SOLUTION: Supply the proof for the following theorem. Theorem: For integers a, b, and c, if a | b and b | c, then a | c. [My “idea”: b = a·r, c = b·s, c = (a·r)·s = a·(r·s). Don't

Algebra ->  Geometry-proofs -> SOLUTION: Supply the proof for the following theorem. Theorem: For integers a, b, and c, if a | b and b | c, then a | c. [My “idea”: b = a·r, c = b·s, c = (a·r)·s = a·(r·s). Don't       Log On


   

Question 440607: Supply the proof for the following theorem.
Theorem: For integers a, b, and c, if a | b and b | c, then a | c.
[My “idea”: b = a·r, c = b·s, c = (a·r)·s = a·(r·s). Don't know the rest.]
Outline: Because a | b and b | c, there exist integers r and s such that
b = a·r and c = b·s. Then you must explain how these two equations can be combined to give c = (a·r)·s (hint: “substitute a·r for b …”). Use the associative property of multiplication to obtain c = a·(r·s). Now c equals a times an integer (why?). Finish the proof as above.
Proof:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
From this line:

[My “idea”: b = a·r, c = b·s, c = (a·r)·s = a·(r·s). Don't know the rest.]

You've essentially proven the statement. a|c if and only if c = a*k, where k is some integer. Since c = a*(r*s), and r*s is an integer, then c is equal to a times some integer, a|c, QED.