Prove if c divides a and c divides b, then c divides any combination an+bm, where n and m are integers.

Since c divides both a and b then there must exist integers j and t such that a=cj and b=ct.

So by substitution into an+bm we get

an+bm = (cj)n+(ct)m = c (jn + tm). So c divides (an + bm) ~~~~

Thus, if c divides a and b, then c also divides any multiple of a and b.

Subscribe to:
Post Comments (Atom)

## No comments:

Post a Comment