If d is a divisor of b, then it would have to be a divisor of any multiple of b. And bq is a multiple of b. For example, if 6 divides 12 (letting d = 6 and b = 12), then 6 divides 12(3) (letting q = 3). More generally, once you know 6 "goes into" 12, you know that 6 "goes into" any multiple of 12. Once you know d "goes into" b, you know that d "goes into" b times any other whole number, so it "goes into" b times q.
Anytime you have d "dividing" a number (i.e. d divides b), then it divides a multiple of that number (so d divides bq). For example, if 6 divides 12, then 6 divides 24, and 36, and 48, etc. So if d|b, then d|bq. Furthermore, if d divides two different numbers, a and bq, then it divides their sum or difference, since if it's a factor of both, you can "factor it out" of the expression. So, if d|a and d|b, then d also divides bq, and therefore it divides a - bq.
By the way, the associative property comes as a byproduct of the other desired properties in this example presented here. A question in my own mind, is why not more generally study loops and not groups? What is so dear to the soul of having an associative table?
you just something I've used 20 minutes struggling to even begin to understand, and explained it in such a simple way in like 5 minutes. Wish more math teachers were like you
At 5:30 when we have our 2 possible groups, A and B, don't we need to check that they are associative? My memory is not exact, but I used this Latin square method of finding all the groups of order 5, and I believe I ended up with a Latin square describing an operation that was not associative, and so the Latin square actually did not describe a group operation.
Thank you so much. I like how you layout the structure of the proof before we began and then kept mentioning that structure throughout. That really helped to connect everything together.
4:31 actually the theorem was proved by using the Euclidean Algorithm, while Euclidean Algorithm is proved using strong induction over the variable a here
You explained it so beautifully. Just loved it. I watched other solution videos for this problem but i just understood the right logic after watching your video. Thank you....
