Proving transitivity, stuck at at algebra

[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.

 

[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.

 

[barylwires]

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

 

[LCKurtz]

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

 

[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!