Proving that (6k+1) is Closed Under Multiplication

[forty]

Show the progression (6k +1) (k is an integer) is closed under multiplication:

Firstly I should check that I remember what this means... If it is closed when you multiply any 2 elements together you get an element that is in the set?

So for this I thought just show (6k+1)(6n+1), where k and n are integers, is of the form (6k+1) but I don't think that works and can't think of any other ways to do it.

Show that the progression (6k+5) contains a prime. Then show it contains infinitely many primes:

Show that it contains a prime 6(1)+5=11 (is that part really that simple :S?) As for showing it has infinitely many...

Any help with this would be greatly appreciated. I'm actually on exchange at Lund Uni in Sweden (I come from Melbourne Uni, Australia) and I'll be stuffed if I can understand my lecturer (on Number Theory) so until I can get my hands on a copy of the textbook this stuff is doing my head in!

 

[Hurkyl]

So for this I thought just show (6k+1)(6n+1), where k and n are integers, is of the form (6m+1) but I don't think that works and can't think of any other ways to do it.

I've changed your second usage of k to m; you shouldn't use the same variable for multiple purposes, you can confuse yourself.

To show something is of the form 6m+1, you just need to find m, right?

 

[forty]

(6k+1)(6n+1)
6(6kn+k+n)+1
m = (6kn+k+n)

is that what you mean?

 

[Hurkyl]

That sounds good. Now all that's left is to check that m is the right kind of object.

Remember, you wanted the form "6m+1 where m is an integer" -- so from "6m+1" we've narrowed things down to one possibility for m, so now we just have to plug into the other condition to see if it works.

(Yes, this check is trivial in this case. But it can be less trival in other problems)