I need homework help :(

I'm trying to prove the following statement:

For all integers a, b, and d, if a and b are both divisible by d then d is divisible by 3a+2b and d is divisible by 2a+b.

I know it's true (duh) but what are step-by-step logically valid steps to proving it?

 

Not entirely sure how to prove it, never liked proofs.

But to make sure its divisible by 3a+2b or 2a+b, all you need to show is that those two can't equal zero.

3a+2b =! 0
2a+b =! 0

Not sure how to show that since a and b have to both be integers, no possible combination of integers can ever satisfy that. Therefore it could never equal 0 and thus divisible for all integers.

 

ehh this is tough, um making them zeros wont really help. try using numbers for the variables that usually helps

 

Since A is divisible by D (A/D) and B iss divisible by D (B/D) we can write that also as (A+B)/D, right?

P.S I think this is set theory right? I don't know the math level of the rest of sythe but it's going to get harder from here, (NOT AN ENDORSEMENT FOR THE FOLLOWING FORUM) I'd post your homework questions at physicsforums.com, I've asked a couple questions and they really help you think through it

 

You don't understand what I was trying to do.

Firstly =! means does not equal to, I just find it easier than =/=

By making it not equal to zero you're trying to show when a and b not equal to zero which will make the expression undefined. If you can show that it can never equal zero then you show that it is defined for all integers which is what he was looking for.

 

:O sorry not that great at math as you say in my other thread lol