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).

You certainly did claim that $f(n^2 - n) = n-1$, you just didn't realize you had done so. When $n \neq 0$, the result of dividing $n^2 - n$ by $n$ is $n - 1$. I have rewritten the answer to spell this out in excruciating detail.

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?

What I presented is a solvable equation. I've made that explicit now. And what do you mean, "If $x=a$ is a solution to (a) than it is also a solution for (b)"? You said $x$ was the value of whatever expression you start with on the left side of an equation, not the name of a variable you're solving for.

Also, "if $x≠0$ is a solution to $n^2−n=n\sin(α)$" is meaningless. There's no $x$ in $n^2−n=n\sin(α)$. You can solve for $n$ in terms of $\alpha$ or $\alpha$ in terms of $n$.

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) ${expression}_3 = {expression}_4$.

Take any equation you want (equation A) and apply any valid operations that you want (equation B, after operations). You will see that any solutions for equation A are also solutions for equation B.

OK, I may have gotten hung up on your use of the word "function" and your use of functional notation. Once you take those out of the picture and look just at the operations that are valid to perform, the question becomes interesting and I'm not sure I have an answer yet. What we need to look at are examples of "proofs" that lose solutions.

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.

This is one reason. I wanted to know if there are any more. I did not understand why you claimed that there is no function that converts $n^2-n$ to $n-1$. I just gave you a function that does it. @justhalf

I was wrong indeed, I stand corrected. I guess we slightly talk past each other, and I missed that you already gave that example before.

@mawaior about whether there are more, it's easy then to generalize your example. Just take any function with restricted domain, like $\log$ or square root. $x^2=1$ has -1 and 1 as solutions, but when you convert it by using $f(x)=\sqrt{x}$ (domain, positive numbers), you lose the negative number solution.

When $f(x)=\sqrt{x}$, $x^2=1 \implies f(x^2) = f(1)$. $\sqrt{x^2}=|x|$ so $|x| = 1$, $x = \pm1$. @justhalf

Hmm, you're right. But you get the idea. Define a function that is undefined in some domain.

There will be a "Automatically" move to chat button, Please use it, Comments are not for extended discussions

@mawaior, Your doubt is very easy to understand to be honest, Let us say $x>0,y>0$ and it is given $x^2=4x$.Consider function $f(t)=sign(t)$, When we apply function $f(t)$ on both sides as you say $f(x^2)=f(4y)$,You will get $1=1$, We have lost both of the solutions of the question when we apply the function, Why is that so? Because $t=z$ implies $f(t)=f(z)$, But $f(t)=f(z)$ does not imply $t=z$, Which means we have lost information

What is $f(t) = sign(t)$? @DheerajGujrathi

@mawaior, it is sign function, Look here en.wikipedia.org/wiki/Sign_function

Is the equation $x^2=4x$ or $x^2=4y$? @DheerajGujrathi

@mawaior,It is indeed $x^2=4x$, That is we need to solve for $x$, Replace all $y$ by $x$, They were a typo

You claim is valid but it doesn't contradict my claim it even supports it. It is the same as using the function $f(x)=0\cdot x$. Some functions are not useful. $x=y \implies f(x) = f(y)$ but $f(x)=(y)$ does not imply $x=y$, this is the reason we get extraneous solutions. In you example we got $1=1$ which means that every $x$ is a solution for this equation. We didn't lose solutions, we gained an infinite amount of solutions. the solutions to the original equation $x=0\lor x=4$ are still solutions for $1=1$. @DheerajGujrathi

Let us continue this discussion in chat.

I have to correct you, we did not lose any solutions. In fact, we gained an infinite amount of solutions. Every $x=a$ is a solution for $1=1$ so the solutions for the original equation are still solutions for $1=1$. This is not the reason we lose solutions because we did not lost any solutions. @DheerajGujrathi

@mawaior,Let us continue this discussion in chat.

This answer is so misleading. Right off the bat, (3) is an application of (4) by taking $f(x) = \frac{x}{n}$ (assuming $n \neq 0$)...

@Adiyah You're making the same mistake as OP. How is $n$ defined in $f(x) = x/n$? If we consider $x = n^2 - n$, then every new value of $x$ might be due to a new value of $n$. A function can only look at the value of $x$ to decide what the output should be. So when $x=2$, for example, the function might output $-1$. There is no function whose input is $n^2-n$ and whose output is $n-1$ for all $n\neq0$.