Need help with Proof

Prove that in an ordered field, if the square root of 2 is a positive number whose square is 2, then the square root of 2 < 3/2.

I tried doing a proof by contradiction, but I wanted to see if there was a more sophisticated proof.
Please Help!

Proof by contradiction would be the best way to do it. Suppose sqrt(2) is not less than 3/2. Then:

sqrt(2) >= 3/2
sqrt(2) * sqrt(2) >= 3/2 * sqrt(2) >= 3/2 * 3/2
2 >= 9/4
8 >= 9

This is a contradiction, so sqrt(2) < 3/2.