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Homework Help: Inequality proof help

  1. Oct 5, 2013 #1
    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. jcsd
  3. Oct 5, 2013 #2

    Ray Vickson

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    There are several versions of the "axioms", so you need to tell us exactly what axioms you are given.
  4. Oct 5, 2013 #3
    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?
  5. Oct 6, 2013 #4
    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
  6. Oct 6, 2013 #5
    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
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