Multiplying any integer with any prime

[ull]

PS. I don't speak or write english very well. I´m doing my best. Is it still ok to post questions?

a = any integer
b = any prime number

a * b = c

Is there a proof that "c" isn't made up by any other prime factors than the prime factors that make up "a" (except for "b")?

 

[Mandelbroth]

PS. I don't speak or write english very well. I´m doing my best. Is it still ok to post questions?

a = any integer
b = any prime number

a * b = c

Is there a proof that "c" isn't made up by any other prime factors than the prime factors that make up "a" (except for "b")?

It's a result of the fundamental theorem of arithmetic.

 

[MathJakob]

Sinse a prime number can only be divided by 1 and itself, no two numbers can ever equal a prime number. I don't know what the proof is called though. I think the OP wants to know the name of the actual law/proof.

 

[Number Nine]

Sinse a prime number can only be divided by 1 and itself, no two numbers can ever equal a prime number. I don't know what the proof is called though. I think the OP wants to know the name of the actual law/proof.

As Mandelbroth said, the proof is the fundamental theorem of arithmetic, from which the OP's desired result follows immediately.

 

[CellCoree]

Yes, there is a proof for this statement. It is known as the Fundamental Theorem of Arithmetic, which states that every integer greater than 1 can be uniquely expressed as a product of prime numbers. This means that when we multiply any integer with any prime number, the resulting product can only be made up of the prime factors that make up the original integer, along with the prime number being multiplied. This is because all other possible combinations of prime factors would result in a different integer, violating the uniqueness of the prime factorization. Therefore, it is safe to say that the product "c" is not made up of any other prime factors than the ones that make up "a" (except for "b").