Do we have to assume $x$ exists before starting to solve algebraic equations?

Question

When solving algebraic equations (for example, $2x+6=10$, $x+5=x$, or $x^2=-1$), do we have to assume that our unknown variable ($x$) exists (i.e., it exists in a specific number system) before solving? Like before we solve or rearrange algebraic equations (or any other equation, such as from calculus or advanced math), do we have to assume that our unknown variable exists (in a number system) (so we can apply algebraic operations to it)? I have added my explanations for my examples below. So, is this correct?

1. $2x+6=10$

For this example, we first have to assume/let $x$ be a member of the real numbers (or any other number system we choose, but this makes the most sense). Then, since we already know that $x$ is a real number, we can apply algebraic operations that are valid in the real numbers to it, and solve the equation to get $x=2$, so we have now proven that $x=2$ is the solution to this equation in the real numbers.

2. $x+5=x$

For this example, we first have to assume/let $x$ be a member of the real numbers (or any other number system we choose, but this makes the most sense). Then, since we already know that $x$ is a real number, we can apply algebraic operations that valid in the real numbers to the equation, and solve the equation to get that $x$ has no solution, so we have proven that no solution from the real number system exists for this equation.

3. $x^2=-1$

For this example, we first have to assume/let $x$ be a member of the real numbers or the complex numbers (or any other number system we choose, but these two make the most sense). So for the real numbers version: we apply algebraic operations that valid in the real numbers to the equation, resulting in $x=\pm\sqrt{-1}$, which is undefined in the real numbers. So we have proven that no solution exists for this equation in the real numbers. For the complex numbers version: we apply algebraic operations that valid in the complex numbers to the equation, also resulting in $x=\pm\sqrt{-1}$, which is defined in the complex numbers and also the same as $x=\pm i$, so we have proven that two soltuions to this equation exist in the complex numbers.

So, my main question is: Is it true that we have to assume our unknown variable, $x$, exists (meaning it exists in a specific number system) before solving (similar to my method that I have explained above)? Any help would be greatly appreciated. Thank you.

Answer 1

It depends on what you mean by "assume."

In common speech people often treat "assumptions" with the same disdain as "leaping to conclusions." That is, they say someone "assumed" something when that person has accepted a statement as an unconditionally true fact. It would be logically unsound to "assume $x$ exists" in that sense.

In mathematical logic, however, we often "assume" something as a provisional fact. It is somewhat like the common phrase, "assume for the sake of argument." It doesn't mean we believe the assumption is true, it just means we want to see what follows from the assumption.

One benefit of this is that if you assume $P$ and in that context you are able to conclude $Q,$ you have established that $P \implies Q$ without needing to know whether either $P$ or $Q$ actually is true.

In order to use the rules of arithmetic over a particular mathematical set of numbers (such as real numbers) you need to work with expressions involving numbers from the set. So we wouldn't get very far solving for $x$ in a typical equation without assuming (in the provisional sense used in mathematical logic) that $x$ belongs to the set. More specifically, we assume there exists a number in the desired set such that the equation is satisfied when $x$ is set to that value. In effect, we assume the equation has a solution.

If we then derive a contradiction, we have $$(\text{"the equation has a solution"}) \implies \mathrm{false},$$ which tells us the equation has no solution. If we instead conclude that $x = 2,$ we have $$(\text{"the equation has a solution"}) \implies (x = 2),$$ which technically doesn't tell us that $x = 2$ is a solution, but it's usually obvious that it is (especially if we're using a familiar technique).

So when solving an equation we absolutely do assume (in the sense endorsed by mathematical logic) that the equation has a solution $x$ in the appropriate set of numbers.

It's a technique of logic, not an unconditional commitment to a belief that might turn out to be false.

Answer 2

okay in the case $2x+6=10$ subtracting $6$ from both sides and then dividing by $2$ . these two operations are independent from the fact whether $x$ exists or not right?

and say imploy one extra condition that $x>10000$ then even by algebraic manipulation we got $x=2$ as a solution in general but this is not a solution with the constraints so no such $x$ exist.

in my opinion we don't assume anything, we just do algebraic manipulation until we reach a point from where we could deduce the result of existence of such $x$, algebraic manipulations are independent of that fact and yet still hold, so in conclusion as far as algebraic manipulations are concerned we don't need to assume anything but ofcourse if we are dealing with modular arithmetic or Limits its not true then, i'm no expert by the way I just wrote what I assumed to be true over time.

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But also say we assumed that such $x$ exist in $ \mathbb{N}$ but later we found a solution which doesn't belong to $ \mathbb{N}$ then that was a contradiction to our assumption, therefore no such $x$ exist. that's it now the other way say we assumed that such $x$ doesn't exist in $ \mathbb{N}$ but later we found a solution which does belong to $ \mathbb{N}$ then that was a contradiction to our assumption, therefore such $x$ exist.

in either of the cases no matter what we assume we should be able to deduce existence/ non existence of $x$.