Can Natural Numbers a, b, c with a Dividing bc Imply a Divides c?

AI Thread Summary
The discussion focuses on proving two mathematical statements involving natural numbers a, b, and c. First, it asserts that if a divides the product bc and a and b are coprime, then a must also divide c. The reasoning involves manipulating the equations and considering the properties of coprime numbers. The second part aims to prove that the least common multiple (lcm) of a and b is given by the formula ab/gcd(a,b), using the relationship between the greatest common divisor and the numbers themselves. The conversation highlights the complexities of these proofs and the importance of understanding coprimality in the context of divisibility.
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Homework Statement


1. If a,b and c are natural numbers and a, b are coprime and a divides bc then prove that a divides c
2. Prove that the lcm of a,b is ab / gcd(a,b)

Homework Equations


if a is a divisor of b then a = mb for a natural number m
if a prime p is a divisor of ab then p is a divisor of a or a divisor of b

The Attempt at a Solution



1.since a is a divisor of bc so am = bc (m is a natural number)
so a = (c)(b/m)
so a/b = c/m
Ok since a,b are coprime so a/b = a number that is not natural
since a/b = c/m so c/m = a number that is not natural so c,m are coprime
back to a = (c)(b/m)
since a = (c)(b/m) which is a natural number, so bc must be a multiplie of m
since c isn't a multiplie of m, b must be so
so b is coprime with m
now a/c = b/m
since b is coprime with m
a is coprime with c
Q.E.D
(Wanna check if my approach is correct or not)

2.Prove that the lcm of a,b is ab/gcd(a,b)
let a = xm , b = ym (m = gcd(a,b))

ab/gcd(a,b) = xmym/m = xmy
It is divisible by a and b so it satisfies being a multiplie
here I gave up.
 
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nmego12345 said:
c/m = a number that is not natural so c,m are coprime
That does not follow. 6/4 is not an integer, but 6 and 4 are not coprime.
Consider some prime divisor of a.
 
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