# Proving transitivity, stuck at at algebra

1. Dec 12, 2009

### barylwires

1. aRb on Z if 5|2a+3b

2. Since 5|2a+3b, 2a+3b=5m, so 2a=5m-3b.

3. Consider 3a+2b=2a+2b+a=5m-3b+2b+a=5m-b+a. I can't show that 5 divides that, but there is another way to wrangle that into a better form. I need help seeing how that works. Thanks.

Last edited: Dec 12, 2009
2. Dec 12, 2009

### LCKurtz

For transitivity you are given aRb and bRc and need to show aRc. You haven't even introduced c yet, nor written in terms of a and c what you have to show.

3. Dec 12, 2009

### barylwires

Re: proving SYMMETRY, stuck at at algebra

Sorry. I'm trying to show symmetry, not transitivity.

4. Dec 12, 2009

### LCKurtz

Hint: What happens if you subtract 5a and 5b from your equation 2a + 3b = 5m?

5. Dec 12, 2009

### barylwires

Good things happen. I think I can write: Assume aRb. Then 5|2a+3b iff 5|-(2a+3b) iff 5|-2a-3b iff 5|-3b-2a+5b+5a iff 5|2b+3a. Now aRb implies bRa, thus R is symmetrical.

I'll sleep on that. Many thanks!