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.

That 5 divides a^{2}+ 4b^{2} if and only if 5 divides a^{2}- b^{2} is a very good hint! Of course, it's clear that it is true: if 5 divides a^{2}- b^{2} then it certainly divides a^{b}- b^{2}+ 5b^{2} and vice versa.

The point is that if a^{2}+ 4b^{2} is not obviously "symmetric" but a^{2}- b^{2} certainly is: if a^{2}- b^{2} is divisible by 5 then b^{2}- a^{2}= -(a^{2}- b^{2}) certainly is!

Once again, that lovely "5 divides a^{2}+ 4b^{2} if and only if 5 divides a^{2}- b^{2}" is key. If 5 divides a^{2}- b^{2}and 5 divides b^{2}- c^{2} then what can you say about a^{2}- c^{2}= (a^{2}- b^{2})- (b^{2}- c^{2})?

a^2 - b^2 is obviously symmetric? I don't see it =\ can you explain this further?

[edit] Ohh... So it's like aRb -> bRa (for all a,b in some set). I see it now. But I don't see why we are using a^2-b^2, where did it come from? What made you say, oh well 5|a^2+4b^2 iff 5|a^2-b^2.

I can follow this really well; but never in a million years would I have used the idea that I could just subtract a mulitple of 5... But yes, this now makes sense.