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 $a\ne b$ then either $a=b+k+1$ for some $k$ or $b=a+k+1$ for some $k$, and the second induction now shows that $10^a\ne 10^b$.