Proving set equivalence Let $S =\{[12],[3]^{-1},[13][4]\}$ and $T= \{[6^{16}],[24]+[67],[-158]\}$ be subsets of $\mathbb Z_{17}$. I am trying to prove $S=T$
So far I have $S =\{[12],[6],[1]\}$ since $[a^{-1}]=[b] \iff ab \equiv 1(mod\,x)$ and $T= \{[1],[6],[12]\}$
But I am not sure if this is sufficient. Does anyone know?

While you arrived at the right conclusion, you may want to be a bit more verbose in your presentation.


*

*Since $6 \cdot 3 \equiv 18 \equiv 1 \mod 17$, we have $[3]^{-1} = [6]$.

*Since $13 \cdot 4 \equiv 52 \equiv 1 \mod 17$, we have $[3]\cdot [14] = [1]$.


This yields $S = \{[12],[6],[1]\}$ and we may also note that these elements are pairwise distinct.
Similarly


*

*$6^{16} \equiv (6^2)^8 \equiv 2^8 \equiv (2^4)^2 \equiv (-1)^2 \equiv 1 \mod 17$ yields $[6^{16}] = [1]$.

*$24+67 \equiv 7 + 16 \equiv 23 \equiv 6 \mod 17$ implies $[24]+[67] = [6]$.

*$-158 \equiv -158 + 170 \equiv 12 \mod 17$ implies $[-158]=[12]$ and thus


$T = \{[1],[6],[12]\}$.
Therefore $S = T = \{ [1], [6], [12] \}$.
