Can we plug a linear sequence of real numbers into the sine function such that the resulting sequence will be a one-to-one sequence?

Question

I was thinking about this problem for some time which can be stated in plain words as follows:

Can we plug a linear sequence of real numbers into the sine function such that the resulting sequence will be a one-to-one\injective sequence?

Translating it into more mathematical language we get that the problem can be stated as:

Can you find real numbers $a,b\in\mathbb{R}$ such that plugging the linear sequence $\{x_n\}_{n=1}^\infty$ defined as $\forall n\in\mathbb{N}, x_n:=an+b$ into the sine function will produce a new sequence $\{y_n\}_{n=1}^\infty$ defined as $\forall n\in\mathbb{N}, y_n:=\sin(x_n)$ that will satisfy $\forall m,n\in\mathbb{N}, y_m=y_n\longrightarrow m=n$
?

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I’ve tried to attack the question by first showing that the condition:

$\forall m,n\in\mathbb{N}, y_m=y_n\longrightarrow m=n$

is equivalent to:

$\forall m,n\in\mathbb{N},\Big( (\exists k_1\in\mathbb{Z},a(n-m)=2\pi k_1) \lor (\exists k_2\in\mathbb{Z}, a(n+m)+2b=\pi+2\pi k_2 )\Big)\longrightarrow m=n$

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Let $m,n\in\mathbb{N}$, Then:

$\begin{gather} y_m = y_n \\ \Updownarrow \\ \sin(x_m)=\sin(x_n) \\ \Updownarrow \\ \sin(x_n)-\sin(x_m)=0 \\ \Updownarrow \\ 2\sin\frac{x_n-x_m}{2}\cos\frac{x_n+x_m}{2} = 0 \\ \Updownarrow \\ \sin\frac{x_n-x_m}{2}\cos\frac{x_n+x_m}{2} = 0 \\ \Updownarrow \\ \sin\frac{(an+b)-(am+b)}{2}\cos\frac{(an+b)+(am+b)}{2} = 0\\ \Updownarrow \\ \sin\frac{a(n-m)}{2}\cos\frac{a(n+m)+2b}{2} = 0 \\ \Updownarrow \\ \sin\frac{a(n-m)}{2}=0 \lor \cos\frac{a(n+m)+2b}{2}=0 \\ \Updownarrow \\ \Big(\exists k_1\in\mathbb{Z}, \frac{a(n-m)}{2}=\pi k_1\Big) \lor \Big(\exists k_2\in\mathbb{Z},\frac{a(n+m)+2b}{2}=\frac{\pi}{2}+\pi k_2\Big) \\ \Updownarrow \\ \Big(\exists k_1\in\mathbb{Z}, a(n-m)=2 \pi k_1\Big) \lor \Big(\exists k_2\in\mathbb{Z},a(n+m)+2b=\pi+2\pi k_2 \Big) \end{gather}$

And therefore, the condition:

$\forall m,n\in\mathbb{N}, y_m=y_n\longrightarrow m=n$

is equivalent to:

$\forall m,n\in\mathbb{N}, \Big((\exists k_1\in\mathbb{Z},a(n-m)=2\pi k_1) \lor (\exists k_2\in\mathbb{Z}, a(n+m)+2b=\pi+2\pi k_2 )\Big)\longrightarrow m=n$

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Thanks...

Note: A sequence $\{x_n\}_{n=1}^\infty$ is called linear if and only if $\exists a,b\in\mathbb{R},\forall n\in\mathbb{N},x_n=an+b$

Answer 1

What about $a=b=1$? You have$$\sin m=\sin n\iff m=n.$$

Answer 2

Well, The answer is yes and based on Santos comment, we can just take $a=1$ and $b=0$, and we will have $\forall m,n\in\mathbb{N}, \sin m=\sin n \longrightarrow m=n$:

Proof:

Let $m,n\in\mathbb{N}$ be such that $\sin m=\sin n$, Then by what was just shown above, We get that there are two possibilities: (1) $\exists k_1\in\mathbb{Z},n-m=2\pi k_1$ or (2) $\exists k_2\in\mathbb{Z}, n+m=\pi+2\pi k_2$.

We will show that case (2) cannot happen, Because if we suppose by contradiction that case (2) happens, The we get that $\exists k_2\in\mathbb{Z}, n+m=\pi+2\pi k_2$, But since $\pi+2\pi k_2$ is an irrational number for any $k_2\in\mathbb{Z}$, We get that $n+m$ is also irrational which contradicts the fact that $n+m$ is rational as it is the sum of two integers. Therefore case (2) cannot happen, And thus it must be the case that case (1) happens and we conclude that $\exists k_1\in\mathbb{Z},n-m=2\pi k_1$, We will show that it must be the case that $k_1=0$ because if we suppose by contradiction that $k_1\neq 0$, The we get that $2\pi k_1$ is an irrational number and thus $n-m$ is also an irrational number which contradicts the fact that $n-m$ is a rational number as it is the difference between two integers. Thus it must be the case that $n-m=0$, And we can conclude that $n=m$ as was to be shown.