# Inequality proof help

1. Oct 5, 2013

### synkk

Using only the axioms of arithmetic and order, show that:

for all x,y satisfy 0≤x, 0≤y and x≤y, then x.x ≤ y.y

I'm really lost on where to start, my attempt so far was this

as 0 <= x and 0 <= y, we have 0 <= xy from axiom (for all x,y,z x<=y and 0<=z, then x.z <=y.z). then we use the same axiom to get y.y >= 0, but not sure where to go from there

2. Oct 5, 2013

### Ray Vickson

There are several versions of the "axioms", so you need to tell us exactly what axioms you are given.

3. Oct 5, 2013

### gopher_p

Do you see how your axiom, when applied to the hypothesis that 0≤x, 0≤y and x≤y, gives you both x.x≤y.x and x.y≤y.y?

Do you see where to go from here?

4. Oct 6, 2013

### synkk

Thanks I got it from there.

How would I go about proving it the other way? I.e. if 0 <= x, 0 <= y, x.x <= y.y then x <= y? How would I prove that? I have gotten to (x-y).(x-y) <= 0 but not sure how to actually prove it using the axioms which are:

A1 (x+y)+z = x+(y+z)
A2 x + y = y + x
A3 x + 0 = x for all x
A3 x + (-x) = 0
A5 x.(y.z) = (x.y).z
A6 x.y = y.x
A7 x.1 = x
A8 x.x^(-1) = 1
A9 x.(y+z) = (x.y) +(x.z)

A10 x<=y or y<=x for all x,y
A11 if x<=y and y <=x then x = y
A12 x<=y and y <=z then x<=z
A13 x<=y then x + z <= y + z
A14 x<=y and 0<=z then x.z <=y.z

5. Oct 6, 2013

### gopher_p

I think the easiest way to prove this is by contradiction.

If you want to try this, I think it'd be easier to start with $0\leq y^2-x^2=(y+x)(y-x)$.

Last edited: Oct 6, 2013