(5) was intended to assume that n∈R≥1, or possibly ∈R≥0 if you want an easy way to prove (6). In that case, how does (8) not follow from (5)? (If f(xy)=yf(x) in general, then f(bn)=nf(b) and f(b)=1⟹f(bn)=n, unless I'm missing something.)
by Nate Soares Jun 29 2016
(5) was intended to assume that n∈R≥1, or possibly ∈R≥0 if you want an easy way to prove (6). In that case, how does (8) not follow from (5)? (If f(xy)=yf(x) in general, then f(bn)=nf(b) and f(b)=1⟹f(bn)=n, unless I'm missing something.)