I have a quick question about (https://math.stackexchange.com/q/921109)'s answer. I asked on there but since the question is old, I feel like I might not get a reply. Anyway, my question was: Why is $\operatorname{lcm}(a,b)\leq drs$ like they said?
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6$drs$ is a common multiple of $a$ and $b$, right ? – Suzet Jun 13 '18 at 00:37
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2There are many common multiples of $a$ and $b$. For example, the common multiples of $6$ and $4$ are ${12,24,36,48,60,72,\dots}$. The least common multiple is a common multiple, but more specifically it is the common multiple which is smaller than any other common multiple. – JMoravitz Jun 13 '18 at 00:44
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2!!!!!!!!!! I really don't understand why out of all abstract algebra, I can never feel like I truly grasp LCM and GCD....thank you so much for the reminder...such an obvious reminder it is haha I wish I could give you both many many upvotes!! – Jun 13 '18 at 00:56
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In this statement it is assumed that gcd(a,b) = d, and d > 1. This means that the greatest common divisor between a and b is greater than 1, therefore they are not relatively prime (they share a factor, namely d).
Since we know that d divides both a and b, we can re-write a and b as
a = d⋅r, b = d⋅s for some integers r and s.
Considering our statement earlier, that a and b share a common factor d, we know that the least common multiple [lcm(a,b)] of a and b is some multiple of d, because both factors share a multiple d. The worst case scenario for the lcm(a,b) is going to be when lcm(a,b) = d⋅r⋅s because this will always be a factor of our two numbers a and b which share a common multiple.
Since we are talking about the LEAST common multiple, there can also be other common factors between them, but only if they are LESS than d⋅r⋅s. Hence the ≤.
We also know that since our lcm(a,b) is at most d⋅r⋅s, and d > 1, that d⋅(d⋅r⋅s) should surely be greater than d⋅r⋅s. Therefore lcm(a,b) ≤ d⋅r⋅s < d⋅(d⋅r⋅s) which we know is equal to a⋅b
Samuel Schmidgall
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