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$$ f(x) = \sin x,$$ where $f$ is defined as $$ f: \mathbb N \to \mathbb R.$$

The exercise is to check if the function was one-one or many-one.

Here is my answer: $$ f(x_1) = f(x_2) $$ $$ \sin x_1 = \sin x_2 $$ $$ x_1 = n\pi + (-1)^nx_2 $$ here we have 2 cases

Case 1: If $n = 0,$ then we will get $$ x_1 = x_2.$$ Case 2: if $x \neq0,$ But this case can be rejected since $x_1$ can't be irrational as it a natural number. This just left us with one case,that is $ x_1 = x_2$ which means that the function is one-one.

Here is my question:

We got $f(x)$ as one-one, but confusingly we can get same value of $y$ for multiple values of $x,$ e.g., both $\sin30$ and $\sin150$ give the same $y=1/2.$ How is this possible?

Gary
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KHAJA OWAIS ALI
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    You need to count the angles in radians for this problem. In degrees, the statement (as you have found out) is not true. –  Oct 04 '21 at 12:47
  • if i take them in degrees how come it is not true, please lemma know – KHAJA OWAIS ALI Oct 04 '21 at 13:20
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    You have already found out that it is not true = $\sin 30^\circ=\sin 150^\circ$. This cannot happen if the angles are in radians, because either the sum or the difference of $x_1, x_2$ will be a multiple of $\pi$ radians (which will be zero only if $x_1=x_2$). –  Oct 04 '21 at 14:31

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We got $f(x)$ as one-one, but confusingly we can get same value of $y$ for multiple values of $x,$ e.g., both $\sin30$ and $\sin150$ give the same $y=0.5.$ How is this possible?

Note that the given sine function $f$ is a restriction of the natural (radian) sine function $\sin_r,$ which is different from the ‘degrees’ sine function $\sin_d:$ $$\sin_d(30^\circ)=\sin_r\left(\frac\pi6\right)=0.5\neq0.988=\sin_d(1718.87^{\circ})=\sin_r(30)=f(30).$$

So, while $\sin_d30^\circ=0.5=\sin_d150^\circ,$ $$f(30)=\sin_r(30)=0.988\neq-0.715=\sin_r(150)=f(150).$$

ryang
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  • if sin_r is restricted then won't it be many-one – KHAJA OWAIS ALI Oct 05 '21 at 08:57
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    @khaja No, it became 1-1 upon being restricted. Aren’t you convinced by your own proof? (It’s a decent proof, even if Case 2’s presentation benefits from more clarity.) – ryang Oct 05 '21 at 09:12