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Show that Mod(x^2-y^2).... watch

1. 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?
2. Surely x=0, y=1, z=1.5 is a counter-example?
3. I would find where the equation equals 2, then plot the graph
4. (Original post by MagicGopher)
I would find where the equation equals 2, then plot the graph
5. (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...
6. if the difference of two integer squares is less than 3, there aren't many possible choices for these integers...
7. (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...
8. What about if x=1 and y=0?
9. (Original post by trickz)
What about if x=1 and y=0?
Ah, oops. 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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Updated: November 20, 2008
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