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

In summary, the conversation discusses how to show that the progression (6k+1) (k is an integer) is closed under multiplication. It is suggested to show (6k+1)(6n+1) is of the form (6m+1) where m is an integer. The conversation also discusses how to show the progression (6k+5) contains a prime and then how to show it contains infinitely many primes. The final conclusion is to check that m is the right form of "6m+1 where m is an integer" by plugging in the previously found value for m.
  • #1
forty
135
0
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!
 
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  • #2
forty said:
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?
 
  • #3
(6k+1)(6n+1)
6(6kn+k+n)+1
m = (6kn+k+n)

is that what you mean?
 
  • #4
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)
 

What does it mean for (6k+1) to be closed under multiplication?

Closed under multiplication means that when any two numbers of the form (6k+1) are multiplied together, the result will also be of the form (6k+1).

Why is it important to prove that (6k+1) is closed under multiplication?

Proving this property is important because it allows us to simplify expressions involving numbers of the form (6k+1) and make predictions about their products without having to perform the actual multiplication.

How do you prove that (6k+1) is closed under multiplication?

To prove that (6k+1) is closed under multiplication, we can use algebraic techniques such as the distributive property and factoring to show that the product of two numbers of the form (6k+1) is also of the same form.

Can you provide an example of (6k+1) being closed under multiplication?

Yes, for example, let's take the numbers 7 (of the form 6k+1) and 13 (of the form 6k+1). When we multiply them together, we get 91, which is also of the form 6k+1 (with k=15). This shows that (6k+1) is closed under multiplication.

How is proving that (6k+1) is closed under multiplication related to number theory?

Proving this property is fundamental to number theory as (6k+1) numbers play a significant role in prime factorization and identifying patterns in prime numbers. Additionally, the property is also related to the concept of congruence in number theory.

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