# Inequality proof help

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

Ray Vickson
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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

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

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

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?

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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

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 think the easiest way to prove this is by contradiction.

I have gotten to (x-y).(x-y) <= 0 but not sure how to actually prove it using the axioms which are:

...

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)##.

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