- Nov 12, 2017
"Multiple" in the phrase "common multiple" means "[positive] integer multiple". What are you claiming?it can just be a common multiple not a "integer multiple"
"Least" in LCM means minimum! So of course they are!What is the minimality they are saying ?
They are implying that LCM is the minimum no . something like that?
ab/GCD is a multiple of both a and b, so it is a common multiple, so you can conclude that it is greater than or equal to the least common multiple. This is one half of proving that it is the LCM.And ab/GCD gives a no which should be at least equal or greater than LCM.
They cannot directly say it can be equal to LCM as they are trying to prove that only.
You could do it that way; they said it differently. So? The conclusion is the same, that the lcm divides ab.I know that as LCM(a,b) and product of a and b are both Common multiples of a and b so I should get r =0 upon dividing ab/LCM as a*b is also a multiple of lcm(a,b) .
That's not useful to say. "Multiple" is taken to mean "multiple by a positive integer"; "LCM" means least positive multiple, not least integer multiple.And as r=0 (0 is a mutliple of every integr)so I know that r is a common multiple of a and b.
But here one can say that then LCM must be 0 as it is the least.
They say this:
What this means is because r = ab-q lcm(a,b), the difference of two integer multiples of both a and b, it must also be an integer multiple of a and b. But since it is less than the least common multiple, it must be zero. (But that doesn't make it the lcm.)By division, we can write
ab = q lcm(a,b)+r where 0≤r<lcm(a,b)Because ab and lcm(a,b) are common multiples of a and b, so is r. By the minimality of the lcm, r=0. Therefore, lcm(a,b) divides ab.
Consequently, ab = q lcm(a,b), which means the lcm divides ab.
Yes, this could also be derived by the fact that the lcm divides every common multiple, and ab is a common multiple; but he is not assuming that to be known.