Understanding surjectivity proof of $f(n)=2^n$.

Question

Working on the book: Richard Hammack. "Book of Proof" (p. 252)

3. Let $B=\{2^n:n \in \mathbb{Z}\}$. Show that the function $f: \mathbb{Z}\to B$, defined as $f(n) = 2^n$ is bijective. Then find $f^{-1}$.

The author proves surjectivity:

The function $f$ is surjective as follows. Suppose $b \in \mathbb{B}$. By definition of $B$ this means $b = 2^n$ for some $n \in \mathbb{Z}$. Then $f(n) = 2^n = b$

Perhaps I'm missing something, but I think this does not proves surjectivity. Instead, would be neccesary to take an arbitrary element $b \in B$, and show there exists an element $a \in \mathbb{Z}$ such that $f(a) = b$. In this case, letting $a = \log_2(b)$, we see $$ f(a)=f(\log_2(b))=2^{\log_2(b)}=b $$

Is my observation correct ?

Answer 1

Perhaps I'm missing something, but I think this does not proves surjectivity. Instead, would be neccesary to take an arbitrary element b∈B, and show there exists an element a∈Z such that f(a)=b.

That is exactly what the author did do!

S/he picked $b$ to be an arbitrary element of $B$. By the definition of $B$ all the elements of $B$ are of the form $2^n$ for some $n\in \mathbb Z$.

So s/he just let $a$ be that $n$. She just didn't want to spend the $25$ cents to buy a second variable. (If she asked me, I know a place where she can get variables wholesale.....)

In this case, letting a=log2(b),

That's one way to get the $a$.... but you have to then prove that $\log_2 b\in \mathbb Z$[1].

But another way to get that $a$ is to say $b = 2^n$ for some $n$, so let $a = n$.

That way you can save on the cost of the function turning handcranks. Variables are cheap but the function turning handcranks are mucho bucks if you don't need them.

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[1] How would you prove $\log_2 b \in \mathbb Z$?

Well, really the only way I see is: ... Let $b \in B$. Then there is an integer $n$ so that $b = 2^n$. So $\log_2 b =\log_2 2^n = n$ which is an integer.

Answer 2

Both proofs are correct. the difference is that you element $ b$ in $ B $, has been written as $2^n$. You can also say :

$ B $ is exactly the range of $ f$, so $ f$ from $\Bbb Z $ to $ B $ is by construction surjective.

on the other hand, $$f(n)=f(m) \implies$$ $$2^n=2^m\implies$$ $$2^{n-m}=1\implies$$ $$n-m=0\implies n=m $$ $ f$ is then injective and bijective and as you wrote $$f^{-1}(b)=\frac{\ln(b)}{\ln(2)}$$