# Common multiple of two numbers

1. Sep 15, 2014

### alpuy

1. The problem statement, all variables and given/known data
I want to demonstrate that the numbers that are multiple of a and b at the same time, are the multiples of ab.

Let a be 2 and b be 3.

In the middle of the proof i get to a point that i have to prove that if 3*k2 is multiple of 2 then k2 is multiple of 2.

2. Relevant equations

3*k2 is multiple of 2 → k2 is multiple of 2

3. The attempt at a solution
As 3 is not multiple of 2, then k2 has to be multiple of 2. I do not know how to prove it formally.

Thanks

2. Sep 15, 2014

### Ray Vickson

The result you state

"the numbers that are multiple of a and b at the same time, are the multiples of ab"

is false in general, but is true for certain a and b (such as your a = 2 and b = 3). To see that the general statement is not true, consider a = 4 and b = 6; their least common multiple is 12, not 4 $\times$ 6 = 24.

What do you think you need to assume about a and b to make the result true?

3. Sep 15, 2014

### Staff: Mentor

Proof by contradiction. If k2 is not a multiple of 2, it is an odd number. What is the product of two odd numbers?

This is related to the unique prime factorization of positive integers, by the way.

See the post of Ray Vickson for a general problem with your question.

4. Sep 16, 2014

### alpuy

Ok, i see that that statement is not true, the statement may be valid for prime numbers, how is the proof?

You are right it can be proven by contradiction. (2m + 1)(2n + 1) = 4mn + 2(m + n) + 1.
and 4mn + 2(m + n) is even so 4mn + 2(m + n) + 1 is odd.

How is it related to the unique prime factorization of positive integers?

Regards

5. Sep 16, 2014

### Staff: Mentor

If a is divisible by 2 and a=b*c, then b or c (or both) is divisible by 2 because the prime factor of 2 has to "be somewhere in the product."