Abstract Algebra: Proof with GCD and Divisibility

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Prove if d = (a,b), a = a0d, and b = b0d, then 1 = (a,b).

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PSA: Take care not to be misled by the statement (around 4:28) that d being equal to a linear combination of a and b "is the same thing as saying" that d is the gcd. It's not! For example, 4*6 + 3*2 = 30 is a linear combination of 4 and 3, but 30 clearly is not the gcd of 4 and 3. We know only that *if* d = gcd(a, b), *then* there exists a linear combination of a and b that equals d. Not every linear combination of a and b equals the gcd. The only time the converse holds is in the case that d = 1. That is, if there is a linear combination of a and b that equals 1, then gcd(a, b) = 1. This is because Bezout's Lemma says that gcd(a, b) is a *divisor* of any linear combination of a and b, and there is only one positive divisor of 1, namely 1.

whomayshebe

a linear combinations must be strictly greater than zero. Hence the minimum possible value is 1. Also, it's a theorem that gcd is a minimum of all linear combinations of some "a" and "b".

muhammedabdullvek

Beth:

Many thanks for your abstract algebra videos. You are a great teacher, and I really like your step-by-step explanations. Could you do some videos on isomorphic groups and quotient groups as these topics are quite challenging for many students of abstract algebra?

Thank you in advance and I look forward to viewing more abstract-alegbra videos from your channel. Have a great day.

> Benny, 6-16-2015

Love_Hope_from_Above

Can someone pls explain me the reason behind the above corollary

mrscooby

If gcd(a, b)=1, how to show gcd(a^2, b^2)=1, following your procedure?

Songvbm

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