the / represents divides or is a divisor of not divide by
2006-09-06 06:35:26 · 5 answers · asked by thamal3 1 in Science & Mathematics ➔ Mathematics
Associativity of multiplication ac / bd = (a/b) * (c/d)
2006-09-06 06:51:33 · answer #1 · answered by Joseph Binette 3 · 0⤊ 0⤋
If you know that a/b and c/d, then you can rewrite a = jb and c = kd, where j and k are some other integers. Our goal is to see if there is some integer q such that ac = q(bd).
a = jb
c= kd
ac = jbkd [then rearrange]
ac=(jk)bd
So ac is the product of bd and some integer jk. Therefore, ac/bd, which was to be demonstrated.
Enjoy!
2006-09-06 13:47:53 · answer #2 · answered by Polymath 5 · 0⤊ 0⤋
Given: a|b, c|d.
By definitions there are integers m, n such that
b = m a
d = n c
Therefore,
b d = (m a) (n c) = (m n) (a c)
which proves that ac | bd.
2006-09-06 13:56:50 · answer #3 · answered by dutch_prof 4 · 0⤊ 0⤋
Polymath has given you a good answer. However, he should have added: q = jk is the integer you seek.
2006-09-06 19:04:08 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I don't understand what it is we are supposed to prove. Is it that a is EVENLY divisible by b, c is EVENLY divisible by d, then ac is EVENLY divisible by bd?
2006-09-06 13:45:10 · answer #5 · answered by Will 6 · 0⤊ 0⤋