<p>I had to look up the relation “|” and some basic facts. I hope this isn’t cheating…</p>
<p>Let a,b,c ∈ Z and m ∈ R and a,m≠0:
- if a|b, then a|(bc)
** if a|b, then (am)|(bm)</p>
<p>Assume 8|16.
By <em>, 8|32 (for c = 2).
By *</em>, 4|16 (for m = 1/2) which is given. ∎</p>
<p>Maybe I should try to prove <em>, and *</em>.
proof of <em>:
a|b iff b/a = k1 for some integer k1 (by definition, really). Multiply both sides by integer c. (bc)/a = k2 for some integer k2 (closure of multiplication among integers). Hence a|(bc). ∎
proof of *</em>:
again, a|b iff b/a = k. Multiply both sides by (m/m). We get (bm)/(am) = k(m/m). (m/m) = 1 (multiplicative inverse). So (bm)/(am) = k for some integer k (multiplicative identity). Hence (am)|(bm). ∎</p>