How to write that the only integers that have a multiplicative inverse are $1$ and $-1$, in symbolic form

Question

Could it be something like for $(1,-1) \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$, such that $xy = 1$?

Answer 1

The units in $\mathbb Z$ are $1,-1$. (Units are just invertable elements in a group).

More formally from wikipedia, a unit of a ring $\displaystyle R$R is any element $\displaystyle u\in R$$\displaystyle u\in R$ that has a multiplicative inverse in $\displaystyle R$R: an element $\displaystyle v\in R$ $\displaystyle v\in R$ such that

$\displaystyle vu=uv=1$$\displaystyle vu=uv=1$, where 1 is the multiplicative identity.

So using this definition, it is quite obvious that the units in $\mathbb Z$ are $1,-1$. (I hope this helped? I am known not to explain things the best).

Answer 2

I have seen $U(R)$ for the group of units in a ring R (I mean it is a group iff the ring contains a unit, so it is not a group if your ring is $2\mathbb{Z}$). I think I have also seen $R^{\times}$. I think the latter is probably the more common one.

Answer 3

For any (unital) ring $R$, the set of units $R^\times$ is the group consisting of $$ \{ x \in R \mid \exists\, y \in R \text{ s.t. } xy = 1 \}. $$

Your (true) claim that the only units in the integers could be written: $$ \{ x \in \mathbb{Z} \mid \exists\, y \in \mathbb{Z} \text{ s.t. } xy = 1 \} = \{ -1, +1 \} $$