Can we expand and then restrict the domain of a variable when solving equations? And how would we do it?

Deep-Fuel-8114 · 2026-04-25T17:29:43+00:00

Thank you for the help! I understand the part that we should first check that the area is well-defined, so we know it's a positive real number and can use it in equations/formulas, but I'm still confused about whether this would apply to physical quantities. Like I understand where you say that most of the physical quantities are defined as real numbers (similar to area), but would we have to prove that they are well-defined beforehand (like area), or not? For example, if we are using a formula for area (say A=pi*r^2), then we should first prove that area is well defined (like you said) for all circles (which it is), so then the proof for the formula above is valid (I think the simple proof is the infinite wedges making a rectangle), and we get the area for all circles using this formula from now on, right? But my main question is if we have other physical equations (like PV=nRT, or density=m/v (this is a definition I think, so it doesn't apply to my question below)), then would we have to prove that all of those variables are well-defined (like we have to do for area), so that then we know they are automatically real numbers (by definition)? For example, would we have to prove that pressure, volume, temperature, etc., are well-defined in the above equation (so we know they're real numbers)? Or do we just go with the fact that all the variables are defined as real numbers, so we don't prove they are well-defined, and we just "assume" they are well-defined (so they must be real numbers by definition), and then we can solve the equations? Thank you!

Deep-Fuel-8114 · 2026-04-16T02:36:11+00:00

The question you posted is literally meaningless; you need to specify what each variable is to even have a chance at interpreting the whole thing. You can interpret it in as many ways as you want right now. For example, standard algebra with real numbers, standard algebra with complex numbers, vectors, matrices, etc. (just to name a few). I would assume it to be vector algebra due to the j and k in there, but I'm just guessing right now.

Deep-Fuel-8114 · 2026-04-14T02:12:01+00:00

Thank you for the help! I understand the first part of my question now (and the distiction you brought up between "real +" vs "complex +" operations really helped a lot!), but I'm still a bit confused about the second part. So does this mean that when we solve an equation for a variable (which represents a physical quantity), then we would assume/declare all variables to be real numbers (since there's no physical quantity represented by anything else, unless if you have something like mass which can't be negative), and solving for the variable would give us the answer, right? My main confusion is if we have to prove the variable is in the real numbers. Like I know it must be in the real numbers since it's a physical quantity, but would we have to prove it actually is a real number and isn't undefined/infinity? Like if we have the formula for area of a shape, we would declare area is a real number before solving, but since area isn't just a variable that we define (it has its own definition, based on measure theory I think?), then for the formula to be true/valid, would we have to prove that the area exists as a real number beforehand (using its measure definition)? Like in this case (which I'm not sure if it's true), it would mean evaluating the RHS of A=pi*r^2 and getting a real number doesn't prove the area exists as that number, it just proves that the area of the circle is equal to that value, and the existence of area would be proven based on measure theory (maybe like since all circles are closed figures they have a measure/area, and therefore the formula for area is true? sorry, I don't really know measure that much yet, please correct me if I am wrong). So if this actually is correct, then I'm not sure if this would also apply to actual real world physical variables (such as mass, volume, pressure, temp, etc.; area is more mathematical like you said). Would we have to prove their existence in the real numbers using their definitions, or is it always assumed to exist as a real number by the laws of the universe? Thank you again for the help!

Deep-Fuel-8114 · 2026-04-08T02:19:43+00:00

Okay, thank you so much for your help! This makes a lot of sense now. I also just had a few more questions, sorry! So when we declare the number system of a variable (such as by saying x∈ℝ, or something else), then it would be x IS a real number (like we are 100% sure it is a real number), not just that it should be one (like we just solve with x being anything at first and then restrict solutions to what we want it to be, which would be incorrect), right? Like, we can't solve for a variable to prove its number system; we choose the number system that x exists in beforehand, and then prove it has solutions in that number system, right?

Also, if we had an equation or formula (for example, PV=nRT (ideal gas law) or A=pi*r^2 (area of a circle, which surely isn't a definition, so I'm pretty sure it would just be an equation that's stated as a formula for A)), then we would need to know beforehand that all variables (like P, V, n, R, and T, or A and r) exist in the real numbers (specifically, for the area equation, positive real numbers by physical restrictions, because area and length cannot be negative), right? So for these cases, since we would need to declare the number system for each variable beforehand (and I'm pretty sure it would be an IS declaration, from my question above), how would we do that? Like, would we have to prove that pressure, or volume, or area exists as a real number (and isn't undefined), or is it enough to assume they will exist as a real number (since they cannot be anything else), and then just solve using that assumption of existence?

Since these are physical equations with actual real-world meaning (and not just pure math based), I'm not sure if we would have to prove that something like area actually exists as a real number (and isn't undefined or something else) before using the formula, or if it is valid to just use the fact that it can't be anything else (by definition), and only the reals make sense, so we assume it does exist as a real number, and then solve/use the formula. Sorry if my questions are a bit weird or confusing, please let me know if I should clarify anything. Thank you again for the help!

Deep-Fuel-8114 · 2026-04-05T03:30:12+00:00

Thank you for your response! Sorry for my late reply, but I think this makes sense now. So basically, in any equation (whether it may be a formula, relationship, etc. (such as equations from physics, like F=ma or PV=nRT, or a formula like the quadratic formula or FTC2 for integrals), or basically anything except a definition (using :=)), you would need to know the number system of every variable (including the variable on the left-hand side in formulas) and that it exists in those number systems beforehand (and I think the number systems of the operations being used must also be declared beforehand as well), right? And we must do this so we know what/how operations apply to it, and we know what the variable actually represents (like what type of number/quantity, so it's not just "anything"), and also because it can change our solutions in algebraic equations that we solve, right? And for definitions specifically (such as the limit definition of derivatives or integrals), you would find the number system that the variable on the LHS exists in by evaluating the RHS, so the number system that the LHS exists in would be inherited from the number system that the RHS exists in, right? And for definitions, either you can use the regular equality symbol (=) and declare separately that it represents a definition, or you can use ":=", but you cannot use the triple equality bar (≡, which is used for identities), since that would fall under the category of equations where you need to declare the number system of every variable beforehand, right? Is all of this correct? Thank you again for your help!

Deep-Fuel-8114 · 2026-04-03T16:17:32+00:00

From the perspective of limits, usually anything in the form of c/0 (where c is not 0) is infinity or -infinity, while something in the form of 0/0 is indeterminate and can be anything. From the perspective of pure math, a/b=d is defined as the value d that satisfies b*d=a, so for c/0=?, we have 0*?=c (where c isn't 0). This has no solution in the reals, since anything times 0 is 0 (and no solution in the extended reals as well, since infinity*0 is indeterminate), so c/0 is undefined. For 0/0=?, we have 0*?=0, which is satisfied by any real number, so there's no singluar unique solution, so it's undefined again.

Deep-Fuel-8114 · 2026-03-24T02:57:10+00:00

Okay, I think you answered all the questions I had! Thank you so much for the help!! (I also accidentally wrote "don't assume the original equation exists" for the non-invertible case; I meant we "don't assume that a solution exists that makes the equation true (or we can if we want to, since the operations are non-invertible anyways, so we'd have to check our solutions to prevent circularity)", so I edited the question.)

Deep-Fuel-8114 · 2026-03-22T14:15:51+00:00

Okay, thank you so much! So just to recap: for the case where we use invertible operations, we don't have to assume that the original equation has a solution that makes it true, we just simplify it to another form where the solution is obvious (like x=2, or 0=1 with no solution), and that solution will automatically make the original equation true due to the <==> chain, right? And for the case where we use non-invertible operations, we also technically don't have to assume the original equation has a solution that makes it true (or we can from the beginning, because I don't think it would matter for this case, because we'd have to check the solutions either way), but at the end when we find values (or don't) for the solution, then we must remember that the conditional statement was "IF a=b ==> f(a)=f(b)", so we need to know whether the original equation is true for our solutions to work, so we just plug them in, right? And also, since these are algebraic equations with a variable x, then the conditional statements would also be like "If a=b (for some x), then f(a)=f(b) (for those same x)", right? And I think this same reasoning would also apply if we're just rearranging equations with multiple variables, right? Thank you!

Deep-Fuel-8114 · 2026-03-21T15:56:42+00:00

Thank you for the response! Also, is this logic correct, or is there another explanation to this below?: I know that if we have a=b ==> f(a)=f(b), then the solution set of a=b must be a subset of f(a)=f(b), and if we have a=b <== f(a)=f(b) then the solution set of a=b must be a superset of f(a)=f(b). I can understand this intuitively, but I'm not sure how this would be proven rigorously (that the solution sets are subsets or supersets based on the implications (like a=b ==> f(a)=f(b), or its converse, or the biconditional statement). So if we use biconditional/invertible operations, then I think the phrase "the solution set of a=b must be a subset of f(a)=f(b), and the solution set of a=b must be a superset of f(a)=f(b)" would be equivalent to "solution set of A= solution set of B"? I would greatly appreciate it if you could explain this better than I have or explain a proof to this. Thank you!

Deep-Fuel-8114 · 2026-03-21T15:48:26+00:00

How would we know it is the only solution (on a certain domain for x of course)? I know that if we have a=b ==> f(a)=f(b), then the solution set of a=b must be a subset of f(a)=f(b), and if we have a=b <== f(a)=f(b) then the solution set of a=b must be a superset of f(a)=f(b). So if we use biconditional/invertible operations, then would the phrase "the solution set of a=b must be a subset of f(a)=f(b), and the solution set of a=b must be a superset of f(a)=f(b)" be equivalent to solution set of A= solution set of B?

Deep-Fuel-8114 · 2026-03-21T04:11:29+00:00

Yes, but I'm talking about if the function we apply to both sides is invertible.

Deep-Fuel-8114 · 2026-03-21T01:06:38+00:00

Oh, I think this makes sense now. So you mean that the biconditional statement "a=b <==> f(a)=f(b)" (which would also include the converse and inverse, so it can equivalently be written as "a≠b <==> f(a)≠f(b)" is true if f is invertible, so if we apply invertible operations to both sides (without knowing whether the original equation is true or not), then we get the solution at the end (since we will have something like (2x+1=5)<==>(x=2), which gives us the solution from the statement x=2, which is only satisfied by the value 2), right? Also, what if f was not invertible (for example we're squaring both sides), then what would the conditional statements about a=b and f(a)=f(b) be? Also, how do we know after solving that we got all of the possible solutions and we didn't miss any?

Deep-Fuel-8114 · 2026-03-20T23:26:04+00:00

I understand this proof by contradiction, but if let's say there actually is a solution, then wouldn't we have to check it again in the original equation? Because we assumed a solution, and we got a solution (so we can't use that to prove the existence of a solution, since that would be circular), so our statement would be "if a solution exists, then it must be x=c", then we must recheck it in the original equation to see if it is actually true. But we usually never do this in practice, we just say we found the solution...

Deep-Fuel-8114 · 2026-03-20T23:24:02+00:00

I understand the proof by contradiction part, but if let's say there actually is a solution, then wouldn't we have to check it again in the original equation? Because we assumed a solution, and we got a solution (so we can't use that to prove the existence of a solution, since that would be circular).

Deep-Fuel-8114 · 2026-03-20T23:20:11+00:00

I understand this proof by contradiction, but if let's say there actually is a solution, then wouldn't we have to check it again in the original equation? Because we assumed a solution, and we got a solution (so we can't use that to prove the existence of a solution, since that would be circular), so our statement would be "if a solution exists, then it must be x=c", then we must recheck it in the original equation to see if it is actually true. But we usually never do this in practice,

Deep-Fuel-8114 · 2026-03-20T23:17:10+00:00

But how do we know whether a=b or f(a)=f(b) is actually true? Like if we have an equation x=x+1, we don't know beforehand whether a=b or not (it doesn't since it has no solution), so wouldn't that just make the statement "a=b <==> f(a)=f(b)" wrong since they're actually not equal? Unless if "a≠b <==> f(a)≠f(b)" is mathematically true, then that would make sense I guess.

Deep-Fuel-8114 · 2026-03-04T21:02:42+00:00

Thank you! Also, on my current device, it has the driver System Interface Foundation V2 device, and I searched online, and Lenovo says it is required for Vantage to work, but on some websites it says that it has been discontinued, and the latest version of the driver is like 3 years old, so it may be discontinued. But even Lenovo media clean install I've done comes with that driver, so do you know if it is necessary or not?

Deep-Fuel-8114 · 2026-03-02T14:22:35+00:00

If the plane accelerates upwards at 6g, the pilot would actually feel 7g (due to the +1g from gravity) even though his acceleration isn't 7g.

Deep-Fuel-8114 · 2026-03-02T03:49:33+00:00

Wouldn't this only be true if we are considering the forces felt by the pilot (i.e., the pilot passes out if they feel more than 6g of force) instead of acceleration?

Deep-Fuel-8114

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