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Spivak Calculus: Problem 1-(iii)

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that if x2 = y2 then x=y or x = -y

    2. Relevant equations

    The 12 Properties of numbers

    3. The attempt at a solution

    I think I should do this case-wise:

    Case (a) if x=y then x2 = x*x=y*y=y2. Simple enough.

    Case (b) if x = -y then x2 = x*x=(-y)*(-y) = ...

    So what is left to show is that (-y)*(-y) = y*y which is where I am getting stuck.

    I was thinking of doing something like:

    (-y)(-y) = (-y)(-y)*1 = (-y)(-y) * [(-y)(-y)]*[(-y)(-y)]-1

    ... but I don't see this going anywhere really. Any thoughts?
  2. jcsd
  3. Jan 31, 2012 #2


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    I think you are actually doing the thing a bit backwards. If x^2=y^2 then x^2-y^2=0=(x-y)(x+y). You should be able to prove that from the 12 properties of numbers, whatever they are. You should also have if a*b=0 then a=0 or b=0. Try it that way.
  4. Jan 31, 2012 #3
    Hmmm..... After some searching around, it seems that my approach might not be the best. The simplest, is to use the given information:

    x2 = y2

    x2 - y2 = y2 - y2 = 0


    x2 - y2 = x2 +xy - xy - y2

    (x+y)*(x-y) = 0

    from here it is easy enough to show that since (x+y) or (x-y) is zero, then x = y or x = -y.


    Does this mean that the original way I was trying never would have panned out? I think it can still be done. I am just having trouble trying to prove that:

    (-a)(-a) = aa

    in a rigorous way. Though I feel like I have done this. Actually, Spivak does prove this on page 7:

    (-a)*a + a*a = a*[(-a) + a] = a*0 = 0

    Adding -(a*a) to both sides:

    (-a)*a + a*a + -(a*a) = -(a*a) so that

    (-a)*a = -(a*a)

    now adding (-a)*(-a) to both sides:

    (-a)*a + (-a)*(-a) = -(a*a) + (-a)*(-a)


    (-a)*[(-a) + a] = -(a*a) + (-a)*(-a) or

    -(a*a) + (-a)*(-a) = 0

    Finally, add a*a to both sides:

    a*a + -(a*a) + (-a)*(-a) = 0 + a*a

    so that:

    (-a)*(-a) = a*a.


    Edit: Hi Dick! By "backwards" do you just mean "not the way I would do it" ? Or do you mean wrongwards? Just curious. Either way, I am satisfied with the answer!
  5. Feb 1, 2012 #4

    Char. Limit

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    Well, what you proved using your original method is that if x=y or x=-y, then x^2 = y^2. It's TRUE, but it's the converse of what you're trying to prove.
  6. Feb 1, 2012 #5


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    What Char. Limit has pointed out is what I meant by backwards. Proving x=y and x=(-y) are solutions doesn't prove there might not be other solutions. Doing it forwards does.
  7. Feb 1, 2012 #6
    I see. Thanks for the tips! I got a lot of mileage out of this one.
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