>It’s asking, what is the set of values that make the statement true?
For this to be the case, there would need to be a statement in the first place. And that statement would involve the =, so you necessarily still have the = symbol representing something other than questioness. I would further say that the equation itself is still just a statement, and any "question" interpretation is based entirely on the context where the equation is presented.
Further, lets take seriously the notion that "x+1" is a polynomial in the formal sense. What does it mean to find the set of values for which x+1=2 is true? Normally, I would say that we are looking for the set of x values which makes that equation true. However, we are insisting that the LHS is a polynomial (again, in the formal sense). This means that there is no variable. The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true. The fact that we give (0,1,0,...) a standard name of x does not suddenly make the question sensical. Nor does the fact x is often used to represent variables.
>I know of no mathematician who thinks x^2-x+1=0 is anything other than a polynomial equation.
To be clear, outside of very particular contexts I would still call x^2-x+1=0 a polynomial equation, because it is extremly useful to talk about polynomials without invoking all of the machinery of formal polynomials.
>And viewing x+3 as an element of C(R) the only reasonable interpretation of x+3=1 is that we are finding the pre-image of 1.
I disagree. Viewing x+3 as an element of C(R), the only reasonable interpretation of x+3=1 is the statement (x↦x+3)=(x↦1).
Viewing x+3 as an element of R would allow us to treat the equation x+3=1 in the "obvious" way. We can prove that the statement x+3=1 implies that x=-2.
Further, I would agree with you that, absent other context, when given an equation which contains an "x" in it, there is some implication that we are supposed to solve for x.
>Your view of how to interpret x^3+4x is too simplistic because the only to way to algebraically manipulate that object is by considering it as an R[x] or R(x).
Why? Suppose I don't know what R[x] or R(x) is. We certainly don't teach highschoolers what either of those are, and they seem to be able to do "algebra" just fine.
Here is a simple approach to dealing with x^3+4x without considering it a member of R[x]. For concreteness, I want to solve x^3+4x=0.
Suppose x \in R such that x^3 +4x = 0.
By the distributive property, this equation is true iff x(x^2 +4x)=0.
By direct calculation, we can verify that x=0 is consistent with this equation, and therefore consistent with the original equation.
Consider the case where x != 0.
Note that the function f(n) = n/x is a bijection. Therefore, we have x(x^2 + 4)=0 iff f(x(x^2+4)) = f(0).
By direct computation, we get that this is true iff x^2+4 = 0.
We know that x^2 >=0, and 4>0.
Therefore, x^2+4 > 0.
This is a contradiction, which means that the case where x!=0 is impossible.
This means that we have proven that x=0.
Now, suppose we were working over C.
Continuing from x^2 + 4 = 0, we can show that:
x^2+4 = 0 iff
(x+2i)(x-2i) = 0 iff
x+2i = 0 OR x-2i = 0 iff
x=2i OR x=-2i
Since the cases x=0 and x!=0 are exhaustive, we have proven the statement x \in {0, 2i, -2i}.
I solved this using the method we teach school children and without invoking any notion of polynomials.
You don't see the problem with this? Saying x=0 and that x is an element base ring means that x is the element 0. You can't later in your problem write "...x=2i..." if you are going to persist in your view that x is an element of the base ring. What you have shown is that the variety of the ideal generated by x^3+4x is the same as the variety of the ideal generated by x, x-2i, x+2i if we are talking about C as the base ring. If the base ring is Q then a different thing is shown.
You can't logically say, in a consistent manner, that x is in R and x^3+4x = 0 and that x is 3 different values. An element of a ring is not three different elements. An element of a ring is itself. If you want to vary the object x then you need to enlarge your ring to an algebraic structure that admits x so that it behaves the way you wish to view it. Your view of what is really happening when solving an equation is not rigorously sound. The proper way to view this is in the context of algebraic geometry.
> Saying x=0 and that x is an element base ring means that x is the element 0.
I never said that x=0. I said that x could be 0. More specifically, I stated that the statement x^3+4x=0 AND x=0 is not inconsistent.
All I have shown in is that (assuming we are working in C), the statement x^3 + x=0 implies that x \in {0, 2i, -2i}.
I suppose you could complain that I have not defined a sense in which {0, 2i, -2i} is correct while {0, 2i, -2i, 7} is incorrect, as it is still a true statement that x^3+4x=0 implies x \in {0, 2i, -2i, 7}.
However, you can easily make this intuition rigourous by saying that the question is to compute the set {x | x^3+4x=0}. Sure, this is invoking machinery not explicitly present in the statement x^3+4x=0. I will even concede that we do not make this machinery explicit when teaching highschool students. However, it is far less machinery than your approach.
I am not claiming that the algebraic approach is not rigourously sound; merely that it is not the only rigourously sound approach.
As far as I can tell, you are claiming that it is the only rigourously sound way of stateing the question.
>You can't logically say, in a consistent manner, that x is in R and x^3+4x = 0 and that x is 3 different values.
I believe I have made this point clear, but I never claimed x is 3 different values. The claim I made was that x is a member of the set {0, 2i, -2i}
Later you have x=2i or x=-2i. Persisting with the idea that you can view x as an element in the base ring and at the same time allowing its value to change or vary indicates you don’t really understand these issues. If x is in R it is a single value. If you want to vary it you need to expand R to include x and add the approroate algebraic structure.
It’s shocking that you think x^3+4x is not a polynomial. The whole discussion I started originally was that math language, like all other human languages, is nuanced and there are lots of abuse of notations. This is ok because math is written for humans by humans. The standard interpretation of x^3+4x is that it’s a polynomial. This is not disputable.
To be clear, by the time I got to this statement, I had changed the question (from a base of Q to C)
Further, the complete statement I was making at that point was:
If ((x^3+4x=0) AND x != 0) then ((x=2i) OR (x=-2i))
I am not allowing x to vary at all here. Suppose, for the sake of arguement, we had x=2i. It would still be true that ((x=2i) OR (x=-2i)).
> Persisting with the idea that you can view x as an element in the base ring and at the same time allowing its value to change or vary indicates you don’t really understand these issues.
Persisting with the idea that I am doing this indicates that your are not reading what I am writing.
Why? Suppose I don't know what R[x] or R(x) is. We certainly don't teach highschoolers what either of those are, and they seem to be able to do "algebra" just fine.
We don’t teach high schoolers what is really going on. We mask what is really going on because making it all precise is not effective or helpful at this stage of development. We teach rules to manipulate equations. We don’t use the language of algebraic geometry because that is too complicated. We don’t say to them the variety of the ideal generated by x+2 is the same as the ideal generated 4x+8 are the same so that x=-2 is equivalent to 3x+1=-x-7.
However, we are insisting that the LHS is a polynomial (again, in the formal sense). This means that there is no variable. The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true. The fact that we give (0,1,0,...) a standard name of x does not suddenly make the question sensical. Nor does the fact x is often used to represent variables.
You are clearly not an algebraist. The polynomial
(3, 1, 0, 0, 0, ....)
Induces a natural map from R to R that is generally called “evaluation”. It maps the number 5 to 8 for instance. We abuse notation and say to begininnng students, “replace x with 5”. We dumb things down. Instead of asking for the pre-image of 8 under this natural map we ask for what values of x do we get a value of 8. The shorthand way of writing this is to say solve the equation:
x+3=8
That you don’t know this is disconcerting since you’ve obscuoulsy had more than an elementary mathematical education. You are confusing the simplistic view of what is taught in basic courses with what is really going on.
Ask a million mathematicians, “is x^2+3x a polynomial” and without hesitation they’ll say yes. Because in standard uasage of that expression it is a polynomial. That’s the default interpretation.
Read the first chapter of any beginning algebraic geometry textbook. x^2+3x-1=0 is an algebraic variety. This is the standard interpretation of that equation.
>The polynomial (3, 1, 0, 0, 0, ....) Induces a natural map from R to R that is generally called “evaluation”.
I am well aware of this. If you look at my comments within this very chain, you will see that I made reference to polynomial evaluation, which I will call Ev(f, x). I Am not denying that the partial application given by f' = x -> f(f,x) is naturally induced by the polynomial f. Nor that this is so natural that it often makes sense to identify f with f' so that we would consider f=f', even though they are different types of objects.
>Ask a million mathematicians, “is x^2+3x a polynomial” and without hesitation they’ll say yes.
As will I, because the distinction between formal polynomials and expressions which can be naturally modeled as polynomials is so unimportant that it is almost never worth thinking about.
Put another way, how would you compute the following sets:
{ x \in C | x^3 - x = 8 }
{ x \in C | x^3 = x + 8 }
{ x \in C | log(x) = x^x }
{ x \in C | sin(x) = .7 }
{ x \in N | exists y \in N such that 5x + 3y = 1 }
Are you really claiming that these questions are ill-posed without stating them in terms of algebraic geometry?
I believe you are the only person trained in mathematics in the world that thinks x^2+3x=0 is not a polynomial equation.
You don’t understand the underlying algebraic theory. This is evidenced by your claim
The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true.
You can’t write such a statement if you really understand that “what values make x+3 the number 5” is a nicer way of conveying the question “under the natural evil map induced by x+3 what is the pre-image of 5”. The text of your I quoted is wrong.
> I believe you are the only person trained in mathematics in the world that thinks x^2+3x=0 is not a polynomial equation.
Then you are not reading my comments.
> To be clear, outside of very particular contexts I would still call x^2-x+1=0 a polynomial equation, because it is extremely useful to talk about polynomials without invoking all of the machinery of formal polynomials.
>You don’t understand the underlying algebraic theory.
My claim is that the underlying algebraic theory is not necessary to rigorously state or solve the problems under discussion.
>You can’t write such a statement if you really understand that “what values make x+3 the number 5” is a nicer way of conveying the question “under the natural evil map induced by x+3 what is the pre-image of 5”.
My claim is that "what values make x+3 the number 5" translates directly into { x \in F | x+3=5}. I further claim that the "x+3" in this interpenetration is not a polynomial in the formal sense
I am not disputing that this is set is the same set as the pre-image of 5 under the natural map induced by x+3.
For this to be the case, there would need to be a statement in the first place. And that statement would involve the =, so you necessarily still have the = symbol representing something other than questioness. I would further say that the equation itself is still just a statement, and any "question" interpretation is based entirely on the context where the equation is presented.
Further, lets take seriously the notion that "x+1" is a polynomial in the formal sense. What does it mean to find the set of values for which x+1=2 is true? Normally, I would say that we are looking for the set of x values which makes that equation true. However, we are insisting that the LHS is a polynomial (again, in the formal sense). This means that there is no variable. The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true. The fact that we give (0,1,0,...) a standard name of x does not suddenly make the question sensical. Nor does the fact x is often used to represent variables.
>I know of no mathematician who thinks x^2-x+1=0 is anything other than a polynomial equation.
To be clear, outside of very particular contexts I would still call x^2-x+1=0 a polynomial equation, because it is extremly useful to talk about polynomials without invoking all of the machinery of formal polynomials.
>And viewing x+3 as an element of C(R) the only reasonable interpretation of x+3=1 is that we are finding the pre-image of 1.
I disagree. Viewing x+3 as an element of C(R), the only reasonable interpretation of x+3=1 is the statement (x↦x+3)=(x↦1).
Viewing x+3 as an element of R would allow us to treat the equation x+3=1 in the "obvious" way. We can prove that the statement x+3=1 implies that x=-2.
Further, I would agree with you that, absent other context, when given an equation which contains an "x" in it, there is some implication that we are supposed to solve for x.
>Your view of how to interpret x^3+4x is too simplistic because the only to way to algebraically manipulate that object is by considering it as an R[x] or R(x).
Why? Suppose I don't know what R[x] or R(x) is. We certainly don't teach highschoolers what either of those are, and they seem to be able to do "algebra" just fine.
Here is a simple approach to dealing with x^3+4x without considering it a member of R[x]. For concreteness, I want to solve x^3+4x=0.
Suppose x \in R such that x^3 +4x = 0.
By the distributive property, this equation is true iff x(x^2 +4x)=0.
By direct calculation, we can verify that x=0 is consistent with this equation, and therefore consistent with the original equation.
Consider the case where x != 0.
Note that the function f(n) = n/x is a bijection. Therefore, we have x(x^2 + 4)=0 iff f(x(x^2+4)) = f(0).
By direct computation, we get that this is true iff x^2+4 = 0.
We know that x^2 >=0, and 4>0.
Therefore, x^2+4 > 0.
This is a contradiction, which means that the case where x!=0 is impossible.
This means that we have proven that x=0.
Now, suppose we were working over C.
Continuing from x^2 + 4 = 0, we can show that:
x^2+4 = 0 iff
(x+2i)(x-2i) = 0 iff
x+2i = 0 OR x-2i = 0 iff
x=2i OR x=-2i
Since the cases x=0 and x!=0 are exhaustive, we have proven the statement x \in {0, 2i, -2i}.
I solved this using the method we teach school children and without invoking any notion of polynomials.