- #1
Servo888
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Ok so here's one of the questions we've been assigned...
So I can graphically see what this relation looks like, and from that I've shown it's reflexive. Now I'm working on proving it as being symmetric, but I can't put it into words.
b) ~ is symmetric. Well we want to show that aRb -> bRa, so [tex]5 | a^2+4b^2[/tex] -> [tex]5 | b^2+4a^2[/tex]. Well I can see graphically once again that [tex]b^2+4a^2[/tex] will always give me a multiple of 5 - this is what I can't put into words... I can't really figure out how to express aRb -> bRa... From another help source I was told this;
[tex]5 | a^2+4b^2[/tex] iff [tex]5 | a^2-b^2[/tex], so then I was told to [tex]a^2+4b^2 - a^2-b^2 = 5b^2[/tex], which is a multiple of 5. But I don't see how the hell that works... But it's based on some equivalance class, modulous 5 (maybe?) that's 4 = -1...
c) ~ is transitive. Graphically this looks true. We need to show that aRb and bRc -> aRc. But I'm not sure where to start on this... I don't see where c fits in.
Let ~ be a relation on th integer such that for every a,b (integers) we have a ~ b iff [tex]5 | a^2+4b^2[/tex] Prove or disprove the following: b) ~ is symmetric, c) ~ is transitive
So I can graphically see what this relation looks like, and from that I've shown it's reflexive. Now I'm working on proving it as being symmetric, but I can't put it into words.
b) ~ is symmetric. Well we want to show that aRb -> bRa, so [tex]5 | a^2+4b^2[/tex] -> [tex]5 | b^2+4a^2[/tex]. Well I can see graphically once again that [tex]b^2+4a^2[/tex] will always give me a multiple of 5 - this is what I can't put into words... I can't really figure out how to express aRb -> bRa... From another help source I was told this;
[tex]5 | a^2+4b^2[/tex] iff [tex]5 | a^2-b^2[/tex], so then I was told to [tex]a^2+4b^2 - a^2-b^2 = 5b^2[/tex], which is a multiple of 5. But I don't see how the hell that works... But it's based on some equivalance class, modulous 5 (maybe?) that's 4 = -1...
c) ~ is transitive. Graphically this looks true. We need to show that aRb and bRc -> aRc. But I'm not sure where to start on this... I don't see where c fits in.