The Opposite of Nothing Is/Isn't Everything

[Laure]

Roger, of course you don't use "0" to "do" things, any more than you use "5" to "do" things. "0" and "5" are symbols; they are nouns. The operators (+,-,x,/) are the verbs; they are what "do" things.

I say again, show me your equality principles and postulates.

[Roger Bissell]

Here are the equality principles and postulates of my math:

Show me yours. I think what you'll find (best-case scenario) is that if you create a new symbol for your special "I can't say 'zero' so let's say 'undefined'", (0*0=#) you will find that that symbol "#" is redundant, and is equivalent to "0". Worst-case, you end up with a system that contains a contradiction.

At first blush, it appears to me that the only postulates I object to are the 3rd, 6th, and 8th ones with a double arrow, which I believe are the first of the commutative law pairs for addition, subtraction, and multiplication. I make an exception to them for z = 0, because (as you are probably way past tired of hearing), 0 does not operate on other numbers additively, subtractively, or multiplicatively, any more than it does divisively (in the 11th one with a double arrow).

Again, you haven't (yet) presented me with a contradiction (in my system) that withstands analysis.

REB

[Roger Bissell]

Here are the equality principles and postulates of my math:

Show me yours. I think what you'll find (best-case scenario) is that if you create a new symbol for your special "I can't say 'zero' so let's say 'undefined'", (0*0=#) you will find that that symbol "#" is redundant, and is equivalent to "0". Worst-case, you end up with a system that contains a contradiction.

At first blush, it appears to me that the only postulates I object to are the 3rd, 6th, and 8th ones with a double arrow, which I believe are the first of the commutative law pairs for addition, subtraction, and multiplication. I make an exception to them for z = 0, because (as you are probably way past tired of hearing), 0 does not operate on other numbers additively, subtractively, or multiplicatively, any more than it does divisively (in the 11th one with a double arrow).

Again, you haven't (yet) presented me with a contradiction (in my system) that withstands analysis.

REB

I also find it amusing that the first of the postulates is labeled "Additon-Subtraction-Oppositing." Shouldn't it be "Additon-Subtractitoff-Oppositing"? :rofl:

REB

[Roger Bissell]

Roger,

Your post #194 is fatally flawed. You tried to mix your premises, ones I disagree with, in with mine. 1*0 = 0. It being undefined is your premise, not mine.

Come on, Merlin, pay attention, fer chrissake!

Laure (whom you were defending) was trying to show how ~MY PREMISES~ lead to a contradiction, and I was showing how they do not, that any attempt to use the distribute law gambit ended up with OTHER expressions that were undefined BY MY PREMISES. In other words, she AND YOU have failed to show a contradiction in my argument so far.

"Fatally flawed" my pasty, white ass.

You people are better than this, aren't you?

REB

[Laure]

Roger,

Your post #194 is fatally flawed. You tried to mix your premises, ones I disagree with, in with mine. 1*0 = 0. It being undefined is your premise, not mine.

Come on, Merlin, pay attention, fer chrissake!

Laure (whom you were defending) was trying to show how ~MY PREMISES~ lead to a contradiction, and I was showing how they do not, that any attempt to use the distribute law gambit ended up with OTHER expressions that were undefined BY MY PREMISES. In other words, she AND YOU have failed to show a contradiction in my argument so far.

"Fatally flawed" my pasty, white ass.

You people are better than this, aren't you?

REB

I was trying to show the contradiction in your saying that 0*1=0 but 1*0 is "undefined". I think what I showed was that for your system to be consistent, 0*1 must also be "undefined". I think if we go through all the symbol manipulations, we will end up with everything in Roger's system being "undefined." Go through my postulates carefully and indicate all the modifications you would make, then we will see how things shake out.

[Roger Bissell]

Roger, of course you don't use "0" to "do" things, any more than you use "5" to "do" things. "0" and "5" are symbols; they are nouns. The operators (+,-,x,/) are the verbs; they are what "do" things.

Really? I always thought that, in performing mathematical operations, you were using numbers to do ~something~ to other numbers, whether to increase them, decrease them, &c. Like, when adding 2 and 3, i.e., in adding 3 to 2, you use 3 in a specific way: to increase 2. The 3 tells you that you are going to do ~something~ with it to the 2, and the + tells you that you use it to ~increase~ the 2 to 5. Just like adding chairs to a room or ingredients to a cake--in a general way of speaking.

I know Merlin thinks that I am butchering terminology by calling the second number in a calculation the "operator," so let me offer what I hope is a helpful clarification of what I mean. The "operators" are ~what~ you do with the second ~operating~ number to the first ~operated upon~ number. ~You~ are the ultimate "doer," of course, but the second number (the operating number) is ~what~ you use, in a certain way (specified by the operator), to do something to the first number (the number operated upon).

E.g., in 6 x 3, I am counting 3 groups of 6 things each. I am using the 3 to operate on the 6 by forming 3 groups of 6 things. Multiplication is the operation, the x or * sign is the "operator" (signifying the operation to be performed BY the second number UPON the first number), and the second number is the "operating number," and the first number is the "number operated upon." If this isn't in any textbook, it should be. :yes:

REB

P.S. -- I need to take a break. I have a rehearsal in Hollywood in a little over an hour, and I need a little time to drive and warmup. I'll be interested to see how the discussion has morphed when I return.

[Roger Bissell]

Roger,

Your post #194 is fatally flawed. You tried to mix your premises, ones I disagree with, in with mine. 1*0 = 0. It being undefined is your premise, not mine.

Come on, Merlin, pay attention, fer chrissake!

Laure (whom you were defending) was trying to show how ~MY PREMISES~ lead to a contradiction, and I was showing how they do not, that any attempt to use the distribute law gambit ended up with OTHER expressions that were undefined BY MY PREMISES. In other words, she AND YOU have failed to show a contradiction in my argument so far.

"Fatally flawed" my pasty, white ass.

You people are better than this, aren't you?

REB

I was trying to show the contradiction in your saying that 0*1=0 but 1*0 is "undefined". I think what I showed was that for your system to be consistent, 0*1 must also be "undefined".

But you didn't. I reject commutativity of arithmetic operations involving 0. The operations are invalid when 0 is the number operating on the other number. But when 0 is the number operated on, the operations are valid. I explained this in my reply to your attempt to use rows and columns to refute me.

I think if we go through all the symbol manipulations, we will end up with everything in Roger's system being "undefined." Go through my postulates carefully and indicate all the modifications you would make, then we will see how things shake out.

You imagine. You haven't shown anything being undefined as yet, except what I've already acknowledged, namely, the consequences of attempted operations using 0 as the number operating upon another number. They are all a consequence of my extending the denial of commutativity involving 0 from division to the other three arithmetic operations.

I'll be very surprised to find anything else that "shakes out" of this modification of your postulate system.

REB

[tjohnson]

Thus, like "i", which was discovered much later in history, the concept "0" is a methodological concept.

It is indeed! Imaginary numbers were invented to solve equations like x^2+1=0 and zero was invented to solve equations like x+1=1.

[Michael Stuart Kelly]

"Fatally flawed" my pasty, white ass.

That's quite a visual.

Is there an integer or symbol for information overload?

Qua referent, if ever there were a need to reify a zero (any old zero will do), that time is now!

Michael

[Dragonfly]

Roger's idea that 0^n = 1 leads to all kinds of absurdities. Many standard formulas would no longer work. Take for example the equation y = -x² + 1, this is a parabola with maximum y = 1 for x = 0 and y = 0 for x = -1 and x = +1. But according to Roger y(0) = 2! In his version the function has a discontinuity at x = 0 and it is no longer differentiable at that point, in other words, it's no longer a parabola. In general every polynomial function would have discontinuities.

The relativistic formula for the energy of a mass m moving with velocity v is E = mc²(1 - v²/c²)^-1/2. If v = 0 (mass at rest) the formula becomes the familiar E = mc², but according to Roger it would be E = mc² (1 - 1/c²)^-1/2. In units in which c = 1, E would even become infinitely large! Or Roger would probably say "undefined". Was Einstein wrong after all?

Another example: the electromagnetic energy density in vacuum = constant*(E² + B²), which would with Roger's math give the wrong result if either the electric field or the magnetic field is zero.

Who is losing touch with reality here?

[tjohnson]

Really? I always thought that, in performing mathematical operations, you were using numbers to do ~something~ to other numbers, whether to increase them, decrease them, &c. Like, when adding 2 and 3, i.e., in adding 3 to 2, you use 3 in a specific way: to increase 2. The 3 tells you that you are going to do ~something~ with it to the 2, and the + tells you that you use it to ~increase~ the 2 to 5. Just like adding chairs to a room or ingredients to a cake--in a general way of speaking.

This is true in applied math but in abstract math all you do is associate a pair of numbers with a third number. Groups can be made on non-numbers as well, it is a purely an associative thing.

[thomtg]

There is no such thing as zero groups.

Here are 2 sets (groups) of numbers: {1, 3, 5} {2, 4, 6, 8}

How many (a number) groups contain the number 7?

Merlin,

The question asked can be unpacked procedurally as,

1. Is 7 an element of the first set {1, 3, 5}? False

2. Is 7 an element of the second set {2, 4, 6, 8}? False

3. How many times was the operation "an element of" performed? 2

4. How many times did it fail to yield the result True? 2

5. How many times did it succeed? 2 - 2 = 0

Would you not agree that it is epistemology that establishes the criteria for postulating anything in any science, including mathematics? If at one time it was thought that concepts could in reality be formed without referentce or linkage to reality, and then at a later time it was rethought otherwise; wouldn't you say that that epistemological change in the standard of the unit would cause a tectonic change in every science that depended on it?

Thom,

My unpacking would be simpler. I count the number of sets containing 7, and the answer is 0. There is no need to count how many sets there are nor to subtract.

My answer to your first question is I agree, tentatively. I add 'tentative' because I don't know what you are trying to achieve by asking the question. What you are trying to achieve with your second question is even less clear. What do "change in the standard of the unit" and "tectonic change" refer to?

Merlin,

I believe we are getting close to an agreement about the nature of mathematics as a science of measurement, for you have agreed that in order to discover any mathematical fact at all, you have to engage in some mental operations (e.g., counting). It is on this basis that I have to disagree with you that you did not have to unpack procedurally the way I did and that you just had to count. Count what in reality? What you counted was not something you perceived in reality. You were counting your mental operations! Introspect, Sir, and see.

On the other question, the "unit" I had in mind is the one Ayn Rand defined in the second chapter of ITOE. With Rand, we have a new standard for this unit. Since, for example, the base of set-theory mathematics is built historically on mapping "set" onto this unit, Rand's new standard necessitates foundational changes to everything related to sets.

[merjet]

I believe we are getting close to an agreement about the nature of mathematics as a science of measurement, for you have agreed that in order to discover any mathematical fact at all, you have to engage in some mental operations (e.g., counting). It is on this basis that I have to disagree with you that you did not have to unpack procedurally the way I did and that you just had to count. Count what in reality? What you counted was not something you perceived in reality. You were counting your mental operations! Introspect, Sir, and see.

You presume way too much. I don't agree that mathematics is simply "the science of measurement." Nor do you know my mental operations and I assure you they did not match yours.

On the other question, the "unit" I had in mind is the one Ayn Rand defined in the second chapter of ITOE. With Rand, we have a new standard for this unit. Since, for example, the base of set-theory mathematics is built historically on mapping "set" onto this unit, Rand's new standard necessitates foundational changes to everything related to sets.

I saw your link. I agree that in ITOE Rand used "unit" in the first two senses -- a member of a set and a standard of measurement. Then she equivocated and arrived at the flawed idea of omitting measurements. If she had stuck simply with treating concepts as sets and instances as members of sets, her theory would have been better, but hardly revolutionary.

[merjet]

I'm sorry I don't know how to show it here properly formatted, but let I2 denote the 2-by-2 identity matrix, which looks like this enclosed in brackets:

1 0

0 1

In linear algebra I2*I2 = I2. But in RogerLand the operation on the left hand-side yields this:

undef. undef.

undef. undef.

Operation blocker, indeed!! Not 0, but Roger.

Edited May 27, 2009 by Merlin Jetton

[tjohnson]

If Roger rejects the commutatibility of operations involving 0 then x+0 ≠ 0+x. But that means 0 ≠ 0, since x=x.

[Dragonfly]

0 does not have a square. 0*0 is undefined. 0^n is NOT 0*0 n times. It is 1 * n factors of zero. There is no number that corresponds to n factors of zero. 0*0 n times is just as undefined as 0*0 is. So 0^n = 1.

This is inconsistent. You say that 0^n is 1 * n factors of zero. Then you say that there is no number that corresponds to n factors of zero, in other words, n factors of zero is undefined. But in that case 0^n must also be undefined, as 1 * [undefined number] = [undefined number] and not 1. However, you state that 0^n = 1, that implies that your 1 * n factors of zero = 1 * 1, or 1 = n factors of zero -> n factors of zero is not undefined, but equals 1 in your system -> contradiction. You assume in fact that multiplying by the undefined number is in your terms "doing nothing" = multiplying by the unit element, but the unit element = 1 is certainly not undefined.

Then the notion that 0 * 1 = 0 but 1 * 0 = undefined. Take 1 * (1 - 1). The distributive law gives 1 * 1 - 1 * 1 = 1 - 1 = 0, all perfectly defined. But if we calculate the expression between brackets first, we get 1 * 0 = undefined according to you -> contradiction.

[Laure]

A basic question, Roger. Do you agree that for all x, x+0=x?

If not, how do you do the algebraic simplification step that gets us from:

x + 0 = y + 1

to

x = y + 1?

If you agree that we can "drop the '+ 0'", how is that in any way different from "substituting 'x' in place of 'x + 0'"? Like my hero Spock says, "A difference that makes no difference is no difference." (He probably got it from some philosopher but I dunno... )

[Roger Bissell]

Roger's idea that 0^n = 1 leads to all kinds of absurdities. Many standard formulas would no longer work. Take for example the equation y = -x² + 1, this is a parabola with maximum y = 1 for x = 0 and y = 0 for x = -1 and x = +1. But according to Roger y(0) = 2! In his version the function has a discontinuity at x = 0 and it is no longer differentiable at that point, in other words, it's no longer a parabola. In general every polynomial function would have discontinuities.

The relativistic formula for the energy of a mass m moving with velocity v is E = mc²(1 - v²/c²)^-1/2. If v = 0 (mass at rest) the formula becomes the familiar E = mc², but according to Roger it would be E = mc² (1 - 1/c²)^-1/2. In units in which c = 1, E would even become infinitely large! Or Roger would probably say "undefined". Was Einstein wrong after all?

Another example: the electromagnetic energy density in vacuum = constant*(E² + B²), which would with Roger's math give the wrong result if either the electric field or the magnetic field is zero.

Who is losing touch with reality here?

Good question. Let's examine your first example. y = - (x)² + 1, for x = 0. You claim that the conventional approach says y = 0, while my approach says y = 2. Oh, really!? Let's do the math ~carefully~, shall we?

y = - (0)² + 1. Remember, on my interpretation of exponents, this means y = -1 (0)². And this means that -1 is NOT multiplied by ANY factors of 0, which means that y = -1 + 1 = 0.

Hey, what the...? I got the SAME ANSWER you did!! Something must have gone wrong. Someone must be losing touch with reality! :poke:

Your second example. The radicals get a bit unwieldy, but you are doing a clever little dipsy doodle there that needs pointing out. You want to saddle me with the consequences of the speed of light, c, being equal to 1. Well, the first thing to notice is that by ~your~ interpretation, E = m. So, if my interpretation shows the same result, then your critique falls flat on this example, too.

OK, the negative fractional radical appears in the denominator (under mc²) as the square root of this expression:

(1 - v²/c²), which can be rewritten as (c² - v²)/c².

Now, if v = 0, then (by my interpretation) v² = 1, and the denominator expression (inside the radical) reduces to (c² - 1)/c².

And you want to let the speed of light be expressed in units so that c = 1? OK, then the denominator expression reduces to the square root of 0, which is undefined. (By my interpretation, you cannot multiply ~any~ number by 0, so 0 * 0 is not permissible, which means that there are no two equal numbers which, when multiplied, produce 0.) Thus, we are asked to divide m by an undefined quantity, which cannot be done, which means no operation occurs, and the numerator remains unchanged by the denominator. This means that when c = 1 and v = 0, then E = m.

Whoa! Two for two! And 0 for 2 for you!

I do not understand the third example, so I'll forgo comment on it.

I get the impression that it's ~me~ that you think is psychotic, Dragonfly, else why would you insinuate I'm not in touch with reality, eh?, but at least I'm in good company!! In fact, you are a great role model, being (in the best case scenario, assuming I flop on dealing with example #3) ~at least~ twice as out of touch with reality in projecting my interpretation of your examples as I am in dealing with them! Congratulations! :cheer:

Let me rephrase that. Feh. You're pretty good at slinging insults, but you can't even plug values into an equation and follow premises accurately.

Next! And please remember, if you're going to show how my premises lead to a contradiction, you must try to APPLY my premises and see if a contradiction results, not omit them or set them aside when it's convenient for you.

Wait, never mind. This is really getting boring. I'm going to try to pick some fights with people about the Liar's Paradox for a while.

REB

[BaalChatzaf]

Wait, never mind. This is really getting boring. I'm going to try to pick some fights with people about the Liar's Paradox for a while.

REB

The new discipline you are are proposing (I will call it $mathematics to distinguish it from what real mathematicians do) produce incorrect answers for physics and engineering and even for bookkeeping. So of what use is $mathematics, as opposed to mathematics?

Ba'al Chatzaf

[merjet]

Good question. Let's examine your first example. y = - (x)² + 1, for x = 0. You claim that the conventional approach says y = 0, while my approach says y = 2. Oh, really!? Let's do the math ~carefully~, shall we?

y = - (0)² + 1. Remember, on my interpretation of exponents, this means y = -1 (0)². And this means that -1 is NOT multiplied by ANY factors of 0, which means that y = -1 + 1 = 0.

Hey, what the...? I got the SAME ANSWER you did!! Something must have gone wrong. Someone must be losing touch with reality! :poke:

No, you did not get the same answer as Dragonfly. He got the correct answer, y=1 when x=0. You got a wrong answer, y=0.

Now who is it that can't plug values into an equation and correctly compute the result? :poke:

Edited May 28, 2009 by Merlin Jetton

[Michael Stuart Kelly]

I'm sorry I don't know how to show it here properly formatted...

Merlin,

If you use a code symbol and it executes instead of displays, you can use the CODE tags (it's the button beside the QUOTE button above). Any text within those quote tags will be monospaced and will not execute.

If you have to use marks that extend over two or more lines, I suggest you do it in Word of another program that allows you to do that, hit the Print Screen button, paste it into an image program, crop the image and save it as a jpg, upload it to Photobucket (or other similar image host), then insert it as an image in your post.

Michael

[Rich Engle]

In the world of normal, balanced people, the subject of this thread would make them like...uh, gotta go wash the car, or something.

Is something everything is everything something nothing, I fergit?

Here: "Yes."

Everyone OK?

Or, work another old vauldevillian O-theme...Does Nothingness Exist?

I like to call this one "Is there shit other than the shit we know we have?"

It's either that or go watch Star Trek: they have the same problems.

r

daaaayyymmmm.....

Edited May 28, 2009 by Rich Engle

[Dragonfly]

Good question. Let's examine your first example. y = - (x)² + 1, for x = 0. You claim that the conventional approach says y = 0, while my approach says y = 2. Oh, really!? Let's do the math ~carefully~, shall we?

y = - (0)² + 1. Remember, on my interpretation of exponents, this means y = -1 (0)². And this means that -1 is NOT multiplied by ANY factors of 0, which means that y = -1 + 1 = 0.

Hey, what the...? I got the SAME ANSWER you did!! Something must have gone wrong. Someone must be losing touch with reality! :poke:

I may have forgotten the minus sign in the calculation according to your method, but as Merlin already pointed out, you still get the wrong answer, namely y = 0 for x = 0, while the standard calculation gives y = 1 for x = 0, so the conclusions about the discontinuity in the function in your system still stand.

Your second example. The radicals get a bit unwieldy, but you are doing a clever little dipsy doodle there that needs pointing out. You want to saddle me with the consequences of the speed of light, c, being equal to 1. Well, the first thing to notice is that by ~your~ interpretation, E = m. So, if my interpretation shows the same result, then your critique falls flat on this example, too.

I think you're missing the point. Also in the conventional system in which c is approximately 3 * 10⁸ m/s your method gives the wrong result. If you claim (incorrectly as I'll show below) that your method gives the correct result for v = 0, then your method gives different results (energy expressed in the same units) for different systems of units, which means that your version is unphysical.

OK, the negative fractional radical appears in the denominator (under mc²) as the square root of this expression:

(1 - v²/c²), which can be rewritten as (c² - v²)/c².

Now, if v = 0, then (by my interpretation) v² = 1, and the denominator expression (inside the radical) reduces to (c² - 1)/c².

And you want to let the speed of light be expressed in units so that c = 1? OK, then the denominator expression reduces to the square root of 0, which is undefined. (By my interpretation, you cannot multiply ~any~ number by 0, so 0 * 0 is not permissible, which means that there are no two equal numbers which, when multiplied, produce 0.) Thus, we are asked to divide m by an undefined quantity, which cannot be done, which means no operation occurs, and the numerator remains unchanged by the denominator.

Now you're making the same error that I indicated in my post #216: dividing a number by an undefined quantity is not the same as "no operation", the result is then also undefined. There are more indications that your method is incorrect: suppose that v = c, or v = 1 in modern units. Then your method gives exactly the same result as when v = 0 (in both cases v² = 1), so then according to your method E = m, in other words a finite value. However, in the standard treatment the result is E = m/0 = ∞, in other words the energy becomes infinitely large for any m > 0, which means that no object with a rest mass > 0 can travel at the speed of light. In your derivation that wouldn't be any problem, however!

I get the impression that it's ~me~ that you think is psychotic, Dragonfly, else why would you insinuate I'm not in touch with reality, eh?, but at least I'm in good company!! In fact, you are a great role model, being (in the best case scenario, assuming I flop on dealing with example #3) ~at least~ twice as out of touch with reality in projecting my interpretation of your examples as I am in dealing with them! Congratulations! :cheer:

Let me rephrase that. Feh. You're pretty good at slinging insults, but you can't even plug values into an equation and follow premises accurately.

When I suggest that it is you who is out of touch with reality, I mean that you bring your method as some grounding in reality of mathematics, while I've shown that you get with your method results that contradict reality. I don't mean it neither see that as an insult, it's a simple conclusion.

Next! And please remember, if you're going to show how my premises lead to a contradiction, you must try to APPLY my premises and see if a contradiction results, not omit them or set them aside when it's convenient for you.

I've done that in post #216, so the ball is in your court. From Saturday I'll be away on vacation for a few weeks, so if you don't hear anything from me, that doesn't mean that I agree...

[Rich Engle]

When I suggest that it is you who is out of touch with reality, I mean that you bring your method as some grounding in reality of mathematics, while I've shown that you get with your method results that contradict reality. I don't mean it neither see that as an insult, it's a simple conclusion.

I love you, but I still have to beat you.

Business as usual.

[Dragonfly]

Merely returning the compliment.