0

I am trying to prove that the sets

A={functions of the form $x\rightarrow16^{ax}$}

B={functions of the form $x\rightarrow2^{ax}$}

for a in the rational numbers

are equal. I know that I must prove that A is a subset of B and that B is a subset of A. I am having the most trouble proving that B is a subset of A.

So far I have

For all $x\in A$ there exists an $x\in Q$ such that $16^{ar} = 2^{as}$ for all r,s such that r and s are inputs of the functions $f(x)=16^{ax}$ and $f(x)=2^{ax}$ respectively. Thus every element of B is also an element of A.

But is that really proof?

Isa.S
  • 5

1 Answers1

0

You have the basic idea, here is the way to make this formal.

Let $f \in B$. Then, $\exists a \in \mathbb{Q}$ such that $f(x) = 2^{ax}$. Let $b = a/4$. Note that since $a \in \mathbb{Q}$, we also have $b = a/4 \in \mathbb{Q}$. Thus, $$ f(x) = 2^{ax} = 2^{4 (a/4) x} = \left(2^4\right)^{bx} = 16^{bx} \in A, $$ since $b \in \mathbb{Q}$.

Consequently, whenever $f \in B$, we also have $f \in A$, thus $B \subseteq A$.

Can you argue the reverse direction?

gt6989b
  • 54,422