Divisibility is a partial order

The definition of a partial order is given. The relation "a divides b" is shown to be a partial order.

[wiggles]

I searched this up because I was curious about the poset of N with the partial order "a divides b", and what the Hasse diagram looks like if you use N = {0, 1, 2, 3, ...}. The least element should be 1, because if x|1, then it's easy to show x=1 (so if x != 1, x is greater than 1 in this partial order). I suppose you could also show 1|x for all x, using the definition directly. The greatest element should be 0, because for all x, x|0 (seen because for any x in N, x0=0). All immediate successors of 1 would be exactly the prime numbers, and I suppose the nth row of the Hasse diagram (the row with 1 being the 0th row) would have all the numbers that have n prime factors. You would then connect the elements appropriately. Wikipedia has a nice little Hasse diagram of this partial order on {0, 1, 2, 3, ...}, but I think they skipped 14 and went straight to 15 in their second row.

The main motivation here is when you want to find gcd(a, b), "greatest" refers to the partial order "a divides b", not the standard total order on {0, 1, 2, 3, ...}, so that gcd(0, a)=a, and gcd(0, 0)=0.

[OvenBakedCookie]

Thank you for the video! It helped me a lot preparing for my Software theory exam!