Question

A few number theory questions about:

If a | c and b | c with GCD (a,b) = 1 -> ab | c

Prove.

Answer 1

if a and b have nothing in common they are relatively prime; and since they both divide into c; they are both products of c

Answer 2

product is prolly a bad choice of words :)

Answer 3

4|20

5|20

4.5|20

Answer 4

You like this sort of question, no Wolfram an' all?

Answer 5

wolfram is for validating my own stupidity :) there aint no signs to drop in this lol

Answer 6

:-)

Answer 7

hey estudier are you sill there?

Answer 8

a|c => ak=c for some integer k
b|c => bi=c for some integer i
gcd(a,b)=1 => ax+by=1
we want to show ab|c => abn=c for some integer n
Since ak=c and bi=c => ak=bi => ak-bi=0
=>ak-bi+1=1=> ak-bi+(ax+by)=1
=> a(k+x)+b(y-i)=1, k+x is an integer and y-i is an integer
=> ac(k+x)+bc(y-i)=c
=> gcd(ac,bc)=c
=>ab|c

Answer 9

i guess we didn't need all this
we have ax+by=1
acx+bcy=c =>gcd(ac,bc)=c => ab|c