Dividing by m to make conclusions for m=0

[A.T.]

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 $m = 0$:

Even if some mathematician would object (which I doubt), it is clear that in physics canceling variables on both sides of an equation is a common and well-established technique. It is part of nearly every non-trivial proof that I have ever seen. So even if a mathematician would object to $mx=my$ for all m implying $x=y$ I don't think that any physicist would object to $ma=mg$ implying $a=g$.

To context is this proof that Newtonian gravity affects particles with $m = 0$:

It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.

Newtonian gravity is a force that exists between masses $(F=G m_1 m_2 / r^2)$. In his equations there would be zero gravitational force acting on zero mass. Since $F=ma$, $F/m=a$, but in this case that is $0/0=a$ which is undefined.

$F=ma$

$GMm/r^2=ma$

$gm=ma$

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

My question is not about physics, which can postulate whatever it wants. I 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 $m = 0$. In my opinion no, because the above step implicitly assumes that $m ≠ 0$, otherwise it would not be a valid operation. So whatever you derive by doing this step cannot be applied when $m = 0$.

 

[pbuk]

My question is not about physics, which can postulate whatever it wants. I would like to hear a mathematicians opinion, on whether you can apply something, that you derived by doing this:

The problem does not arise - physics tells us that $a = F/m$ and this clearly does not hold where $m = 0$ so when you write $gm = ma$ this already only holds when $m \ne 0$.

 

[Fredrik]

If I understand the question correctly, the answer is simply this: If mg=ma for all m, then g=1g=1a=a.

One can argue that "for all m" is ambiguous if we haven't specified what set we're talking about. So the statement should be something like "for all m in S, we have mg=ma". But what set S should we use? It doesn't really matter. All we need for the conclusion g=a to be valid is that S contains the real number 1.

Note that Newtonian mechanics doesn't even attempt to describe zero-mass particles, so the most natural choice for S is the set of positive real numbers, $S=\{x\in\mathbb R|x>0\}$.

 

[A.T.]

The problem does not arise - physics tells us...

As I said, I'm asking about the purely mathematical correctness of the above derivation, not about physics.

...that $a = F/m$ and this clearly does not hold where $m = 0$

But that is not used in the above derivation. What is used is $F=ma$, which is a valid statement for $m=0$, in which case it merely implies that $F=0$ regardless what $a$ is.

so when you write $gm = ma$ this already only holds when $m \ne 0$.

I disagree. Mathematically $gm = ma$ is a valid (and trivially true) statement for $m = 0$. The restriction to $m \ne 0$ is introduced by dividing both sides by $m$ to obtain $g=a$. That's why we cannot use $g=a$ for cases where $m = 0$ based on this derivation.

 

[pbuk]

I have read the thread I think you are referring to: DaleSpam's derivation is clearly flawed, however note Vanadium 50's contribution.

Of course Newtonian mechanics is based only on observations of massive objects and algebraic manipulation of its formulae cannot tell us anything about massless particles (which is just as well otherwise we would have trouble keeping their velocity at $c$ while accelerating at $g$): the only way to see if, for instance $a = g$ holds for massless particles is to observe them. Which of course we have done, and we see that it is true, and that both this phenomenon and Newtonian mechanics are explained by general relativity where we see that $a = F / m$ is not even true in general when $m \ne 0$

 

[A.T.]

If mg=ma for all m, then g=1g=1a=m.

Is that last 'm' supposed to be an 'a'? How do you go from mg=ma to g=a without dividing by m, which is not valid for all m.

 

[Fredrik]

I don't think I understood the question correctly. I'm still not sure I understand it. I tried to edit my post, but my computer bluescreened. Since several new posts have appeared, I will write a new post instead.

If we naively apply Newton's theory of gravity to a mass M=0, then it says that the acceleration of an object with mass m>0 is given by $ma=GMm/r^2=0$. Clearly this implies that a=0. But what about the acceleration of the object with mass M? We can't cancel M from $Ma=GMm/r^2=0$. But there's no physics in that equation anyway. It's only telling us that 0=0, and that doesn't tell us anything about either of the objects.

So a "theory" that tells us that that's the equation from which the acceleration is to be obtained, fails to make predictions and can't be considered a theory. This means that Newton's theory of gravity, in its usual form, is not a theory of zero mass particles.

However, there's nothing that prevents us from changing the "force law" to an "acceleration law" that says that the accelerations of the two masses m and M are given respectively by $a_m=GM/r^2$ and $a_M=Gm/r^2$. If all we're interested in is gravity, then there's no need to even introduce the concept of "force". What is "force" anyway? The way I see it, classical mechanics is built up around the theorem that says that a differential equation of the form $x''(t)=f(x'(t),x(t),t)$, where f is a nice enough function, has a unique solution for each initial condition. The "force" on an object with mass m>0 is then defined as $f(x'(t),x(t),t)/m$. Clearly, it wouldn't make sense to use this formula when m=0.

So I would say that the issue here isn't whether Newtonian gravity makes sense when one of the masses is 0. It's whether the concept of "force" does. And I would say that it doesn't.

 

[A.T.]

DaleSpam's derivation is clearly flawed, however note Vanadium 50's contribution.

Vanadium 50 misunderstood what the controversy was about. Nobody in the other thread had a problem with postulating $g=a$ as a fundamental physical law, which also applies when $m=0$. The controversy was about the math in DaleSpam's derivation based on $gm=am$ and it's applicability to a case where $m=0$.

 

[Fredrik]

Is that last 'm' supposed to be an 'a'? How do you go from mg=ma to g=a without dividing by m, which is not valid for all m.

Sorry about that typo. I have edited it in my post. I don't understand the question. I didn't divide by m to get g=a. I just used the defining property of the number 1, and the fact that the assumption was a "for all" statement.

 

[pbuk]

As I said, I'm asking about the purely mathematical correctness of the above derivation, not about physics.

No, you are asking about the correctness of the interpretation, which has nothing to do with mathematics.

But that is not used in the above derivation. What is used is $F=ma$

We have observed that $a = F/m$ for massive objects traveling at low speed. This enables us to conclude that $F = ma$ for massive objects traveling at low speed. It does not enable us to make any statements about massless objects, or objects which are traveling at high speed. Attempting to use it to predict the motion of a massless, light-speed photon is therefore doubly wrong.

I disagree. Mathematically $gm = ma$ is a valid (and trivially true) statement for $m = 0$. The restriction to $m \ne 0$ is introduced by dividing both sides by $m$ to obtain $g=a$. That's why we cannot use $g=a$ for cases where $m = 0$ based on this derivation.

I didn't state myself clearly - I agree with what you are saying.

 

[A.T.]

This means that Newton's theory of gravity, in its usual form, is not a theory of zero mass particles.

That was the point of those objected to DaleSpam's derivation. That you cannot algebraically derive the acceleration of zero mass particles from the Newton's force law of gravity.

However, there's nothing that prevents us from changing the "force law" to an "acceleration law"...

That was never contended in the other thread, and is not the topic of this thread. The issue was purely with the mathematical correctness of the derivation for m=0 from the "force law".

 

[Fredrik]

That was never contended in the other thread, and is not the topic of this thread. The issue was purely with the mathematical correctness of the derivation for m=0 from the "force law".

I still don't understand what this thread is about. Did someone say that the mg=ma law predicts that zero mass objects would have acceleration g?

mg=ma is based on the empirical observation that every time we drop something, the acceleration is approximately g. So it makes more sense to think of "mg=ma" as "g=a multiplied by m" than to think of "g=a" as "mg=ma divided by m".

 

[pbuk]

Better format for vectors

I think we are all saying the same thing - you cannot derive $\mathbf a = \mathbf g$ for massless objects using $\mathbf F = m \mathbf a$.

Frederik and I are going one stage further, by saying that $\mathbf F$ (and therefore $\mathbf F = m \mathbf a$ doesn't have any meaning for massless objects.

 

[A.T.]

We have observed that ...

Please note that I'm not asking about consistency with observation here. That's why I posted that under math not physics.

The restriction to $m \ne 0$ is introduced by dividing both sides by $m$ to obtain $g=a$. That's why we cannot use $g=a$ for cases where $m = 0$ based on this derivation.

I agree with what you are saying.

Thanks, that was the core of my question.

 

[A.T.]

I think we are all saying the same thing - you cannot derive $\mathbf a = \mathbf g$ for massless objects using $\mathbf F = m \mathbf a$.

Frederik and I are going one stage further, by saying that $\mathbf F$ (and therefore $\mathbf F = m \mathbf a$ doesn't have any meaning for massless objects.

Yes, thanks for the clarification.

 

[A.T.]

Did someone say that the mg=ma law predicts that zero mass objects would have acceleration g?

That is how I interpret DaleSpam's post. He responds to CKH, who states that the force law leads to undefined a for m=0, by deriving a=g from mg=ma, and concluding that a is well defined.

It seems to me that Newtonian gravity as described by Newton could not predict an effect on something mass-less.

Newtonian gravity is a force that exists between masses $(F=G m_1 m_2 / r^2)$. In his equations there would be zero gravitational force acting on zero mass. Since $F=ma$, $F/m=a$, but in this case that is $0/0=a$ which is undefined.

$F=ma$

$GMm/r^2=ma$

$gm=ma$

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

 

[Nugatory]

Family disputes should happen indoors, not in the yard in front of the neighbors. This thread is evolving into something that does not belong in the math sub forum.

Moving, and closing pending moderation.

 

[Dale]

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 $m = 0$:

A.T., you are again misstating my position. Setting aside the physics and concentrating on the math only, the mathematical question is the following: Given the proposition “$mx=my$ for all m”* what can you conclude about the relationship between x and y?

Note that the proposition contains two parts, “$mx=my$” and “for all m”. It is important to remember both parts of the proposition, since both parts must be satisfied. A.T., you consistently move the second part "for all m" to the conclusions instead of to the givens which results in an incorrect inference and which I am NOT claiming (it is a strawman argument). That is partly my fault since I did not consistently write it in the previous thread, I wrote it often, but not consistently every time.

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$, however that procedure is under dispute because of the concern about division by zero. So we seek another way to determine the relationship without using division at all. One other way to determine the relationship is to simply plug in all possible values for x, y, and m, and see which values satisfy the proposition.

Try, x=1 and y=1. That satisfies $mx=my$ for m=0, m=1, m=2, and in fact for all m, therefore it satisfies the proposition “$mx=my$ for all m”. Similarly, x=3 and y=3, satisfies $mx=my$ for m=0, m=1, m=2, and for all m, and therefore it also satisfies the proposition “$mx=my$ for all m”.

In contrast, x=1 and y=3 satisfies $mx=my$ for m=0, but not for m=1, nor m=2, nor any other m, therefore x=1 and y=3 does not satisfy the proposition “$mx=my$ for all m”.

A little thought shows that the only values for x and y which satisfy the proposition “$mx=my$ for all m” are the values $x=y$. Therefore, given the proposition “$mx=my$ for all m” we can indeed conclude that the relationship between x and y is unambiguously and without further qualification $x=y$. This is shown by examination of possible solutions and testing them directly against the proposition, without any use of division.

* we should also specify the set of admissible values for m, i.e. "for all m" should technically be "for all m in S" where S is some set appropriate for the problem. In the question that spawned this thread S would be the non-negative real numbers, but similar proofs would still be valid for other S.

 

[Dale]

OK, the thread is reopened for now. A.T. do you understand the two proofs given, how both make use of the "for all m" part of the proposition, and how neither is subject to criticism from division by 0?

 

[pbuk]

A.T., you are again misstating my position. Setting aside the physics and concentrating on the math only, the mathematical question is the following: Given the proposition “$mx=my$ for all m in the real numbers” what can you conclude about the relationship between x and y?

We have observed that $F = ma$ for massive objects traveling slowly. We can therefore make the proposition $mg = ma \ \mathrm{for} \ m > 0$. We CANNOT make the proposition that $mg = ma \mathrm{\ for\ } m \in \mathbb R$ any more than we can make the proposition that momentum of a massless particle is given by $P = mv = 0c = 0$.

 

[Dale]

I disagree, but the OPs question is only about the validity of the reasoning, not about the validity of the proposition.

 

[pbuk]

A little thought shows that the only values for x and y which satisfy the proposition “$mx=my$ for all m” are the values $x=y$.

That is true if x and y are values independent of m, but what if x (or in this case a) is a function of m: $a(m) = g \ (m > 0), 0 \ (m = 0)$? Your proof has to exclude that possibility - so in order to prove that a = g holds independent of mass, you have to assume that a is independent of mass.

 

[pbuk]

I disagree, but the OPs question is only about the validity of the reasoning, not about the validity of the proposition.

You cannot separate the reasoning from the proposition; in particular if the proposition is the equivalence of two functions of $m$ whose domains include $m = 0$ then division by $m$ is not a valid step of reasoning.

 

[Fredrik]

Is there really anything to discuss here other than this?

Theorem: For all $x,y\in\mathbb R$ such that $mx=my$ for all $m>0$, we have $x=y$.

Proof: Let $x,y$ be arbitrary real numbers such that $mx=my$ for all $m>0$. Since $1>0$ and $mx=my$ for all $m>0$, we have $x=1x=1y=y$.

This is the theorem that A.T. asked about, and its complete proof. Sure we can consider different theorems, but then we're going off topic again.

 

[Dale]

if the proposition is the equivalence of two functions of $m$ whose domains include $m = 0$ then division by $m$ is not a valid step of reasoning.

Neither of the proofs used division by m in their reasoning. They are not subject to that criticism.

 

[Dale]

Is there really anything to discuss here other than this?

The only other thing that I can see to discuss is the generalization of the proof to handle $\forall m \in S$ where S is different from "positive reals". Your same proof would also hold for S is "non-negative reals" (which is of particular interest to the OP), "all reals", "all complex numbers", "all integers", ... I think that the only requirement is that $1 \in S$.

 

[A.T.]

We have observed that...

DISCLAIMER: Consistency with observation is irrelevant here. This thread is not about physics. It was moved from math to physics sub forum against the wish of the topic starter.

 

[A.T.]

Theorem: For all $x,y\in\mathbb R$ such that $mx=my$ for all m>0, we have $x=y$.

No, that is not the claim by DaleSpam. His proposition contains "for all m" not "for all m > 0":

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$

Therefore, given the proposition “$mx=my$ for all m” we can indeed conclude that the relationship between x and y is unambiguously and without further qualification $x=y$.

 

[A.T.]

About Misunderstandings:

Note that the proposition contains two parts, “$mx=my$” and “for all m”.

I always understood it like that, even when “for all $m$” wasn't stated explicitly. Unless stated otherwise, I assume no constraints on the value of $m$.
About dividing by m:

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$,...

Unless stated otherwise, I assume no constraints on the value of $m$. So I assume that the last $x=y$ means "$x=y$ for all $m$". Is that correct?

...however that procedure is under dispute because of the concern about division by zero.

Do you dispute the following?:

Dividing an equation by m is only valid under the assumption that m≠0. So whatever is derived by dividing by m is also only valid under the assumption that m≠0.

Explicitly written that means:

$mx=my$ for all $m$ | divide both sides by $m$

$x=y$ for all $m≠0$
​

About the workaround that avoids dividing by m:

Therefore, given the proposition “$mx=my$ for all m” we can indeed conclude that the relationship between x and y is unambiguously and without further qualification $x=y$.

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 $m$.

 

[Fredrik]

No, that is not the claim by DaleSpam. His proposition contains "for all m" not "for all m > 0":

OK, but what does "for all" mean? In mathematics based on set theory, it should mean "for all sets" unless we say otherwise, but it can't mean "for all sets" here, because multiplication isn't defined for arbitrary sets. So the person who writes the statement must have some specific set in mind. Let's denote that set by S. When he writes "for all m", it means "for all m in S".

It makes no difference if S is the set of positive real numbers or the set of non-negative real numbers, because if S is any subset of $\mathbb R$ that contains the number 1, we get the result x=y (without using division):

Theorem: $\forall x,y\in\mathbb R\ ((\forall m\in S\ mx=my)\ \Rightarrow\ x=y))$

Proof: Let x,y be arbitrary real numbers. We want to prove the implication $(\forall m\in S\ mx=my)\ \Rightarrow\ x=y$, so suppose that mx=my for all m in S. Since $1\in S$, this assumption implies that $x=1x=1y=y$. So the implication is true. Since x,y are arbitrary real numbers, this proves the theorem.

 

[Fredrik]

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. 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 $x=y$.

 

[pbuk]

Neither of the proofs used division by m in their reasoning. They are not subject to that criticism.

The proof referred to in the original post did as you confirmed:

I claim that given "$mx=my$ for all m" you can simply divide out the m and get $x=y$

... and I think we are all agreed that the operation of dividing out the m (and therefore that proof) is invalid when $m = 0$. You have now presented a new argument at the crux of which is this statement:

the only values for x and y which satisfy the proposition “$mx=my$ for all m” are the values $x=y$.

against which I provide a counter-example:

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

 

[Fredrik]

the only values for x and y which satisfy the proposition “$mx=my$ for all m” are the values $x=y$.

against which I provide a counter-example:

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

I think it's clear that what DaleSpam meant here is that if we want to assign values to x and y that make "mx=my for all m" a true statement, we have to assign the same value to both x and y. I would interpret x and y as variables that represent real numbers, since nothing else was explicitly stated. If that's the intended interpretation, then DaleSpam isn't wrong (at least not about this).

A counterexample to such a statement should assign values to both variables, not just describe a relationship between them. It looks like you would use something like this: x=y=1 when m≠0, and x=0, y=1 when m=0. But if x and y are variables that represent real numbers, this isn't an assignment of values. It's an infinite number of assignments (one for each m). So it's not a counterexample to the claim that DaleSpam made there.

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.

 

[Dale]

I always understood it like that, even when “for all $m$” wasn't stated explicitly. Unless stated otherwise, I assume no constraints on the value of $m$.

"For all m" doesn't mean that there are no constraints on m. In fact, there can be constraints on m in the form of defining the admissible set membership for m. What "for all m" means is that, whatever the set of admissible m is, the relationship holds for every single member of that set.

Personally, I was taking the set of admissible m to be all real numbers, but my proof works for all admissible sets which include at least one non-zero number, and Fredrik's proof works for all admissible sets which include the number 1.

Unless stated otherwise, I assume no constraints on the value of $m$. So I assume that the last $x=y$ means "$x=y$ for all $m$". Is that correct?

I don't know what "$x=y$ for all m" means since m is not in the expression $x=y$. To me, it doesn't make any sense to append the "for all m" to that, but I suppose that there is no harm in doing it (although it seems strange and I certainly see no need to do it either).

 

[Dale]

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 $m$.

against which I provide a counter-example:

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

These do not satisfy the proposition. If x depends on 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.

 

[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.