Is $|\sin(n)|\leq1$ or $|\sin(n)|<1$ for integer $n$? $\pi$ is irrational, therefore there exist no finite integers $m,n$ such that $n=(m+\frac{1}{2})\pi$, therefore there is no $\sin(n)=\pm1$. So if n defined to be a finite integer, I am comfortable saying $|\sin(n)|<1$. But what if $m,n$ are only specified to be integers? Can integers be infinite? Then could $n=(m+\frac{1}{2})\pi$ and $\sin(n)=\pm1$?
I realize that the answer lies in the subtleties of the precise definition of irrational.
 A: An integer or a real number cannot be infinite. There are no integers $n, m$ such that $n = \left(m+\frac 12\right)\pi$. Hence we are safe concluding
$$\left|\,\sin n \,\right| < 1 \quad\text{ for all integers } n$$
A: No, nothing subtle about it. An integer is finite. There are no integers $n$ and $m$ with the property you mention, and yes it follows that $|\sin(n)|<1$ for every integer $n$. Exactly as you said.
A: Assume that
$$|\sin(n)| \not \lt 1$$
Where $n$ is an integer. Then,
$$\sin(n)=1$$ or $$\sin(n)=-1$$
Must be true for some $n$. Solving for $n$, we have,
$$n=\pm \left(m+{1 \over 2} \right) \cdot {{\pi} \over 2}$$
Where $m$ is an integer. If we isolate the equation for ${{\pi} \over 2}$, we get,
$$\pm {{\pi} \over 2}={{n} \over {m+1/2}}$$
However, $\pi$ is irrational, so this equation can never be true for any combination of $m$ and $n$. Therefore, we have a contradiction. If the assumption is false, then,
$$|\sin(n)| \lt 1$$
Must be true.
A: If there exist an $n$ such that $|\sin(n)|=1$, then $\pi$ would be rational, which is a contradiction. Therefore $|\sin(n)|<1
