Bounds of sin(x) where x is rational Hello all I am starting an analysis course at my university, and we were asked to justify why Sup(S) and Inf(S) is 1 and -1 respectively for the set
$$\ S = \{\sin(x): x \in \mathbb{Q}\} $$
I try to prove by contradiction that there cannot exist an upper bound M that is smaller than 1 however I don't know where to proceed from saying there exists a $\sin(\frac{p}{q}) > M$ that is in S.

Assume $(M)<(1)$ is the UPPER BOUND.
Hence all rational numbers $(r)$ have $\sin(r)<(M)<(1)$
Hence $(r)<\sin^{-1}(M)<\sin^{-1}(1)$
Between two real numbers, we can always get a rational number.
There Exists $P/Q$ where $(r)<\sin^{-1}(M)<(P/Q)<\sin^{-1}(1)$
Hence $\sin(r)<(M)<\sin(P/Q)<(1)$
Contradiction :
$(M)<\sin(P/Q)$
Hence Assumption is not true.
Hence $(M)=(1)$ is the UPPER BOUND.
