The function "divide by n" is $f(x)=\frac{x}{n}$. $n^2-n=n \sin(\alpha) \implies f(n^2-n)=f(n\sin(\alpha))$ I didn't understand what you wrote. If I have an equality I can do the same operation on both sides of the equation. And any operation can be expressed as a "function". $2x-3=5 \implies 2x-3+3=5+3, 2x=8 \implies \frac{2x}{2}=\frac{8}{2}, x=4$

I think that you didn't understand what I meant. The "functions" work on the left and right sides of the equation. I meant that the equality between the left and right side is preserved for any "function". So the function "divide by n" does preserve the equality. I didn't mean that $n^2-n = n-1$. $x=y \implies f(x)=f(y)$ but not $x=y \implies x=f(x)$

Example: $2x-3=5,\quad f(x)=x+3,\quad g(x)=\frac{x}{2}$. I know that $2x-3=5$ we can use $f(x)$ to get $f(2x-3)=2x, \; f(5)=8$. Now we can use $g(x)$ to get $g(2x)=x,\;g(8)=4$. $g(f(2x-3))=g(f(5)),\; x=4$

Why isn't "divide by n" a function? A function doesn't have to be defined for all real numbers. If you divide by $n$ you assume that $n \neq 0 $, it is a function. Can you give an example of an operation which is not a function?

I didn't claim that $f(n^2-n)=n-1$. $x=y \implies x=f(x)$ is not true, but $x=y \implies f(x)=f(y)$ is true. The equality preserves. If you know that $x=a$ is a solution to $x=y$ than it is a solution to $f(x)=f(y)$. If the right side of the equation is equal to the left side of the equation than for any function, f(left side of the equation) = f(right side of the equation).

No, I did not claim that $f(n^2 -n) = n - 1$. I claimed that if $x \neq 0$ is a solution to $n^2 -n = n\sin(\alpha)$ than it is also a solution for $n-1=\sin(\alpha)$. I claimed that the equality is preserved. If you know that $n \neq 0$ you can divide by $n$ and the function which is used on both sides is $f(x)=\frac{x}{n}$. For any function $f$ if $x=y$ than $f(x)=f(y)$ I did not mean that $x=f(x)$.

If this is not true, how can we solve equations?

$x=2 \implies 2x=4$, $2 \neq 4 \land x \neq 2x$. But $x=2$ is a solution to $x=2$ than it is a solution for $2x=4$.

Take any equation (a) and do any valid operations you want on both sides of the equation until you have a new equation (b). If $x=a$ is a solution to (a) than it is also a solution for (b).

Can you give an example of a solvable equation which supports your claim?

1) I meant $n$ or $\alpha$. $x$ is a "name" for a solution. 2) I have an equation (A) ${expression}_1 = {expression}_2$ I do operations on both sides of the equation and get equation (B) ${expression}_3 = {expression_4}$. If there is a solution for (A) ${expression}_1 = {expression}_2$ it is also a solution for (B) ${expression}_3 = {expression}_4$.

Agree with DavidK here. I think OP is conflating operation and function. The one that has the property defined in the first sentence of the question (x=y -> f(x)=f(y)) are functions. Operations may or may not have the same property, but that hasn't been established in the question (and they don't)

Can you give me an example of an operation which is not a function? Operations are defined using functions. @justhalf

The usage of the word "function" is correct. Operations are defined using functions so every operation is a function. @DavidK

@mawaior In mathematics (which is this what this stack is about) the term "function" is not used the way you describe. It's evident when you say you're not claiming $f(n^2-n) = n-1$ in DavidK's example. You're doing an operation to convert $n^2-n$ into $n-1$. So you're doing an operation which takes the argument $n^2-n$ then outputs $n-1$. If it's also a function in mathematical terms, you would write it as $f(n^2-n) = n-1$ (which is read as: the function $f$ when given the value $n^2-n$ returns the value $n-1$), but you already said it's not the case.

Since your argument in the main question depends on the property $x=y \Rightarrow f(x)=f(y)$, please provide an example of $f$ for converting $n^2-n=n\sin(\theta)$ into $n-1=\sin(\theta)$. Then we can talk on common ground. =)

You are wrong, in mathematics operations are defined using functions. I did not said that $x=y \implies x=f(x)$, this is not correct. But $x=y \implies f(x)=f(y)$ is correct. If I have an equation $f(x)=g(x)$ I can apply a function $h(fx)$ on both sides to get $h(f(x))=h(g(x))$. If $x=a$ is a solution for $f(x)=g(x)$ it is also a solution for $h(f(x))=h(g(x))$. That's all I claimed. @justhalf

The function to convert $n^2-n=n\sin(\alpha)$ to $n-1=\sin(\alpha)$ is $f(x)=\frac{x}{n}$. Note that $n \neq 0$. If there is a solution for $n^2-n=n\sin(\alpha)$ it is also a solution for $n-1=\sin(\alpha)$ @justhalf

Just curious, who said $x=y \Rightarrow x=f(x)$?

@mawaior About your function, now it's clear how it loses a solution, no? The domain of the conversion function excludes $n=0$, which happens to be one of the solutions to the original problem.