Results 1 to 3 of 3

- May 25th 2011, 04:48 PM #1

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 8

- May 25th 2011, 06:28 PM #2

- Joined
- Mar 2010
- Posts
- 715
- Thanks
- 2

- May 31st 2011, 03:30 PM #3

- Joined
- May 2011
- From
- Florida
- Posts
- 17

## This was not quite clear to me, so I found a way that might be easier

* c|b notation reads c divides b

Let (b,a) = c and (a,r) = d

c|b and c|a, then c|b-aq

r = b - aq, so c|r

c (less than or equal to) d

d|a and d|r, then d|aq + r

b = aq + r, so d|b

d (less than or equal to) c

Therefore d=c