1
$\begingroup$

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

$\endgroup$
4

1 Answer 1

Reset to default
0
$\begingroup$

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 $m$.

Using this principle, it is pretty straightforward to interpret greatest common divisor and least common multiple in terms of prime factorizations.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .