Dividing by m to make conclusions for m=0

[pbuk]

These do not satisfy the proposition. If x is a function of m as described then the pair x=5, y=5 does not satisfy the proposition because x=5, y=5 does not exist for m=0.

No, in my counter-example if m = 0 then x = 0 (and y can take any value, 5 if you like). The proposition is $mx = my$, which is satisfied by $0 \times 0 = 0 \times 5$.

We are running into problems because, despite A.T.'s wishes, and Fredrik's efforts this thread has not proceded with anything like mathematical rigour. We need to go back to the starting point and note that x and y are not independent variables free to take any value in R, they are real valued functions x(s) and y(s) over some unspecified domain S. So when we write mx = my what we are asserting is that for all s in S and all m in R, mx(s) = my(s) is a theorem.

DaleSpam asserts that if mx(s) = my(s) for all m in R and s in S [Theorem 1] is a theorem then x(s) = y(s) for all S [Proposition 2] is a theorem. I provide a (better) counter-example which is consistent with Theorem 1 but contradicts Proposition 2 as follows:

$$ x(s) =
\left\{
\begin{array}{ll}
y(s) & \mbox{if } m \ne 0 \\
1 + y(s) & \mbox{if } m = 0
\end{array}
\right. $$

 

[Dale]

If we interpret x and y as variables that represent elements of some other set, e.g. a set of functions, then things are certainly different. But I don't think that's the intended interpretation.

You are correct, I had intended x and y to be real numbers. But your proof actually works if x and y are elements of any vector space. Since the set of functions is a vector space, it works there also.

The problem comes that they are trying to treat x and y as functions in choosing the admissible set of x and y and then as something else for evaluating $mx=my$. Whatever set they choose for x and y they have to treat them consistently, and your proof works for x and y in any vector space.

 

[Dale]

We are running into problems because, despite A.T.'s wishes, and Fredrik's efforts this thread has not proceded with anything like mathematical rigour.

You are correct, I have not been rigorous at all and have assumed that people would understand what was in my head without my explicitly writing it down. Obviously a poor strategy for communication. My apologies for that.

I am not sure how to write it rigorously, but Fredrik's proof holds for x and y elements of any vector space over some field and m an element of any subset of that field that contains the multiplicative identity, 1.

x and y are not independent variables free to take any value in R, they are real valued functions x(s) and y(s) over some unspecified domain S. So when we write mx = my what we are asserting is that for all s in S and all m in R, mx(s) = my(s) is a theorem.

DaleSpam asserts that if mx(s) = my(s) for all m in R and s in S [Theorem 1] is a theorem then x(s) = y(s) for all S [Proposition 2] is a theorem. I provide a (better) counter-example which is consistent with Theorem 1 but contradicts Proposition 2 as follows:

$$ x(s) =
\left\{
\begin{array}{ll}
y(s) & \mbox{if } m \ne 0 \\
1 + y(s) & \mbox{if } m = 0
\end{array}
\right. $$

As written x is not a real-valued function over S. It is a set of many such functions, one for each m. The real-valued functions over S is a vector space and Fredrik's proof holds for them.

 

[pbuk]

Fredrik's proof holds for them.

Just so that I can keep track, are you still asserting that your orginal proof including division by m where m = 0 is valid? Are you still asserting that {m = 0, y = 5, x = 0} cannot be a solution to mx = my? Or is Fredricks proof the only one that you now support?

If that is the case, note that Fredricks proof started with the assumption "I would interpret x and y as variables that represent real numbers, since nothing else was explicitly stated." and stated "If we interpret x and y as variables that represent elements of some other set, e.g. a set of functions, then things are certainly different." so I don't think that you can state without proof that it holds for for a vector space of functions.

Looking at it another way, if $mx = my$ is to be interpreted as a theorem schema $mx_m(s) = my_m(s)$, then when m = 0 we have the theorem $0x_0(s) = 0y_0(s)$ from which it is impossible to conclude that $x_0(s) = y_0(s)$ and therefore also impossible to conclude that $x_m(s) = y_m(s)$ is a theorem schema.

 

[DrStupid]

MI would like to hear a mathematicians opinion, on whether you can apply something, that you derived by doing this:

gm=ma
g=a

to a situation where $m = 0$.

This is allowed both in math and physics. You are right that m/m is not valid operation for m=0 but the singularity is removable. The correct mathematical operation is

$a = \lim\limits_{m \to 0} \frac{m \cdot g}{m} = g$

 

[pbuk]

This is allowed both in math and physics. You are right that m/m is not valid operation for m=0 but the singularity is removable. The correct mathematical operation is

$a = \lim\limits_{m \to 0} \frac{m \cdot g}{m} = g$

We don't know anything about the nature of the singularity; in particular if m = 0 => a = 0 there is a jump discontinuity and the limit is not defined.

 

[Dale]

Just so that I can keep track, are you still asserting that your orginal proof including division by m where m = 0 is valid? Are you still asserting that {m = 0, y = 5, x = 0} cannot be a solution to mx = my? Or is Fredricks proof the only one that you now support?

I think that they are all valid, but this thread is too confused to chase down so many different proofs. Let's stick with Fredrick's, and its associated generalizations.

If that is the case, note that Fredricks proof started with the assumption "I would interpret x and y as variables that represent real numbers, since nothing else was explicitly stated." and stated "If we interpret x and y as variables that represent elements of some other set, e.g. a set of functions, then things are certainly different." so I don't think that you can state without proof that it holds for for a vector space of functions.

It seems clear to me that it does, but let me try to formalize it.

For $x,y \in V$ where $V$ is a vector space over field $F$ and for $1 \in S \subseteq F$.

Given: $mx=my, \forall m \in S$
$x=1x$ by identity element of scalar multiplication in $V$
$1x=1y$ since $mx=my, \forall m \in S$ is given and $1 \in S$
$1y=y$ by identity element of scalar multiplication in $V$
therefore $x=y$ by transitivity of equality

Since the space of functions is a valid vector space, $V$, over the reals, $F$, then the proof applies.

I think that the proof could even be further generalized to apply for S with any non-zero element, but the notation would be unnecessarily cumbersome, something like $S\subseteq F| k \in S | k\ne 0$

 

[DrStupid]

We don't know anything about the nature of the singularity

Maybe you forgot that we are talking about $m \cdot a = \frac{{G \cdot M \cdot m}}{{r^2 }}$.

 

[pbuk]

Well m is a scalar so $m \notin S$ but that is easily fixed.

I agree that your proof is valid for vector spaces over a field. However I think if you include my counter-example function x(s) then the set is no longer a field.

 

[DrGreg]

This is true, but now x and y aren't independent real numbers. That's why the theorem doesn't apply. Suppose e.g. that you make x and y functions and make the claim that mx(m)=my(m) for all m≥0. (This is what MrAnchovy suggested and then deleted, because he felt that it was off topic, but now it seems that it isn't). Now we can't just choose m=1, because that would prove x(1)=y(1), not x=y. All we can do is this: Let m be an arbitrary non-negative real number. If m≠0 then the assumption implies that x(m)=y(m). If m=0, the equality just says 0=0, and gives us no information. So the conclusion is that x(m)=y(m) for all m>0. If we let $x_+$ and $y_+$ denote the restrictions of x and y to the positive real numbers, the conclusion is that $x_+=y_+$, not that $x=y$.

I think this addresses the source of A.T.'s confusion. Using this notation, you can't prove that $x(0) = y(0)$ unless you make the additional assumption that the functions $x$ and $y$ are both continuous at zero, and then you can.

(For readers unfamiliar with the definition of "continuous", it means $x(m) \rightarrow x(0)$ as $m \rightarrow 0$.)

 

[A.T.]

You are right that m/m is not valid operation for m=0 but the singularity is removable. The correct mathematical operation is

$a = \lim\limits_{m \to 0} \frac{m \cdot g}{m} = g$

Yes, in the previous thread we all agreed that taking the limit is the mathematically correct way to deal with this.

 

[Fredrik]

If that is the case, note that Fredricks proof started with the assumption "I would interpret x and y as variables that represent real numbers, since nothing else was explicitly stated." and stated "If we interpret x and y as variables that represent elements of some other set, e.g. a set of functions, then things are certainly different." so I don't think that you can state without proof that it holds for for a vector space of functions.

I was a bit sloppy with the statement about functions. The theorem in post #30 holds even if we replace $\mathbb R$ with an arbitrary vector space, as long as we take S to be a set that contains the multiplicative identity of the associated field.

What I had in mind was situations like the one described in post #31:

Suppose e.g. that you make x and y functions and make the claim that mx(m)=my(m) for all m≥0.
[...]
Now we can't just choose m=1, because that would prove x(1)=y(1), not x=y. All we can do is this: Let m be an arbitrary non-negative real number. If m≠0 then the assumption implies that x(m)=y(m). If m=0, the equality just says 0=0, and gives us no information. So the conclusion is that x(m)=y(m) for all m>0. If we let $x_+$ and $y_+$ denote the restrictions of x and y to the positive real numbers, the conclusion is that $x_+=y_+$, not that $x=y$.

 

[BruceW]

Looking at it another way, if $mx = my$ is to be interpreted as a theorem schema $mx_m(s) = my_m(s)$, then when m = 0 we have the theorem $0x_0(s) = 0y_0(s)$ from which it is impossible to conclude that $x_0(s) = y_0(s)$ and therefore also impossible to conclude that $x_m(s) = y_m(s)$ is a theorem schema.

yeah. this is how I would interpret it. And so from $ma_m(s)=mg_m(s)$, we don't get $a_m(s)=g_m(s)$, unless we also specify that $a_m(s)$ and $g_m(s)$ is the same for any permissible choice of $m$. Which I suppose we can do, but if you are trying to logically derive some physics, I don't think it is a good idea usually.

 

[A.T.]

I disagree that it's unambiguous, because aside of the relationship

$$R_1: \hspace{10 mm} x=y$$

there is an infinite number of relationships between $x$ and $y$, that satisfy the proposition. For example:

$$R_2: \hspace{10 mm} x =\begin{cases} y & m \neq 0 \\ 3y & m = 0 \end{cases}$$

Your objection in the previous thread was that $x = 3y$ doesn't satisfy the proposition for all $m$, but that is irrelevant, because $x = 3y$ only comes into play for $m=0$. The relationship $R_2$ as a whole does sastify $mx=my$ for all $m$.

This is true, but now x and y aren't independent real numbers.

You mean the relationship between $x$ and $y$ depends on $m$ in the $R_2$ case? That is true, but is this ruled out by the preposition: $mx = my$ for all $m$ ?

If m≠0 then the assumption implies that x(m)=y(m). If m=0, the equality just says 0=0, and gives us no information. So the conclusion is that x(m)=y(m) for all m>0.

So if the relationship between $x$ and $y$ depends on $m$, we can only prove $x=y$ for the $m > 0$ case.

Using this notation, you can't prove that $x(0) = y(0)$ unless you make the additional assumption that the functions $x$ and $y$ are both continuous at zero, and then you can.

Yes, I obviously chose functions that are not continuous at zero, to demonstrate that you have to make that additional assumption.

 

[Dale]

Yes, I obviously chose functions that are not continuous at zero, to demonstrate that you have to make that additional assumption.

The proof in 42 works even for discontinuous functions, which are still a vector space. You just have to make sure that you are treating them consistently, I.e. if x and y are functions then treat them as functions everywhere in the proof, not as functions in one place and reals in another.

 

[Dale]

I know that this thread is about the math, but we should look at the class of mathematical objects of interest to the physics, even if we don't specifically examine the physics itself.

The acceleration of a classical point object is a vector valued function of time a(t). The gravitational field experienced by a classical point object is also a vector valued function of time g(t). This does not place any restrictions on the continuity or differentiability of a or g.

The set of all vector valued functions of time, V, forms a vector space over the reals, F. The non-negative reals is a subset, S, of F which contains 1. Therefore, if $ma=mg, \forall m \in S$ then $a=g$ per the proof in 42 (which is just a generalization of Fredrik's proof).

Yes, there are some objects which are not part of a vector space and some quantities which are not subsets of a field on that vector space. Those would not be covered by the proof, but the case of physical interest is covered by the proof.

 

[Fredrik]

You mean the relationship between $x$ and $y$ depends on $m$ in the $R_2$ case? That is true, but is this ruled out by the preposition: $mx = my$ for all $m$ ?

If the relationship depends on m, then at least one of x and y depends on m. So I would say YES to that question, because if x and y depend on m in any way, it would be absurd to hide this dependence behind a notation like "mx=my". This notation suggests extremely strongly that the short "mx=my for all m" is a shortened version of a statement of the form

Let x,y,m be arbitrary. Suppose that $x,y\in\mathbb R$ and that the following implication holds: $m\in S\ \Rightarrow\ mx=my$.​

There's some ambiguity about what set S is, but we know that if it contains 1, then the assumption is strong enough to imply that x=y, as I've shown several times above.

So if the relationship between $x$ and $y$ depends on $m$, we can only prove $x=y$ for the $m > 0$ case.

Before we can prove anything, we have to take a different statement than "mx=my for all m" as the starting point, because I can't interpret "mx=my for all m" very differently from how I interpreted it above. So let's take "mx(m)=my(m) for all m" as the starting point. This would be a shortened version of the following statement.

Let x,y,m be arbitrary. Suppose that x and y are real-valued functions with domain S, and that the following implication holds $m\in S\ \Rightarrow\ mx(m)=my(m)$.​

This assumption implies that $x(m)=y(m)$ for all $m\in S-\{0\}$.

 

[A.T.]

You mean the relationship between $x$ and $y$ depends on $m$ in the $R_2$ case? That is true, but is this ruled out by the preposition: $mx = my$ for all $m$ ?

If the relationship depends on m, then at least one of x and y depends on m. So I would say YES to that question, because if x and y depend on m in any way, it would be absurd to hide this dependence behind a notation like "mx=my".

What do you mean by "hide the dependence"? We honestly don't know if the relationship between x and y depends on m or not, because we don't know what that relationship is. We are trying to derive that relationship, and at some point in that derivation we arrive at "mx=my for all m". Why should we assume that the relationship between x and y doesn't depend on m?

 

[pbuk]

Before we can prove anything, we have to take a different statement than "mx=my for all m" as the starting point, because I can't interpret "mx=my for all m" very differently from how I interpreted it above. So let's take "mx(m)=my(m) for all m" as the starting point. This would be a shortened version of the following statement.

Let x,y,m be arbitrary. Suppose that x and y are real-valued functions with domain S, and that the following implication holds $m\in S\ \Rightarrow\ mx(m)=my(m)$.​

This assumption implies that $x(m)=y(m)$ for all $m\in S-\{0\}$.

Yes, this is the essence of the problem. We cannot assume that x is independent of m because this is what DaleSpam is attempting to prove.

 

[Dale]

We cannot assume that x is independent of m because this is what DaleSpam is attempting to prove.

No, it isn't. I am only attempting to prove that if $mx=my,\forall m \in S$ then $x=y$. Fredrick did so in post 3 for x and y in the reals and S the positive reals and I rather trivially generalized it in post 42 for x and y in any vector space and S a subset of the field. What I wanted to prove is proven and I have no desire to prove independence.

 

[Fredrik]

Yes, this is the essence of the problem. We cannot assume that x is independent of m because this is what DaleSpam is attempting to prove.

Is it? I didn't get that impression, but I must admit that I haven't read all his posts in detail. Edit: OK, I see now that DaleSpam has explicitly denied this in the post above this one.

What do you mean by "hide the dependence"? We honestly don't know if the relationship between x and y depends on m or not, because we don't know what that relationship is. We are trying to derive that relationship, and at some point in that derivation we arrive at "mx=my for all m". Why should we assume that the relationship between x and y doesn't depend on m?

I thought you started this thread to ask about what can be derived from the statement "mx=my for all m". As I said in my previous post, there no way I can interpret this as allowing x to depend on m. It's possible that I have misunderstood what you meant to ask when you started this thread, so maybe your question is something else entirely, but the question about the consequences of that statement is trivial.

I am pretty confused by all this. You started this thread in a math forum (it was later moved here), and at some point the notation got changed from g and a to x and y, presumably to further distance the discussion from the physics discussion where the statement "mg=ma for all m" was first made. These things seem to support that the thread is about that statement. But apparently it's not.

So what is it about? About proving that the empirically determined constant g=9.81 m/s² doesn't depend on m? No, that can't be it. I'll tell you what it looks like to me. This thread is about the question "How do you find both the solution and the correct statement of a problem, when you only have a vague idea what the statement and the solution should be?"

 

[Dale]

Regarding independence. The dependence that A.T. and MrAnchovy are introducing makes it so that x and y are not elements of a vector space with m an element of the field (at least not in any way that I can tell). So the proof in 42 doesn't apply to those x, y, and m.

Fortunately, "vector spaces" is not too restrictive, and so, although the proof is not universal, it is useful for a broad category of practical problems. In particular, it is sufficiently broad to cover the physics that spawned the question.

That there exist x, y and m where it doesn't hold is fine, we at least know some of the places it does hold and we can use that relation in those cases.

 

[A.T.]

I have no desire to prove independence.

I had the impression that this was the whole point of your original derivation. After all, it is the conclusion that you state at the end:

$F=ma$

$GMm/r^2=ma$

$gm=ma$

$g=a$The acceleration is independent of mass, and is well defined.

The dependence that A.T. and MrAnchovy are introducing makes it so that x and y are not elements of a vector space with m an element of the field (at least not in any way that I can tell). So the proof in 42 doesn't apply to those x, y, and m.

So in order for your proof of x=y to work, you have to assume that such a dependence on m doesn't exist. And then you conclude from your x=y that in fact it doesn't exit.

 

[Fredrik]

It's been a long discussion. Two long discussions actually, but I have only partipated in one of them. None of us has been able to make it perfectly clear what we meant in every single post. It seems to me that what you're doing is to ignore the posts where DaleSpam made it perfectly clear what he meant, and focus only on the ones where he didn't.

I don't think it's unreasonable to describe the theorem

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\in S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

as "the acceleration is independent of mass". It's obviously not as clear a statement as the theorem, but it sounds like an attempt to summarize the content of the theorem.

Edit: Corrected $1\subseteq S$ to $1\in S$.

 

[Dale]

I had the impression that this was the whole point of your original derivation. After all, it is the conclusion that you state at the end:

Remember, you are trying to separate the math from the physics. The whole point of the math derivation is only to show that given $mx=my,\forall m \in S$ you do indeed get $x=y$. That was the point of the physics derivation where you objected to the math, and that is the step which is justified here.

Since we were not trying to separate them in the other thread, the independence I mentioned there is obtained not only from this mathematical step but also from other knowledge of the physics (specifically that g is independent of mass). If you are satisfied with the validity of the math in 42 then we can go back and talk about the physics and see if you agree that it is legitimate to apply 42 in the specific case of the physical question that spawned this mathematical discussion.

I would recommend a separate thread for that. One which takes the math as an agreed upon starting point and returns to the physics. Do you have any objections to the proof in 42? Do you agree with its validity as stated?

 

[pbuk]

I don't think it's unreasonable to describe the theorem

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\subseteq S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

as "the acceleration is independent of mass". It's obviously not as clear a statement as the theorem, but it sounds like an attempt to summarize the content of the theorem.

Are you saying that that theorem holds even where a = a(m) may depend on m?

 

[A.T.]

The whole point of the math derivation is only to show that given $mx=my,\forall m \in S$ you do indeed get $x=y$.

You get it only under the assumption of independence of m, as you explained in post #57.

Since we were not trying to separate them in the other thread, the independence I mentioned there is obtained not only from this mathematical step...

Why "not only"?. Shouldn't that be a simple "not"? A mathematical step that assumes independence from m, cannot be used at all to show the independence from m.

...but also from other knowledge of the physics (specifically that g is independent of mass).

So "acceleration is independent of mass" was a premise of your derivation, not its result?

 

[Dale]

Are you saying that that theorem holds even where a = a(m) may depend on m?

It obviously holds as long as a is a vector and m is an element of its field. Did you not understand post 42?

That said, I cannot think how to make a function a vector over its own domain. It seems like the equivalent operation of vector multiplication would convert a function to a real, an inconsistency in your approach that I have mentioned several times.

 

[Dale]

So "acceleration is independent of mass" was a premise of your derivation, not its result?

"g is independent of mass" was an unstated premise of my derivation. "a is independent of mass" was a conclusion, obtained through the fact that a=g.

A.T., in 42 I posted a solid proof of the mathematical topic under discussion for THIS thread. 22 posts later, you have never even acknowledged it even though it is central to the topic of this thread. Instead of commenting on that proof you have begun to reach back to a different discussion in a different thread that is (unsurprisingly) different from the one here. You point out the obvious differences in my comments in what seems like an attempt to discredit the proof in 42 by discrediting me (for the heinous crime of having different comments in different threads on different topics).

Instead of avoiding the issue, please directly examine the proof in 42 and directly respond to it. If the math in 42 is solid then it is solid regardless of my own inconsistencies. Do you agree with it or disagree with it? If you disagree, on what grounds do you do so? Address the proof on its own merits, regardless of what my personal inconsistencies and failings may be.

 

[Alien8]

In the physics section there is doubt, that a mathematician would object to deriving something by dividing an equation by a variable $m$, and then applying that derived result to all values of $m$ including $m = 0$:

$F= G \frac {m1m2}{r^2}$ resulting in a number greater than zero when mass m1 or m2 is zero seems about as likely as $F= k_e \frac {q1q2}{r^2}$ resulting in a number greater than zero when charge q1 or q2 is zero.

 

[Fredrik]

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\in S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.

Are you saying that that theorem holds even where a = a(m) may depend on m?

I find it difficult to process that question. If a depends on m, then we're not talking about that theorem. Apparently you want to change something in the assumptions, but it's not entirely clear what. So all I can say is that the following two statements are both theorems with trivial proofs.

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\in S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

For all $g,a,S$ such that $S\subseteq \mathbb R$ and $g,a:S\to\mathbb R$, if $mg(m)=ma(m)$ for all $m\in S$ then $g(m)=a(m)$ for all $m\in S-\{0\}$.​

Feel free to quote, copy, paste and edit if you want to ask about a different statement.

Edit: Corrected $1\subseteq S$ to $1\in S$.

 

[pbuk]

I find it difficult to process that question. If a depends on m, then we're not talking about that theorem. Apparently you want to change something in the assumptions, but it's not entirely clear what. So all I can say is that the following two statements are both theorems with trivial proofs.

For all $g,a,S$ such that $g,a\in\mathbb R$ and $1\subseteq S\subseteq \mathbb R$, if $mg=ma$ for all $m\in S$ then $g=a$.​

For all $g,a,S$ such that $S\subseteq \mathbb R$ and $g,a:S\to\mathbb R$, if $mg(m)=ma(m)$ for all $m\in S$ then $g(m)=a(m)$ for all $m\in S-\{0\}$.​

Feel free to quote, copy, paste and edit if you want to ask about a different statement.

Thank you, that's exactly what I wanted to clarify.

 

[Fredrik]

I see that I've been writing $1\subseteq S$ in a few places where I meant $1\in S$. I hope that hasn't caused any confusion. I have edited my recent posts above to correct this mistake.

 

[A.T.]

Instead of avoiding the issue, please directly examine the proof in 42 and directly respond to it.

I have no problem with the proof in #42, under the assumption you state in post #57: That the relationship between x and y doesn't depend on m.

"a is independent of mass" was a conclusion, obtained through the fact that a=g.

But you have derived that a=g by doing a step which assumes that the relationship between a and g doesn't depend on m. Aren't you just concluding your previous assumption?

 

[Dale]

I have no problem with the proof in #42, under the assumption you state in post #57: That the relationship between x and y doesn't depend on m.

If x and y could depend on m and still be elements of a vector space then the proof in 42 would still hold. It is the vector nature that is important to the proof, not the independence.

That said, if you agree with the proof in 42, then I think that the math discussion is complete, and the rest is necessarily a physics discussion.