Question

help! It said the Fundamental Theorem of Arithmetic is used to prove. Let m = p1^e1 * p2^e2 ... ps^es, where pi is a prime. m|n if and only if pi^ei | n for all i.

Answer 1

all numbers can be written as a product of primes

Answer 2

primes have factors of 1 and themself

Answer 3

now this means that if m divide n
then every prime^exponent divides n

Answer 4

does it make sense?

Answer 5

maybe u need to say this statement too
if m|n
then n=k*m
there fore
k*(p1^e1*p2^e2....)

Answer 6

ok, m | n means mk = n for some integer k

p1^e1 ( p2^e2 ... ps^es * k) = n implies p1^e1 | n
p2^e2 * (p1^e2 * p3^e3 ... ps^es k) = n implies p2^e2 | n
and so on to ps^es.


How do you prove the other direction?

Answer 7

what do u mean

Answer 8

it's an if and only if statement

Answer 9

okay since u saw that
n=k*m
then
n=k*p1^e1*p2^e2...
therefore
p1^e1,p2^e2... all have to be factors

Answer 10

and u say primes cannot be decomposed into other primes so u are done

Answer 11

there is no other prime representation for m, so it goes both ways

Answer 12

well, we just proved that direction. The other direction is

if pi^ei | n for all i, then m|n

Answer 13

hmm to me its the same thing lol

Answer 14

ok how about saying it like this

Answer 15

if pi^ei | n for all i then
(p1^e1)(p2^e2)(p3^e3).....(pn^en) | n
so m|n

Answer 16

because if* p1|n and p2|n then p1*p2|n if p1 and p2 are prime

Answer 17

this has to be true as a 2 different primes cannot share factors

Answer 18

so a lemma was used a long the way
Given p1 and p2 are primes. If p1| and p2|n, then p1*p2 | n

Answer 19

I think m * gcd(k1, k2, ... ks) = n will prove the result.