Prove that in an ordered field , if $\sqrt 2$ is a positive number whose square is 2, then $\sqrt 2

Question

Prove that in an ordered field , if $\sqrt 2$ is a positive number whose square is 2, then $\sqrt 2<3/2 $
Please, help

loser66 · Answer

@SithsAndGiggles

loser66 · Answer

Suppose $\sqrt 2>3/2$ then $\sqrt 2−3/2>0$
 square a positive number gives us a positive number also. So that $\sqrt2−3/2)^2>0$ it gives me $2−3\sqrt 2+9/4>0$
 that means $17/4>3\sqrt 2$ or $17/12>\sqrt 2$
 since both sides are >0, we square it , the sign doesn't change and I get 2.0006 > 2
I go nowhere. ;(

loser66 · Answer

However, I don't see my mistake, please, point it out.

loser66 · Answer

Oh, I got it, friend. :) Thanks for being here