 < √2 < 2.

If you subtract all numbers by 1, you get:

0 < √2 - 1 < 1.

If √2 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √2. Therefore, √2n must be an integer, and n must be the smallest multiple of √2 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply √2n by (√2 - 1). This gives 2n - √2n. Well, 2n is an integer, and, as we explained above, √2n is also an integer; therefore, 2n - √2n is an integer as well. We're going to rearrange this expression to (√2n - n)√2 and then set the term (√2n - n) equal to p, for simplicity. This gives us the expression √2p, which is equal to 2n - √2n, and is an integer.

Remember, from above, that 0 < √2 - 1 < 1.

If we multiply this inequality by n, we get 0 < √2n - n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √2p < √2n. We've already determined that both √2p and √2n are integers, but recall that we said n was the smallest multiple of √2 to yield an integer value. Thus, √2p < √2n is a contradiction; therefore √2 can't be rational and so must be irrational.

Q.E.D.