Numerus “Numerans-numeratus”

AndyDora

You misunderstand. I was not talking about your notation. I was talking about the intuition behind it. Your notation is trying to give a representation to a division that is not happening. The idea that the result of the division is that you place the numerical quantity into a dimensional space is how you are representing the fact that you are assigning division by zero the result of the original numerical quantity. The result is equivalent to the division simply not happening. That is incontrovertible, regardless of what your notation accomplishes or the actual process by which you arrive at that result.
I accept and appreciate you conclusion that this is not intuitive and clunky....I have thought considerably about how to address this....however...

0-0...is an acceptable act. That does occur.

The act of placing a symbol denoting the absence of value into a dimensional space....is possible.

(0) = the absence of a numerical quantity

(_) = a dimensional unit

(0) = an absence of numerical quantity "placed" into a dimensional unit...then all subtracted except one.

It did occur....."What am I missing" this seems perfectly plausible to me?

Thanks...I apologize for continually misunderstanding you....

SlipEternal

MHF Helper
I accept and appreciate you conclusion that this is not intuitive and clunky....I have thought considerably about how to address this....however...

0-0...is an acceptable act. That does occur.

The act of placing a symbol denoting the absence of value into a dimensional space....is possible.

(0) = the absence of a numerical quantity

(_) = a dimensional unit

(0) = an absence of numerical quantity "placed" into a dimensional unit...then all subtracted except one.

It did occur....."What am I missing" this seems perfectly plausible to me?

Thanks...I apologize for continually misunderstanding you....
You are looking only at your process and not at the result. The result is equivalent to having no division take place at all. You are describing a process by which you arrive at the same result as if you had not done the division.

Consider multiplication by 1. The result of the multiplication is equivalent to had no multiplication taken place at all. That was my point. That was my only point. I was not commenting on your chosen method for describing this interaction. I was commenting on a more philisophical understanding of what the result might signify, and how I interpret the result.

1 person

AndyDora

You are looking only at your process and not at the result. The result is equivalent to having no division take place at all. You are describing a process by which you arrive at the same result as if you had not done the division.

Consider multiplication by 1. The result of the multiplication is equivalent to had no multiplication taken place at all. That was my point. That was my only point. I was not commenting on your chosen method for describing this interaction. I was commenting on a more philisophical understanding of what the result might signify, and how I interpret the result.
In the case of (a * 1_)

(1_) = the dimensional unit quantity of (1)
(a) = (a) placed into the dimensional unit quantity of (1_), then all (a) added.

In this case the end result is shown to be specifically because of the definition of multiplication.
In this case the operation does occur.

Perhaps then the idea that all operations are considered to occur: is enough of a "need" for the application of this idea?

Would it be possible for you to suggest a way in which I might clean this up...that is make it more intuitive and less clunky.

I understand you have already decided to move on to something else...I appreciate any comments and directions you have or may yet give.

SlipEternal

MHF Helper
In the case of (a * 1_)

(1_) = the dimensional unit quantity of (1)
(a) = (a) placed into the dimensional unit quantity of (1_), then all (a) added.

In this case the end result is shown to be specifically because of the definition of multiplication.
In this case the operation does occur.

Perhaps then the idea that all operations are considered to occur: is enough of a "need" for the application of this idea?

Would it be possible for you to suggest a way in which I might clean this up...that is make it more intuitive and less clunky.

I understand you have already decided to move on to something else...I appreciate any comments and directions you have or may yet give.
One thing that I will say is that you really need to look beyond your notation. I brought up a point that had nothing to do with your notation, and in every response you made, you ignored my point and tried to bring it back to your notation. You basically eluded to the idea that my point was incorrect. You are in a situation where you are too close to the project, and it is making it difficult to see any point of view other than your own. I recognize that you are trying, which is why I am continuing to help, but I urge you to put your notation aside for a while and try to think about the bigger picture. Here is an example:

Let's give your number system a symbol. We have $\mathbb{R}$ is real numbers. Let's refer to your number system as $\mathfrak{N}$ for the title you have given it. Consider the function $f:\mathbb{R}\setminus \{0\} \to \mathbb{R}\setminus \{0\}$ defined by $f(x) = \dfrac{1}{x}$. This function is continuous everywhere. Now, consider the same function, but over the one-point compactification (the real numbers union the point at infinity, division by zero is defined to be the point at infinity, although 0/0 is still undefined, and division by infinity gives 0): $f:\overline{\mathbb{R}} \to \overline{\mathbb{R}}$ defined by $f(x) = \dfrac{1}{x}$. This function is also continuous (and includes division by zero). Next, consider this same function over your number system: $f:\mathfrak{N} \to \mathfrak{N}$ defined by $f(x) = \dfrac{1}{x}$. This is discontinuous at $x=0$. This is an important distinction if we are to use your number system for any analytic mathematics, and while not a flaw in your system, it is definitely inconvenient. An approach that would push me away would be to try to find some notation that will smooth this over. An approach that would draw me in would be a recognition of this mathematical inconvenience, and either a point that it will remain inconvenient or perhaps even an attempt to minimize its inconvenience.

You might also view the work of Mahāvīra, an Indian mathematician who in 850 AD (or 830 AD, depending on the source you look at) came up with a similar definition for division by zero. He said, "A number divided by zero remains unchanged." His work was proven to yield algebraic inconsistencies. While I have not studied his work, you may want to. Find what he did wrong. Perhaps your work corrects it. Perhaps not.

Anyway, for your notation, I recommend learning more mathematics. Once you have seen how other mathematicians have generated their notations, you will find some notation you like and some you do not like. Perhaps your notation will be a continual work in progress until you finally feel you have perfected it. But, I urge you to keep an open mind and try to think beyond your current notation.

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AndyDora

One thing that I will say is that you really need to look beyond your notation. I brought up a point that had nothing to do with your notation, and in every response you made, you ignored my point and tried to bring it back to your notation. You basically eluded to the idea that my point was incorrect. You are in a situation where you are too close to the project, and it is making it difficult to see any point of view other than your own. I recognize that you are trying, which is why I am continuing to help, but I urge you to put your notation aside for a while and try to think about the bigger picture. Here is an example:

Let's give your number system a symbol. We have $\mathbb{R}$ is real numbers. Let's refer to your number system as $\mathfrak{N}$ for the title you have given it. Consider the function $f:\mathbb{R}\setminus \{0\} \to \mathbb{R}\setminus \{0\}$ defined by $f(x) = \dfrac{1}{x}$. This function is continuous everywhere. Now, consider the same function, but over the one-point compactification (the real numbers union the point at infinity, division by zero is defined to be the point at infinity, although 0/0 is still undefined, and division by infinity gives 0): $f:\overline{\mathbb{R}} \to \overline{\mathbb{R}}$ defined by $f(x) = \dfrac{1}{x}$. This function is also continuous (and includes division by zero). Next, consider this same function over your number system: $f:\mathfrak{N} \to \mathfrak{N}$ defined by $f(x) = \dfrac{1}{x}$. This is discontinuous at $x=0$. This is an important distinction if we are to use your number system for any analytic mathematics, and while not a flaw in your system, it is definitely inconvenient. An approach that would push me away would be to try to find some notation that will smooth this over. An approach that would draw me in would be a recognition of this mathematical inconvenience, and either a point that it will remain inconvenient or perhaps even an attempt to minimize its inconvenience.

You might also view the work of Mahāvīra, an Indian mathematician who in 850 AD (or 830 AD, depending on the source you look at) came up with a similar definition for division by zero. He said, "A number divided by zero remains unchanged." His work was proven to yield algebraic inconsistencies. While I have not studied his work, you may want to. Find what he did wrong. Perhaps your work corrects it. Perhaps not.

Anyway, for your notation, I recommend learning more mathematics. Once you have seen how other mathematicians have generated their notations, you will find some notation you like and some you do not like. Perhaps your notation will be a continual work in progress until you finally feel you have perfected it. But, I urge you to keep an open mind and try to think beyond your current notation.
I must admit I am not following you. I do not understand why my version is discontinuous. I realize you have done your best to explain this to me. I will study it further on my own. I will due as you suggest and continue on my journey. Glad to hear this is not a inconsistency, but clearly you are right I must acknowledge this "discontinuance" thing, and be able to address it is some for or another.....thank you again very much.

1 person

SlipEternal

MHF Helper
I must admit I am not following you. I do not understand why my version is discontinuous. I realize you have done your best to explain this to me. I will study it further on my own. I will due as you suggest and continue on my journey. Glad to hear this is not a inconsistency, but clearly you are right I must acknowledge this "discontinuance" thing, and be able to address it is some for or another.....thank you again very much.
The definition of continuity is that $f(x)$ is continuous at $x=a$ if $\displaystyle \lim_{x \to a} f(x) = f(a)$.

Let's consider the sequence $a_n = \dfrac{1}{n}$. It should be clear that $\displaystyle \lim_{n \to \infty} a_n = 0$. Example, $\dfrac{1}{2}>\dfrac{1}{3}> \cdots > \dfrac{1}{50}$, etc. You are getting closer and closer to zero. Now let's consider the division you propose:

$\displaystyle \lim_{n \to \infty} f(a_n) = \lim_{n \to \infty} \dfrac{1}{\tfrac{1}{n}} = \lim_{n \to \infty} n = \infty \neq f(0) = 1$. Therefore, $f:\mathfrak{N} \to \mathfrak{N}$ is discontinuous at zero.

Edit: I should probably explain a little more why this is inconvenient. Suppose you are doing geometry. You try to calculate the slope of the line $x=4$. You choose two points: $(4,0)$ and $(4,1)$. You calculate that the slope of the line is $\dfrac{1}{0} = 1$. Someone else comes along and chooses the points $(4,0)$ and $(4,2)$ and calculates the slope of the line is $\dfrac{2}{0} = 2$. This is the same line, but it has every real number as a possible slope.

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AndyDora

The definition of continuity is that $f(x)$ is continuous at $x=a$ if $\displaystyle \lim_{x \to a} f(x) = f(a)$.

Let's consider the sequence $a_n = \dfrac{1}{n}$. It should be clear that $\displaystyle \lim_{n \to \infty} a_n = 0$. Example, $\dfrac{1}{2}>\dfrac{1}{3}> \cdots > \dfrac{1}{50}$, etc. You are getting closer and closer to zero. Now let's consider the division you propose:

$\displaystyle \lim_{n \to \infty} f(a_n) = \lim_{n \to \infty} \dfrac{1}{\tfrac{1}{n}} = \lim_{n \to \infty} n = \infty \neq f(0) = 1$. Therefore, $f:\mathfrak{N} \to \mathfrak{N}$ is discontinuous at zero.

Edit: I should probably explain a little more why this is inconvenient. Suppose you are doing geometry. You try to calculate the slope of the line $x=4$. You choose two points: $(4,0)$ and $(4,1)$. You calculate that the slope of the line is $\dfrac{1}{0} = 1$. Someone else comes along and chooses the points $(4,0)$ and $(4,2)$ and calculates the slope of the line is $\dfrac{2}{0} = 2$. This is the same line, but it has every real number as a possible slope.

I am trying...but this one will take time.......thank you, very much.

Warning Speculation...

I feel I have a philosophical answer for this. Probably not worth your time...but if it happens to interest you...

*I understand "philosophy is not the proper dialogue to exchange here"...but perhaps you might be inclined...after you hear this.

I have thought about this many times before. I however did not have any idea how to discuss this mathematically...as you have just shown me.

So...philosophically and mathematically speaking....

It is with this system that I gave, that approaching zero is unique. Zero is defined as empty space...so then as we "move" on a number line....we "approach" only infinitesimals....that is to reach "zero", philosophically or mathematically, you must change the "perspective" of the number line....examples....

If I am "approaching" zero on the number line I will reach a level of infinitesimals no longer measurable by numerical quantity or dimensional quantity....However if I denote varying amounts of "zero" in some fashion.....I might say that the number line "resets".
That is the number infinitesimal (.0...1)...is what is yielded by the equation we are discussing...and that:

(.0...1)from zero2 = 1 from zero1

a continuum if you will.....also this ...

So consider a black whole...the mathematics of which yields divsion by zero....if I am asking "what goes into the black whole"...

So that anything going in is on a number line approaching zero...it reaches this point of un-measurable numerical quantity and dimensional quantity...but it does not stop existing....it still exists...but the "change" of the dimensional space and numerical quantity is now viewed from a a new dimension notated with zero2.

On the last note...any slope in this form of mathematics can be considered vertical...so no matter the number given (a/0) : a=/=0, then the result is the same...

a slope of 2/0 = vertical
a slope of 1/0 = vertical

This last bit probable not yet sufficient.

SlipEternal

MHF Helper
If you wind up with a more complex process to give the same geometric results as the standard number system, the obvious question is, what value is your number system. You have a few options. You can give up on slopes and say that your number system will not give value to calculations of slope (one of the more common instances where division by zero occurs). Then the obvious question is, when does your division by zero become useful? When will $\dfrac{a}{0}=a$ have some sort of physical meaning that will be useful to a mathematician, philosopher, or anyone else? If this is nothing more than a theoretical construction, is it worth anyone's time? I do not have the answer to that.

Another possibility is to force the math to work with your system. It will still yield the same result as more traditional mathematics when it comes to slope, but it will still use your division. Define $\mathcal{L}(\mathfrak{N}^2)$ to be the set of all lines in two dimensions. Given any line $L \in \mathcal{L}(\mathfrak{N}^2)$ define slope to be $\displaystyle \lim_{\begin{matrix}(x_1,y_1),(x_2,y_2) \in L \\ d((x_1,y_1),(x_2,y_2)) \to \infty \end{matrix}}\dfrac{y_2-y_1}{x_2-x_1}$. This is probably not the best notation, but it is the best I can come up with this late at night. It basically says that you are defining the slope of the line to be the limit as you take the slope between points that are increasingly far apart on the line. For non-vertical lines, it is easy to show that because slope is constant, the limit will equal the constant slope. It is unchanging no matter how far apart the two points you choose are. For vertical lines, you wind up with $x_1=x_2$ no matter how far apart the two points are, and you wind up with the numerator getting increasingly large in the positive or negative direction. In order for this limit to exist, you need to define a single point at infinity (basically causing your numbers to wrap around). For the purposes of slope, vertical lines would wind up with a slope of infinity.

But, this has its own weirdness. If you allow an infinite slope, and you are writing lines in the form $y=mx+b$ where $b$ is the value of the $y$-intercept, can you write $y=\infty x + b$ for a vertical line? Not really, because there is no $y$-intercept. So how about writing the line as $y-y_1 = m(x-x_1)$. This becomes $y-y_1 = \infty(x-x_1)$, which again is a rather meaningless notation. If you cannot really do algebra with infinity, what purpose does it have? Are you basically saying that your number system does not work with vertical slopes?

Whichever way you go, this is not really different from existing (and simpler) algebras that already exist without dimensional units. So, again, the question becomes, what value does your system provide that these other systems do not?

You bring up infinitesimals. There is already a rigorously defined theory that includes infinitesimals. It is the theory of Hyperreal numbers: https://en.wikipedia.org/wiki/Hyperreal_number

Before we progress any further, perhaps come up with some concrete examples of how your theory might be useful. Without some solid examples of when it is useful to have dimensional units, I cannot justify to myself any further work on this. If you want to continue thinking about it, that is fine, but without some understanding of how it might further mathematical knowledge, it seems like a waste of time.

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2 people

AndyDora

If you wind up with a more complex process to give the same geometric results as the standard number system, the obvious question is, what value is your number system. You have a few options. You can give up on slopes and say that your number system will not give value to calculations of slope (one of the more common instances where division by zero occurs). Then the obvious question is, when does your division by zero become useful? When will $\dfrac{a}{0}=a$ have some sort of physical meaning that will be useful to a mathematician, philosopher, or anyone else? If this is nothing more than a theoretical construction, is it worth anyone's time? I do not have the answer to that.

Another possibility is to force the math to work with your system. It will still yield the same result as more traditional mathematics when it comes to slope, but it will still use your division. Define $\mathcal{L}(\mathfrak{N}^2)$ to be the set of all lines in two dimensions. Given any line $L \in \mathcal{L}(\mathfrak{N}^2)$ define slope to be $\displaystyle \lim_{\begin{matrix}(x_1,y_1),(x_2,y_2) \in L \\ d((x_1,y_1),(x_2,y_2)) \to \infty \end{matrix}}\dfrac{y_2-y_1}{x_2-x_1}$. This is probably not the best notation, but it is the best I can come up with this late at night. It basically says that you are defining the slope of the line to be the limit as you take the slope between points that are increasingly far apart on the line. For non-vertical lines, it is easy to show that because slope is constant, the limit will equal the constant slope. It is unchanging no matter how far apart the two points you choose are. For vertical lines, you wind up with $x_1=x_2$ no matter how far apart the two points are, and you wind up with the numerator getting increasingly large in the positive or negative direction. In order for this limit to exist, you need to define a single point at infinity (basically causing your numbers to wrap around). For the purposes of slope, vertical lines would wind up with a slope of infinity.

But, this has its own weirdness. If you allow an infinite slope, and you are writing lines in the form $y=mx+b$ where $b$ is the value of the $y$-intercept, can you write $y=\infty x + b$ for a vertical line? Not really, because there is no $y$-intercept. So how about writing the line as $y-y_1 = m(x-x_1)$. This becomes $y-y_1 = \infty(x-x_1)$, which again is a rather meaningless notation. If you cannot really do algebra with infinity, what purpose does it have? Are you basically saying that your number system does not work with vertical slopes?

Whichever way you go, this is not really different from existing (and simpler) algebras that already exist without dimensional units. So, again, the question becomes, what value does your system provide that these other systems do not?

You bring up infinitesimals. There is already a rigorously defined theory that includes infinitesimals. It is the theory of Hyperreal numbers: https://en.wikipedia.org/wiki/Hyperreal_number

Before we progress any further, perhaps come up with some concrete examples of how your theory might be useful. Without some solid examples of when it is useful to have dimensional units, I cannot justify to myself any further work on this. If you want to continue thinking about it, that is fine, but without some understanding of how it might further mathematical knowledge, it seems like a waste of time.
How about it simply being a mathematical construct to solve for division by zero errors in computer programming?

Also here is an example of how it might help psychology: given more work....

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3114072/

That link is quite ironic considering the verbiage.....

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SlipEternal

MHF Helper
Meadows are being investigated for avoiding division by zero errors. I'm asking for concrete examples of when $\dfrac{a}{0}=a$ is a desirable result. What does it mean? Just because it is possible to do something does not mean that there is value in doing it. It is possible to learn Klingon. Some people do learn Klingon. But, in general, it is not a valuable language. There are very few speakers and very little literature. Other than as an intellectual curiosity, it serves little purpose.

Is division by zero meaningful? If Bob has five friends and he divides them into zero groups, what does that mean? This is where my mind was when I said it is as if no division happened. But, I am not sure that answer is desirable. There is value in a computer generating a division by zero error. I actually work as a software developer. I've never been in a situation where I thought, I wish I could divide by zero! It has never had value in my work. Under what conditions would software be better if division by zero were possible?

3 people