Common multiple of two numbers

[alpuy]

Homework Statement

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.

Homework Equations

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

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

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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

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

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?

 

[mfb]

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

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.

 

[alpuy]

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?

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

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.

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

 

[mfb]

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

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."