What members of $\Bbb{Z}_2^{\times}$ don't have a multiplicative inverse?
Is it just $0$?
My motivation is to better understand the multiplicative group.
Do all positive integers i.e. $1,5,7,9$?
What about negative ones such as $-5, -17, -25,-37$?
And fractions such as $\frac{125}{47}, \frac{211}{47}$
I ask not really knowing where to start I'm afraid. Any answer should be pitched LOW.
The ring $\mathbb{Z}_2$ is the closed unit ball (aka the completion of $\mathbb{Z}$) sitting inside $\mathbb{Q}_2$, or $$\mathbb{Z}_2=\{x\in\mathbb{Q}_2:|x|_2\leq 1\}.$$ Then $x\in\mathbb{Z}_2$ has an inverse $x^{-1}\in\mathbb{Z}_2$ if and only if both $|x|_2\leq 1$ and $|x|_2^{-1}=|x^{-1}|_2\leq 1$, which is the case if and only if $|x|_2=1$. Therefore $$\mathbb{Z}_2^\times=\{x\in\mathbb{Q}_2:|x|_2=1\}=\mathbb{Z}_2\setminus\{x\in\mathbb{Q}_2:|x|_2<1\}.$$ But the open unit ball we are removing above is exactly $$2\mathbb{Z}_2=\{2x\in\mathbb{Q}_2:|x|_2\leq 1\}=\{x\in\mathbb{Q}_2:|x|_2\leq 1/2\}=\{x\in\mathbb{Q}_2:|x|_2<1\}$$ because $|\cdot|_2$ can only take values in the set $$2^{\mathbb{Z}}\cup\{0\}=\{0\}\cup\{\dots,1/4,1/2,1,2,4,\dots\}.$$ That is, $\mathbb{Z}_2^\times=\mathbb{Z}_2\setminus 2\mathbb{Z}_2$ is the "unit circle" (aka, the difference between the closed and open unit balls) in $\mathbb{Q}_2$. From this it is not hard to see that $$\mathbb{Q}\cap\mathbb{Z}_2^\times=\{a/b\in\mathbb{Q}:2\nmid a\text{ and }2\nmid b\},$$ which contains all odd integers and fractions thereof (like the ones you mentioned).