MHB Define $f: \mathbb{Z} \to \mathbb{Z}: f^{-1}(\left\{0,1,2\right\})$

[Guest2]

Define $f: \mathbb{Z} \to \mathbb{Z}$ via $f(n) = n^2$ for all $n \in \mathbb{Z}$. Why does $f^{-1}(\left\{0,1,2\right\}) = \left\{0,-1,1\right\}$? The definition I'm using is $f^{-1}{(T)} = \left\{a \in A: f(a) \in T \right\}$ so we have $f^{-1}({ \left\{0,1,2\right\} }) = \left\{n \in \mathbb{Z}: n^2 \in \left\{0,1,2\right\} \right\}$. So I could think about this as all the integers whose squares is the set $ \left\{0,1,2\right\}$ but nothing I square would give me $2$.

 

[I like Serena]

Define $f: \mathbb{Z} \to \mathbb{Z}$ via $f(n) = n^2$ for all $n \in \mathbb{Z}$. Why does $f^{-1}(\left\{0,1,2\right\}) = \left\{0,-1,1\right\}$? The definition I'm using is $f^{-1}{(T)} = \left\{a \in A: f(a) \in T \right\}$ so we have $f^{-1}({ \left\{0,1,2\right\} }) = \left\{n \in \mathbb{Z}: n^2 \in \left\{0,1,2\right\} \right\}$. So I could think about this as all the integers whose squares is the set $ \left\{0,1,2\right\}$ but nothing I square would give me $2$.

Hi Guest,

Indeed, no integer squares to $2$, which is why there is no inverse of $2$.
It's not the set of integers whose squares is the set $ \left\{0,1,2\right\}$, but the set of integers that have squares in the set $ \left\{0,1,2\right\}$.