What does this syntax mean: "$f^{-1} : N_{10} \Rightarrow N_b $ is the inverse of $f: N N_{b} \Rightarrow N_{10}$?"

Question

I'm trying to solve this but I haven't seen syntax like this before. Can someone please explain the syntax?

https://i.stack.imgur.com/XHkvS.png

The image is

Show that the one-to-one function $f^{-1} : N_{10} \Rightarrow N_b $ is the inverse of $f: N N_{b} \Rightarrow N_{10}$.

Hint: Show that $f^{-1} (f (n_b) ) = n_b$

I'm not sure how to format math on stackexchange either, so if someone could add what's in the png that would be great as well.

This is from a worksheet of things I should know before entering an assembly language class next semester.

Thanks!

Answer 1

When you define a function $f$ in higher mathematics, you first write $$f:A\rightarrow B$$ Here $A$ will denote the set the function is from (the domain) and $B$ will denote the section the function is going to (the codomain).

After this, you (generally) give a formula expressing how the function is defined. For example we could write "$f:\mathbb{N}\rightarrow\mathbb{N}$ defined by $f(n)=n+1$."

In the image you've attached, you've got a function called $f^{-1}$ which goes from whatever $N_b$ is (probably the natural numbers in base $b$) to $N_{10}$ (probably the natural numbers in base $10$). I would imagine that they have given you a definition for $f$ in a previous problem, or in the preceding chapter, or something.

It seems like there is a mistake in the problem - an extra $N$ before $N_b$ in the definition of $f$.

The problem should be very easy. If $f$ is one-to-one, then there is a unique element of $N_{10}$ associated with each element of $N_{b}$ in the image of $f$. (Which is probably all of $N_b$, as if not $f^{-1}$ is not well defined.)

Answer 2

To prove this statement, let's clarify what it means for $f_{a \rightarrow b}: \mathbb{N_{a} } \rightarrow \mathbb{N_{b}}$ (arbitrary integer bases). If $n_a = \overline {a_k a_{k-1} \ldots a_0}_a$ and $0 \leq a_m < a$ (i.e. in base a expansion, $n_a = a_k a^ k + a_{k-1} a^{k-1} + \ldots + a_0$), then $f(n_a) = \overline{b_j b_{j-1} \ldots b_0}_b$, where $a_k a^ k + a_{k-1} a^{k-1} + \ldots + a_0 = b_j b^j + b_{j-1} b^{j-1} + \ldots + b_0 b^0$ and $0\leq b_m < b$

Now, the map $f^{-1} _{a \rightarrow b} f_{a\rightarrow b}(n_a)$ would yield $\overline{c_i c_{i-1} \ldots c_0}_a$, where $b_j b^j + b_{j-1} b^{j-1} + \ldots + b_0 b^0 = c_k a^ k + c_{k-1} a^{k-1} + \ldots + c_0 $ and $0 \leq c_m < a$. To prove the function is actually an inverse, it remains to show that $a_m = c_m$, which is left to you.