ruveyn
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 Posted: Sun Dec 25, 2011 9:52 pm Post subject: Re: Real Number Operations Caveat |
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| jackmt wrote: |
You cannot seem to think beyond what you have been taught. If you disagreed that would be one thing, but clearly you cannot comprehend; in your 3 responses you have not even once addressed what I have said. I'm done. |
What you have said makes no mathematical sense.
ruveyn |
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jackmt
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| Posted: Sun Dec 25, 2011 10:19 pm Post subject: Re: Real Number Operations Caveat |
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| ruveyn wrote: | | jackmt wrote: |
You cannot seem to think beyond what you have been taught. If you disagreed that would be one thing, but clearly you cannot comprehend; in your 3 responses you have not even once addressed what I have said. I'm done. |
What you have said makes no mathematical sense.
ruveyn |
Again, clearly you cannot comprehend. Nor can you even address the issues I have raised. |
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ruveyn
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| Posted: Mon Dec 26, 2011 4:03 am Post subject: Re: Real Number Operations Caveat |
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| jackmt wrote: | | ruveyn wrote: | | jackmt wrote: |
You cannot seem to think beyond what you have been taught. If you disagreed that would be one thing, but clearly you cannot comprehend; in your 3 responses you have not even once addressed what I have said. I'm done. |
What you have said makes no mathematical sense.
ruveyn |
Again, clearly you cannot comprehend. Nor can you even address the issues I have raised. |
That is right. I deal with mathematics, not nonsense.
ruveyn |
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Ahaseurus2000
My Great-Granddady tried to blow up Hitler!
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| Posted: Mon Dec 26, 2011 5:49 am Post subject: |
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when a number is raised to a power, and the power is odd, a negative number will have a negative solution, as there are an odd amount of factors being multiplied. e.g. -4 raised by the power of 3 is equal to -4 x -4 x -4. the first multiplication gives +16 (as negative times negative always gives a positive) and that times the remaining -4 will always give a negative.
but any number raised by an even power always has a positive solution, because of what happens with the first multiplication mentioned above. e.g. -4 to the power of 6 is -4 x -4 x -4 x -4 x -4 x -4. imagine we pair off the -4s, and multiply to get 16 x 16 x 16, which gives 4096. and if you miltiplied the -4s other ways you always get the same result.
This is what the OP's statement means: any negative real number raised to an even power always and only gives a positive real solution. Therefore only a positive real number will give a real solution when an even root is applied.
if a negative real number is lowered by an even root, it think it becomes an imaginary number. The simpliest to remember is i which is from the square root of -1. I am unsure about what happens with higher even roots and whether imaginary solutions occur, though.
Also, while any real number divided by zero has no solution and is undefined, the limit (the value it seems to tend towards) of any real number divided by zero is infinity. As infinity is not a defined number, a divide by zero is undefined. This is why when you graph y=1/x, you get an asymptote at x=0, and the limits either side approach infinity. The asymptote shows where an undefined region or line occurs.
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lau
Really nice person to know. :)
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| Posted: Mon Dec 26, 2011 8:39 am Post subject: Re: Real Number Operations Caveat |
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| jackmt wrote: | | lau wrote: | | jackmt wrote: |
Math is abstracted from language. It is a language with its own syntax, logical operators (verbs), substantives (e.g. numbers), pronouns, (variables), negation, etc. It also has implicit, i.e., non-overt elements: 1 for example, is deleted as multiplier because it is the identity element of multilplication. Similarly with zero and addition. This is not dreck.
And I know that 0^n = 0. You answer as if I had made some simple grade school math error, but you do not even address what I have said. Please check your attitude and look again. And what is nonsensical about what I presented? If it is wrong, show me. But don't insult me and dismiss me. It makes me think you can't answer me. You haven't so far. |
None of what you say has any weight.
"0^n" is zero, provided "n" is not zero.
With "z" in the complex domain, even...
"0^z" is zero, provided "z" is not zero.
You have made a "simple grade school math error".
"z^0" is unity, provided "z" is not zero.
0^0 is undefined. |
You cannot seem to think beyond what you have been taught. If you disagreed that would be one thing, but clearly you cannot comprehend; in your 3 responses you have not even once addressed what I have said. I'm done. |
I went back over what you have been saying, starting from:
| jackmt wrote: | ...
Concerning division by 0; I wrote a linguistic proof that 0^0 (zero raised to the zero power) = 1 and is therefore a legitimate divisor. Any one agree? disagree? |
(I've no idea exactly what you mean by a "legitimate divisor".)
Here you made the assertion that you had a "linguistic proof". I vaguely thought that you would eventually present one of the usual fallacies. Your subsequent posts just skirted around with words and gave no coherent "proof" - just a series of bald statements with no attempt at structuring them into a proof of any kind. Viz:
| jackmt wrote: | | Division by zero is undefined. Agreed? |
No. Division is defined for all elements of Z, with the exception of zero. Division by zero is not undefined. It is perfectly well defined. It does not exist. In your terminology, zero would be the unique value which is not a "legitimate divisor".
| jackmt wrote: | | In my original claim I said I have a linguistic proof; ... |
We're still waiting for this.
| jackmt wrote: | | ... i.e., 0^0=1 for grammatical reasons. |
Just stating that your assertion is true "for grammatical reasons", without making any attempt to give those grammatical reasons, doesn't make it true. (Unless you say it three times?)
| jackmt wrote: | | ... 1 for example, is deleted as multiplier ... |
I have no idea what you mean by "deleted" in this.
| jackmt wrote: | | And I know that 0^n = 0. |
Which is where you are going wrong. If you start with a premise which is not true, you can get any answer to any question.
....
If you would like a good book to read, " Surreal Numbers: How Two Ex-Students Turned on to Pure Mathematics and Found Total Happiness" is rather lovely.
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jackmt
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| Posted: Mon Dec 26, 2011 10:24 pm Post subject: Re: Real Number Operations Caveat |
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| lau wrote: | | jackmt wrote: | | lau wrote: | | jackmt wrote: |
Math is abstracted from language. It is a language with its own syntax, logical operators (verbs), substantives (e.g. numbers), pronouns, (variables), negation, etc. It also has implicit, i.e., non-overt elements: 1 for example, is deleted as multiplier because it is the identity element of multilplication. Similarly with zero and addition. This is not dreck.
And I know that 0^n = 0. You answer as if I had made some simple grade school math error, but you do not even address what I have said. Please check your attitude and look again. And what is nonsensical about what I presented? If it is wrong, show me. But don't insult me and dismiss me. It makes me think you can't answer me. You haven't so far. |
None of what you say has any weight.
"0^n" is zero, provided "n" is not zero.
With "z" in the complex domain, even...
"0^z" is zero, provided "z" is not zero.
You have made a "simple grade school math error".
"z^0" is unity, provided "z" is not zero.
0^0 is undefined. |
You cannot seem to think beyond what you have been taught. If you disagreed that would be one thing, but clearly you cannot comprehend; in your 3 responses you have not even once addressed what I have said. I'm done. |
I went back over what you have been saying, starting from:
| jackmt wrote: | ...
Concerning division by 0; I wrote a linguistic proof that 0^0 (zero raised to the zero power) = 1 and is therefore a legitimate divisor. Any one agree? disagree? |
(I've no idea exactly what you mean by a "legitimate divisor".)
Here you made the assertion that you had a "linguistic proof". I vaguely thought that you would eventually present one of the usual fallacies. Your subsequent posts just skirted around with words and gave no coherent "proof" - just a series of bald statements with no attempt at structuring them into a proof of any kind. Viz:
| jackmt wrote: | | Division by zero is undefined. Agreed? |
No. Division is defined for all elements of Z, with the exception of zero. Division by zero is not undefined. It is perfectly well defined. It does not exist. In your terminology, zero would be the unique value which is not a "legitimate divisor".
| jackmt wrote: | | In my original claim I said I have a linguistic proof; ... |
We're still waiting for this.
| jackmt wrote: | | ... i.e., 0^0=1 for grammatical reasons. |
Just stating that your assertion is true "for grammatical reasons", without making any attempt to give those grammatical reasons, doesn't make it true. (Unless you say it three times?)
| jackmt wrote: | | ... 1 for example, is deleted as multiplier ... |
I have no idea what you mean by "deleted" in this.
| jackmt wrote: | | And I know that 0^n = 0. |
Which is where you are going wrong. If you start with a premise which is not true, you can get any answer to any question.
....
If you would like a good book to read, " Surreal Numbers: How Two Ex-Students Turned on to Pure Mathematics and Found Total Happiness" is rather lovely. |
I'll try again. Please forgive my attitude. I will try to keep it in check.
No, I have not presented my proof. It is rather lengthy for a forum such as this. I had hoped to begin a discussion of the topic, but I just got responses that showed my claim was not properly understood. The first response alone assumed that I claimed division by zero is allowable.
And it has been a while since I've use the jargon, so I'm a little rusty. Feel free to correct my usage or suggest the term I am looking for. I thought I learned that division by zero is undefined, or did I forget the right term, or has it been replaced with another?
Math is a mental construct, abstracted from everyday language and formalized , as is formal logic. In written form it has its own syntax, substantives, logical operators, etc., as enumerated in an earlier post. Its mental representation includes all those elements and more and is much more complex. As a theoretical linguist specializing in logic and semantics, I am interested in the mental representation that makes the language of math logically and semantically sound. Its representation on the blackboard is misleading as is everyday language. There is a maxim of linguistic efficiency that says "Do what you have to, get by with what you can." Redundant information, understood elements, etc. are deleted from expressions routinely.
Men are mortal.
All men are mortal.
Logically speaking these 2 statements are equivalents; true under exactly the same conditions. The universal quantifier is suppressed in the first and overt in the second. In the underlying mental representation or Logical Form (LF) it is necessarily present in both. Other necessary elements of both representations are absent as well.
Where 1 exists in an equation as a factor, it may be deleted or made non-overt (implicit). And where 0 exists as an addend, it may be deleted or need not be made explicit.
1(x+2) = x+2 We may delete it once, or, after distribution, twice.
X^n+0 = X^n
Please provide your terms if these don't work for you and I will use them, instead.
Upon reflection, it seems I did not make it clear that my assertion was theoretical, not pragmatic. I know the rules of operations as generally taught. It later occurred to me that maybe ruveyn was giving me the theoretical basis for x^0=1:
X^0 = X^(1-1) = X^1/X^-1 = 1 for all non-zero x.
I remember now that this is how it is explained in the textbooks. And this is exactly what I object to on theoretical grounds. It is a stop-gap measure in an otherwise (nearly) perfect paradigm.
My objections stand.
1. We are defining the simple in terms of the complex by saying that 0 = (1-1). We are further complicating things by requiring the substitution of equal values for only zero, and only in the context of exponents, and for the sole purpose of filling a gap in the paradigm and then making an arbitrary restriction on that, to fill another gap. Ockham is chomping at the bits to get his razor on this.
2. Because we define the exponent 0 as (1-1) and require its substitution to satisfy the problem of X^0 = 1, we get:
0^n = 0^(n+0) = 0^n+(1-1) = 0^n+1/0^1 = undefined (until you give me the correct term)
Thus, all 0^n is undefined. Clearly, we don't want that. I suppose we could restrict the obligatory substitution further to keep the paradigm happy. (def.: paradigm - n. a multi-headed monster that thrives on exceptions to rules, growing uglier and more unreasonable with every feeding.) Further, we define negative exponents in terms of division. That leaves us with the problem of explaining where we get the 1 in
X^-2 = 1/ X^2
from a theoretical perspective.
X^3 is an overt, convenient notation (one that causes no problems pragmatically) for what is, on paper, X*X*X. I require, for reasons of syntax, identity, and truth, that X^3 is, minimally, (1*X*X*X) underlyingly. "X" must be represented as something other than "X," as well. And other implicit elements must be made explicit. "Times" is a 2 place verb; it requires 2 arguments in specific positions to satisfy its semantic grid; #times# where "#" denotes an argument position, with numbers being arguments.
Much more needs to be said, but I need to know if I am beginning to make the problem clear, or if I am wasting my time. I will continue if this makes sense to anyone. |
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jackmt
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| Posted: Tue Dec 27, 2011 8:34 pm Post subject: Re: Real Number Operations Caveat |
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| ruveyn wrote: | | jackmt wrote: | | ruveyn wrote: | | jackmt wrote: | | This is not quite right, but let's accept it. 0^0 says that zero is used zero times. How is it then division by zero? |
0^0 = 0^1 / 0^1 = 0 / 0 = undefined.
ruveynb |
This equation is illegitimate for at least 3 reasons. First, on this model all 0^n is undefined as it may be represented as
0^n = 0^((n+1) -1) = 0^(n+1)/0^1
And we don't want to say that all 0^n is undefined, but that it is 0 for all n not equal to zero. So 0^0 is already an exception to a rule. My claim is that it an exception to a different rule; namely, that it is the only value of 0^n that may be used as a divisor.
Second, it defines the simple in terms of the complex and in terms of itself, as if we were to define 1 as:
1 = 1-(1-1).
True enough, but what have we gained?
Third, it comes dangerously close to, if not actually, committing the fallacy of petitio principii: division by zero is already proscribed.
There are other problems as well.
The problem most mathematicians seem to have with my claim is the feeling that the outward expression, being composed of only zeroes, must represent zero underlyingly somehow. But a positive number may be expressed as a double negative.
There is much more to say on the matter. Where does the '1' come from in x^0=1, for instance? The standard answer to this that it behaves as 1. This is weak. It behaves as 1 because it is 1.
In my original claim I said I have a linguistic proof; i.e., 0^0=1 for grammatical reasons. It completely neatens the paradigm, leaving only the one above mentioned exception. |
In mathematics linguistic proofs are dreck.
What you have presented is mostly nonsense.
For example 0^1 = 0, In fact for n > 0 0^n = 0. Why? 0*0*...*0 (n terms) = 0.
ruveyn |
Sorry about my attitude before. I'd like to try again.
This is an example of your not addressing what I have said. I said nothing about 0^n, n>0.
I am asking about 0^0 and what it expresses underlyingly. It is a theoretical question.
7^3 = 7*7*7.
0^0 = ?
The response you gave earlier is exactly what I object to on theoretical grounds: this is a makeshift solution to what I see as not a problem.
0^0 = 0^(1-1) = 0^1/0^1
I believe this to be illegitimate for reasons expressed in my post above. I think I have stated my case better there. |
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ruveyn
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| Posted: Tue Dec 27, 2011 8:58 pm Post subject: Re: Real Number Operations Caveat |
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Let n != 0.
0^0 = 0^(n - n) = 0^n * 0^(-n) = 0^n/0^(-n) = 0/0 which is not defined.
Q.E.D.
I don't do verbal arguments. I do mathematics. What do you do?
ruveyn |
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jackmt
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| Posted: Tue Dec 27, 2011 9:39 pm Post subject: Re: Real Number Operations Caveat |
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| ruveyn wrote: |
Let n != 0.
0^0 = 0^(n - n) = 0^n * 0^(-n) = 0^n/0^(-n) = 0/0 which is not defined.
Q.E.D.
I don't do verbal arguments. I do mathematics. What do you do?
ruveyn |
Q.E.N.D.
Exponents are convenient notation for underling multiplication.
7^3 =7^4-1 =7^5-2 etc. only after manipulation
7^3 = 7*7*7 simply.
0^0 = 0^1-1 only after manipulation
0^0 = ? without manipulation, i.e., simply? If that can't be answered, how do we justify saying 7^3= 7*7*7?
As stated in my post subsequent to our last round, which appears above, I do theoretical linguistics, emphasis in logic and semantics. Please read it. I think I have stated the problem more clearly and some of my thinking on it. I am not the idiot you seem to think me. You have failed to understand my claim; you keep answering things I haven't asked and not answering things I have. I am asking for theoretical justification for 0^0 = undefined. Let us start with 'What does the expression 0^0 conveniently notate?
The standard response of 0^0=0^1-1 is the basis for the "proof" that 1=2.
Last edited by jackmt on Tue Dec 27, 2011 9:45 pm; edited 1 time in total |
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ruveyn
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| Posted: Tue Dec 27, 2011 9:42 pm Post subject: Re: Real Number Operations Caveat |
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| jackmt wrote: | | ruveyn wrote: |
Let n != 0.
0^0 = 0^(n - n) = 0^n * 0^(-n) = 0^n/0^(-n) = 0/0 which is not defined.
Q.E.D.
I don't do verbal arguments. I do mathematics. What do you do?
ruveyn |
Q.E.N.D.
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Cut the crap and tell me which step that I did is wrong and why.
If you want the law of exponents preserved then 0^0 must be undefined.
Just as 0/0 is undefined.
ruveyn |
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jackmt
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| Posted: Tue Dec 27, 2011 9:51 pm Post subject: Re: Real Number Operations Caveat |
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| ruveyn wrote: | | jackmt wrote: | | ruveyn wrote: |
Let n != 0.
0^0 = 0^(n - n) = 0^n * 0^(-n) = 0^n/0^(-n) = 0/0 which is not defined.
Q.E.D.
I don't do verbal arguments. I do mathematics. What do you do?
ruveyn |
Q.E.N.D.
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Cut the crap and tell me which step that I did is wrong and why.
If you want the law of exponents preserved then 0^0 must be undefined.
Just as 0/0 is undefined.
ruveyn |
I guess we'll have to end this. Again, you seem unable to comprehend what I'm asking.
I just thought of something. Maybe this will make some of the problem clear.
You insist that 0^0 = 0^1-1 = 0^1/0^1. So 0^0 is undefined.
So, 0^3 = 0^4-1 = 0^4/0^1
Thus all 0^n = undefined. If we disallow this because we don't want all 0^n undefined, we must find another justification for 0^0= undefined. This is what I am after. Clear? |
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lau
Really nice person to know. :)
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| Posted: Tue Dec 27, 2011 10:21 pm Post subject: Re: Real Number Operations Caveat |
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Thanks for staying in with this. I think part of the reason you have trouble with the concept of 0^0, division by zero, and so on, is that you try to make an argument via making the expression more complex.
Actually, ruveyn also is guilty of this, although, to be fair, his "proof" that 0^0 is undefined is not particularly rigorous. What he showed was that, if the notation 0^0 is meaningful, at all, then a sequence of manipulations of the expression can result in obtaining a value that it more clearly(?) undefined (viz 0/0).
When I earlier suggested that a non-zero value divided by zero was not "undefined", I was trying to be careful. There are two distinct cases here. This one (non-zero divided by zero) does not have any "real number" associated with it. I.e. in does not result in a value which is in the set of real numbers. The result can be defined as "infinity", extending the reals by this element, but all the rules of arithmetic then need to be augmented with caveats about what it is valid to do when (plus or minus?) infinity is involved.
The other case, division of zero by zero, results in an indeterminate value. I.e. although it is possible to say "0/0", division by zero NEVER results in an identifiable real number. Should any calculation apparently result in 0/0, then the value is completely arbitrary.
All of the above applies to the standard real (or complex, etc) number systems. There is nothing to stop you developing a number system where "0/0 = 1" and/or "0^0 = 1". However, once you do, you will have to build in a whole load of exceptions to your arithmetic rules, in order to avoid contradictions.
---
Mathematics is not a subset of linguistics. Mathematics struggles to be expressed by linguistics.
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jackmt
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| Posted: Tue Dec 27, 2011 11:08 pm Post subject: Re: Real Number Operations Caveat |
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| lau wrote: |
Thanks for staying in with this. I think part of the reason you have trouble with the concept of 0^0, division by zero, and so on, is that you try to make an argument via making the expression more complex.
Actually, ruveyn also is guilty of this, although, to be fair, his "proof" that 0^0 is undefined is not particularly rigorous. What he showed was that, if the notation 0^0 is meaningful, at all, then a sequence of manipulations of the expression can result in obtaining a value that it more clearly(?) undefined (viz 0/0).
When I earlier suggested that a non-zero value divided by zero was not "undefined", I was trying to be careful. There are two distinct cases here. This one (non-zero divided by zero) does not have any "real number" associated with it. I.e. in does not result in a value which is in the set of real numbers. The result can be defined as "infinity", extending the reals by this element, but all the rules of arithmetic then need to be augmented with caveats about what it is valid to do when (plus or minus?) infinity is involved.
The other case, division of zero by zero, results in an indeterminate value. I.e. although it is possible to say "0/0", division by zero NEVER results in an identifiable real number. Should any calculation apparently result in 0/0, then the value is completely arbitrary.
All of the above applies to the standard real (or complex, etc) number systems. There is nothing to stop you developing a number system where "0/0 = 1" and/or "0^0 = 1". However, once you do, you will have to build in a whole load of exceptions to your arithmetic rules, in order to avoid contradictions.
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Mathematics is not a subset of linguistics. Mathematics struggles to be expressed by linguistics. |
Thank you for your rational and civil response. I have never had a problem with the concept of division by zero and why it is disallowed. And your response still does not answer my question. You say, "Should any calculation apparently result in 0/0 then the value is completely arbitrary." Behold:
If we insist that
0^0 = (by definition) 0^1-1 = 0^1/0^1 = error (whatever it is called)
then we must insist that
0^3 = (by instantiation) 0^3+0 = (by substitution) 0^3+(1-1) = (by association) 0^(3+1)-1 = (by addition) 0^4-1 = (by definition) 0^4/0^1 = 0/0 = same error, and so for all 0^n .
For whatever reason we disallow this, we must also disallow the definition of 0^0 for consistency's sake. Or come up with a good reason for disallowing it.
Now, I ask again, what is the theoretical reason that 0^0 is undefined? If 7^3 is a convenient representation of the simpler 7*7*7, at least this is what it decomposes into at the last. Of what is 0^0 the convenient representation? What mathematical expression does it denote in simpler terms or in its decomposition? If, by definition, X^y means X is used y times as a factor, then 0^0 means that 0 is used 0 times in the expression, i.e., it is not used in the expression. How then is it division by zero? What is the underlying representation? Where in the underlying representation is 0? I can't think of any other ways to ask the question. And while we're at it, where is the 1 in x^0 underlyingly?
BTW, I think you need to edit your 2nd paragraph. your meaning is unclear. A wrong word or two, I think.
Logic and math are said by many philosophers of science to be one, having (nearly) identical systems of rules. And as math is a mental construct, and theoretical linguistics seeks to map the mind as regards language, and as math is an abstract language, abstracted from human language, I say it is in the domain of linguistics. It exists in the mind before it is put on the blackboard. |
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ruveyn
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| Posted: Wed Dec 28, 2011 8:17 am Post subject: Re: Real Number Operations Caveat |
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| jackmt wrote: |
I guess we'll have to end this. Again, you seem unable to comprehend what I'm asking.
I just thought of something. Maybe this will make some of the problem clear.
You insist that 0^0 = 0^1-1 = 0^1/0^1. So 0^0 is undefined.
So, 0^3 = 0^4-1 = 0^4/0^1
Thus all 0^n = undefined. If we disallow this because we don't want all 0^n undefined, we must find another justification for 0^0= undefined. This is what I am after. Clear? |
The justification is simple. either 0^0 = 0 or it = 1 IF IT IS DEFINED. Either one will lead to a contradiction. Therefore it is left undefined. It is undefined for the same reason the 0/0 is undefined. There is no single number X such that 0/0 = X. Any value of X satisfies 0 = 0*X. Since on one value will do, it is undefined.
ruveyn |
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lau
Really nice person to know. :)
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| Posted: Wed Dec 28, 2011 4:00 pm Post subject: Re: Real Number Operations Caveat |
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| In his second paragraph, lau wrote: | | Actually, ruveyn also is guilty of this, although, to be fair, his "proof" that 0^0 is undefined is not particularly rigorous. What he showed was that, if the notation 0^0 is meaningful, at all, then a sequence of manipulations of the expression can result in obtaining a value that it more clearly(?) undefined (viz 0/0). |
| jackmt wrote: | | BTW, I think you need to edit your 2nd paragraph. your meaning is unclear. A wrong word or two, I think. |
Not that I can see.
You are still approaching things backward. In mathematics (and hopefully any other rigorous discipline), something is not defined until it is defined.
Within the real (or complex) number systems. we define the meaning (and value) of "x^0" for all values of x other than zero. There in no overarching reason for 0^0 to be undefined. It is just not defined. If you choose to extend the reals by merely defining a value for 0^0, you will find yourself able to "prove" that any value equals any other value. This does not result in a very useful number system.
By applying arithmetic rules, as you have done, to 0^0, you have shown exactly that.
E.g. defining 0/0 as unity, one of the easiest proofs that "1 = 2" would be: | Code: | 1*0 = 2*0
therefore 1 = (2*0)/0
therefore 1 = 2*(0/0)
but 0/0 = 1
therefore 1 = 2*1
therefore 1 = 2 |
As I am saying, you can choose to define 0/0 and 0^0 if you wish. However, unless you then make a lot of other modifications to your number system, you would not have anything very useful.
The IEEE floating point system does such things: http://en.wikipedia.org/wiki/IEEE_754-2008
It defines "values" for +/- infinity. It also has its NaN (Not a Number) values. It then has various rules for combining infinities, zeroes, and all the rest of its panoply of clever stuff. It works extremely well, in the domain (practical numerical calculations) is is used for.
There are other numerical computation programs/systems that go even further, by carrying along with a value an estimate of how accurate that value might be. They are very good at recognising when a result has so wide an error range that the value should not be taken seriously.
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