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

Answer 1

@SithsAndGiggles

Answer 2

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. ;(

Answer 3

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

Answer 4

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