So, yesterday at work a student asked me why you can't distribute constants over parentheses when you have exponents (e.g. why doesn't a(x + y)2 = (ax + ay)2) and it got me thinking. Since a(x + y) = ax + ay, it's very tempting to just distribute blindly. So, what do you do if there are exponents involved? I couldn't find anything in the texts that we had there, so I came up with this; let me know what you think:
Let a, x, y ∈ ℝ, n ∈ ℕ. Then a(x + y)n = (a1/nx + a1/ny)n.
By the Binomial Theorem, (a1/nx + a1/ny)n = Σ nk=0 (nk) (a1/nx)n - k (a1/ny)k. Rewriting,
Σ nk=0 (nk) a1/n(n - k)xn - k a1/n(k)yk
= Σ nk=0 (nk) a(1 - k/n)xn - k ak/nyk
= Σ nk=0 (nk) a1 - k/n + k/nxn - kyk
= Σ nk=0 (nk) axn - kyk
= a Σ nk=0 (nk) xn - kyk
= a(x + y)n
So, a(x + y)n = (a1/nx + a1/ny)n □
Listening to: Tank!