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    Show that Mod(x^2-y^2) <=2 is transitive.

    If mod(x^2-y^2)<=2 and mod(y^2-z^2)<=2 then mod(x^2-y^2)+a=2 and mod(y^2-z^2)+b=2.

    Is this the right thing to do? Do I set them equal to each other and then can I do anything further?
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    Surely x=0, y=1, z=1.5 is a counter-example?
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    I would find where the equation equals 2, then plot the graph
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    (Original post by MagicGopher)
    I would find where the equation equals 2, then plot the graph
    Either you're misunderstanding the question, or I'm misunderstanding your answer...
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    (Original post by DFranklin)
    Surely x=0, y=1, z=1.5 is a counter-example?
    Sorry I am supposed to be showing the relation V on the set of integers defined by setting xVy if Mod(x^2-y^2)<=2 is an equivalence relation. I've shown that it's reflexive and symmetric but I'm stuck with showing it's transitive. I think what I've written so far will hold for the integers though...
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    if the difference of two integer squares is less than 3, there aren't many possible choices for these integers...
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    (Original post by trickz)
    Sorry I am supposed to be showing the relation V on the set of integers defined by setting xVy if Mod(x^2-y^2)<=2 is an equivalence relation. I've shown that it's reflexive and symmetric but I'm stuck with showing it's transitive. I think what I've written so far will hold for the integers though...
    If we're working over the integers, it should be relatively painless to prove that "mod(x^2 - y^2) <= 2" is the same condition as "x = y". And equality is an equivalence relation...
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    What about if x=1 and y=0?
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    (Original post by trickz)
    What about if x=1 and y=0?
    Ah, oops. :p: I thought we were talking about the positive integers, for some reason. Even so, my reasoning still more or less holds. The only cases in which this is allowable are: (a) |x| = |y|, (b) one of them is zero, the other is +/- 1. So now we have potentially lots of cases to check:

    (i) |x| = |y|, and |y| = |z|
    (ii) |x| = |y|, and y = 0, |z| = 1
    (iii) |x| = |y|, and |y| = 1, z = 0
    (iv) x = 0, |y| = 1 and |y| = |z|
    (v) ...

    ...except all we need to do is notice that, in all of these cases, we have mod(x^2 - y^2) <= 1, and so by the triangle inequality, mod(x^2 - z^2) <= mod(x^2 - y^2) + mod(y^2 - z^2) <= 2. That's enough to establish transitivity.
 
 
 
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