Question

I have a proof of a theorem for abstract algebra, but I want to ensure that it is valid. The theorem states that gcd(a,bc)=1 iff gcd(a,b)=1 and gcd(a,c)=1. My proof goes as follows:
Assume that gcd(a, bc)=1 for some integers a, b, and c. Then, there must exist integers s and t such that as+bct =1. Since 1 can be written as a linear combination of a and b, gcd(a,b)=1. 1 can also be written as a linear combination of a and c, so gcd(a,c)=1.

Now assume gcd(a,b)=1 and gcd(a,c)=1. There must exist integers s, t, u, and v such that 1=as+bt and 1=au+cv. Multiplying both equations, we have 1*1 = (as+bt)(au+cv). From this, we have 1=a^2su+acv+btau+btcv. Factoring out an a from the first three terms, we have that 1=a(asu+cv+btu)+bc(tv). Since 1 has been written has a linear combination of a and bc, we have gcd(a,bc)=1. QED.

Answer 1

post it, I tag someone, when they came online, they can check
@oldrin.bataku @pgpilot326 @Wallach

Answer 2

how can you say that is a linear combination, isnt it now dependant on a and b

Answer 3

no, dan, it's linear. because gcd(a,bc) =1

Answer 4

\(asu+cv+btu=m\in \mathbb{Z}\text{ and } tv=n\in\mathbb{Z}\)

proof looks good to me

Answer 5

why medal me? i did'n do the proof.

did'n... ha ha. i mean didn't.

Answer 6

aw

Answer 7

give it to him

Answer 8

oops, didn't mean to hit undo

Answer 9

por que?

Answer 10

i mean why the medals?

Answer 11

give em to @Loser66 it's his proof.

Answer 12

hahahah, not me, I didn't start abstract algebra yet, next week, not now

Answer 13

i haven't taken it either...

Answer 14

if gcd (a,bc)=1
if gcd(a,c)=1
and gcd (a,b)=1
then (a)+(bc) = n(a1+b1c2) suppose n not equal to 1
then
gcd(a,c) .. a+c = n(a1+bc2), therefore gcd(a,c) not equal =1
similarly gcd(a,b)
proof by contradiction , therefore
if gcd(a,b)=1 and gcd(a,c)=1, then gcd(a,bc)=1

Answer 15

where a1,b1,and c2 E Z