$y\mapsto 10^y$ Bijection between integers and powers of 10 Say $y$ and $x \in \mathbb N$ such that $10^y = x$ MSE research gives me this: $y\mapsto 10^y$ For some of you to see the bijection between the integers and powers of 10 will be obvious, but I really can't find a way to show it formally. I'm aware this is related to other questions on MSE but I'm pretty desperate. The only thing I can think of is that every power of ten maps to a decimal...or something :/ Normally this is the part where I would insert what I have tried - but after hours of trawling the internet I'm afraid I've learned very little. _Looking for a bijection (or just injection?) to show it is countably infinite._ Thanks a lot guys.

You'll need to prove that the function is _injective_.

Now, "the powers of 10" needs to be defined. The most natural definition of that phrase is actually that "powers of 10" means _the range of the function you have just defined_.

Then you just need to appeal to the general fact that an injective function is a bijection onto its range.

* * *

Why is it injective? That depends on what definition you have for $10^y$, but if we assume a recursive definition along the lines of $$ 10^0 = 1 \\\ 10^{n+1} = 10^n\cdot 10 $$

then you can prove by induction on $n$ that $10^n\ge 1$ and then by induction on $k$ that $10^n < 10^{n+k+1}$.

In particular, if you have $a\
e b$ then either $a=b+k+1$ for some $k$ or $b=a+k+1$ for some $k$, and the second induction now shows that $10^a\
e 10^b$.