# Prove that there are prime numbers such as $p_{i}$ such that $A = p_{1}^{a_{1}}*p_{2}^{a_{2}} * ... * p_{k}^{a_{k}} , a_{i}> 0$ [duplicate]

## A and B are natural numbers.

Prove that there are prime numbers such as $$p_{i}$$ such that $$A = p_{1}^{a_{1}}*p_{2}^{a_{2}} * ... * p_{k}^{a_{k}} , a_{i}> 0$$

Then we consider $$B = p_{1}^{b_{1}}*p_{2}^{b_{2}}*...*p_{k}^{b_{k}}$$ , prove that $$(a,b) = p_{1}^{c_{1}}*...*p_{k}^{c_{k}} --- c_{i} = min (a_{i} , b_{i})$$ , and $$[a,b]=p_{1}^{d_{1}}*...*p_{k}^{d_{k}}--- d_{i}= max(a_{i},b_{i})$$ . for all : 1<= i <= k

notice: The first part of the proof originates directly from the Fundamental Theorem of Arithmetic, but I have no idea for the second part of the prove, thanks for your help

• Search: unique factorization, gcd, and lcm.
– user672528
Commented Jan 8, 2021 at 0:56
• Does this answer your question? how to prove the existence and uniqueness of the prime factorization of any natural number.
– user672528
Commented Jan 8, 2021 at 0:59
• Does this answer your question? Induction hypothesis misunderstanding and the fundamental theorem of arithmetic. Commented Jan 8, 2021 at 1:22
• The proof that you are looking for is not short. Further, it is a standard result in number theory, involving such intermediate results as that any common divisor $r$ of $(a,b)$ is also a divisor of the gcd$(a,b)$. In my opinion, the way to nail down the validity of your assertion is by finding a number theory book and starting on page 1. Commented Jan 8, 2021 at 2:51

For each prime $$p$$, you can talk about the multiplicity with which the prime occurs in a given natural number. For example $$8$$ has multiplicity $$3$$ at the prime $$2$$, and it has multiplicity $$0$$ at every other prime.
It's not difficult to see that a natural number $$n$$ is divisible by another natural number $$m$$ if and only if at each prime $$p$$, the multiplicity of $$p$$ in $$n$$ is greater than or equal to the multiplicity of $$p$$ in $$m$$.