# Trouble understanding injection problem

Went over a problem today that $$f:\mathbb{Q}\rightarrow \mathbb{R}$$ defined by $$f(x)=\sin x$$ is injective.Can someone explain to me the reason, and clear up confusion?

Proof:Here was the proof given. Assume $$f(x_1)=f(x_2)$$ $$\implies \sin (x_1) = \sin (x_2)$$ $$\implies x_2=2n\pi +x_1$$ or $$x_2=(2n+1)\pi-x_1$$ for $$n\in \mathbb{Z}$$

Case1:$$x_2=2n\pi +x_1$$ Since$$x_1,x_2$$ are rational $$n$$ must be $$0$$ thus $$x_2=2(0)\pi +x_1 \implies x_1=x_2$$

Case2:$$x_2=(2n+1)\pi-x_1$$

But since $$(2n+1)\pi$$ is never rational this case cannot happen.

Here is where my confusion begins:First, it does not seem like we evaluated the sine function $$f(x)$$ at any particular x value($$x_1$$ and $$x_2$$).We just determined the intervals where any particular $$x$$ is periodically equal to $$f(x)$$.It seems like equal values in the domain,were shown to prove injectivity.Values in the range were never used. Second I cannot wrap my head around why the function is injective since it appears to me $$\sin (0)=0$$ would be repeated, unless the domain is restricted.

• Unfortunately, the proof is incorrect (it is not true that if $f(x_1)=f(x_2)$, then $x_2=2n\pi + x_1$....). But that does not appear to be the main source of your confusion. – Arturo Magidin Nov 9 '19 at 1:44
• Can you explain to me if this is a possible proof.Is the sine function really injective from the rationals to the reals? – user707991 Nov 9 '19 at 1:47
• I can address your misconceptions regardless of the bad argument you were given. – Arturo Magidin Nov 9 '19 at 1:49
• Thank you for this – user707991 Nov 9 '19 at 1:50
• @ArturoMagidin I am going to add something to the end of the write up, since it seems to make sense,looking at my notes. – user707991 Nov 9 '19 at 1:55

For instance, showing that $$f\colon\mathbb{R}\to\mathbb{R}$$ given by $$f(x)=x^3$$ is injective is not about evaluating $$x^3$$ at any particular value, but rather about showing that if $$f(x_1)=f(x_2)$$, then $$x_1=x_2$$.
Second, don’t confuse inputs with outputs! While it is true that $$\sin(0)=0$$, the zero on the left hand side is playing a very different role from the zero on the right hand side: the zero on the left is an input, the zero on the right is an output. Remember what injectivity means: either “different inputs yield different outputs” or “if ‘two’ inputs yield the same output, then they are actually the same input.”
So you don’t compare the input $$0$$ with the output $$0$$. It doesn’t matter that the output is the same as the input, because we don’t compare inputs with outputs: we compare outputs with outputs, inputs with inputs.
Second: the “proof” you are given is just plain wrong.. While it is true that if $$x_2=2n\pi + x_1$$ then $$\sin(x_2)=\sin(x_1)$$, this is not the only way in which you can have $$\sin(x_2)=\sin(x_1)$$. And while it is true that for some values of $$x_1$$, $$\sin(x_1)$$ will have the same value as $$\sin(x_2)$$ with $$x_2=(2n+1)\pi-x_1$$, for arbitrary real numbers it does not follow that th is is the only way. For instance, $$\sin(\frac{\pi}{4}) = \sin(\frac{3\pi}{4})$$, but $$x_1=\frac{\pi}{4}$$ and $$x_2=\frac{3\pi}{4}$$ do not have the given form.
For real numbers, it is not true that $$\sin(x_1)=\sin(x_2)$$ implies that either $$x_2=2n\pi+x_1$$ or else that $$x_2 = (2n+1)\pi + x_1$$. So for the given argument to work, you would need to somehow show that this implication holds if we further assume that $$x_1$$ and $$x_2$$ are rational... but that is essentially what you are trying to prove in the first place (because the final step here is trivial). I doubt very much that this has been established before this problem. As such, I just say “Bad exercise, false proof.”