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Oh, the ambiguity! 0^0 = ??? (Final draft)
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Dan Christensen
Nov 19, 2019, 11:56:10 PM11/19/19
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Full text in HTML format at https://dcproof.wordpress.com/ (originally posted Oct. 9, 2013)
Introduction
Most of us learned in high school that 0^0 is somehow undefined or ambiguous. In college or university, your calculus professor will confirm this, citing the ambiguity resulting from different limits. We have Lim (x --> 0): x^0 = 1. But we also have Lim (y --> 0): 0^y = 0. There is an obvious discontinuity in the function z = x^y at (0, 0). (See "0^0 with Continuous Exponents" at https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero#Continuous_exponents)
Your algebra professor, on the other hand, may tell you that you can assume that 0^0 = 1 on the natural numbers--for convenience mostly. They may justify it with analogies to various conventions, e.g. usually the convention of so-called empty products--the product of no numbers? Many simply define it to be 1. It’s apparently not something you can actually prove. As for 0^0 being undefined on the real numbers, and exponentiation being entirely consistent in both domains, that is a mere coincidence. We are talking about entirely different functions here, they will say. If that strikes you as being just a bit too, well, “hand wavy” for your liking, read on!
Here, I will develop the exponentiation function on the natural numbers with 0^0 undefined, given only the operations of addition and multiplication on N. I use what I believe to be a novel approach that looks at all possible functions that satisfy the usual requirements for an exponentiation function on N. In so doing, we can justify leaving 0^0 undefined, as it is on the set of real numbers R. I will also look at some implications for the usual laws of exponents on N for undefined 0^0.
Exponentiation Defined as Repeated Multiplication on N
When you were first introduced to exponents in elementary or high school, you probably started with the exponents greater than or equal to two. After all, you need at least 2 numbers to multiply. For all a ε N, we have:
a^2 = a.a
a^3 = a.a.a = a^2. a
a^4 = a.a.a.a = a^3. a
a^5 = a.a.a.a.a = a^4. a
and so on.
This infinite sequence of equations can be recursively summarized in just two equations for all a, b ε N as follows:
1. a^2 = a.a
2. a^(b+1) = a^b . a
These two equations, by themselves, do not, however, tell us anything about exponents 0 or 1. It turns out that there are infinitely many such exponent-like functions on N that satisfy these equations. Proof (21 lines) at http://dcproof.com/PowFunctionInfinitelyManyBase2.htm
Fortunately, these infinitely many functions differ only in the value assigned to 0^0. Proof (194 lines) at http://dcproof.com/PowFunctionEquivBases2.htm
This suggests that, in our definition of exponentiation on N, we should simply leave 0^0 undefined. To this end, we can construct (i.e. prove the existence of) a unique partial function for exponentiation on N. Proof (618 lines) at http://dcproof.com/PowPartialFunction.htm
Thus we can define exponentiation on N as follows:
1. a^b ε N (for a or b ≠ 0)
2. 0^1 = 0
3. a^0 = 1 (for a ≠ 0)
4. a^b+1 = a^b . a (for a or b ≠ 0)
The Laws of Exponents on N for undefined 0^0
Using the above definition, we can derive the 3 Laws of Exponents on N:
1. The Product of Powers Rule: a^b. a^c = a^(b+c) (for a ≠ 0 OR both b, c ≠ 0) Proof at http://dcproof.com/ProductOfPowersV3.htm
2. The Power of a Power Rule: (a^b)^c = a^(b.c) (for a ≠ 0 OR both b, c ≠ 0) Proof at http://dcproof.com/PowerOfAPowerV2.htm
3. The Power of a Product Rule: (a.b)^c = a^c . b^c (for c ≠ 0 OR both a, b ≠ 0) Proof at http://dcproof.com/PowerOfAProductV4.htm
Interestingly, these restrictions would not apply if 0^0 was defined to be either 0 or 1. So, adding these laws to the requirements for exponentiation would narrow down the infinitely many possibilities to only two. But we would still be left with some ambiguity—is 0^0 equal to 0 or 1? Oh, the ambiguity!
My Recommendation
If, as in most well used, centuries-old textbook applications where an informal proof will usually suffice, it is convenient and probably safe to assume 0^0=1 on the natural numbers, if not on the reals. If, however, you are pushing the limits of number theory and only a formal proof will do, you should take the extra care and avoid assuming any particular value for 0^0 in both the real numbers and the natural numbers. There are easy work-arounds in many cases. Applications of the binomial theorem to obtain the usual expansions of (x+y)^n, for example, will result in terms including 0^0 when either x or y are zero. In such cases, however, there is a simple work-around, e.g. (x+0)^n = x^n for x or n ≠ 0.
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Introduction
Most of us learned in high school that 0^0 is somehow undefined or ambiguous. In college or university, your calculus professor will confirm this, citing the ambiguity resulting from different limits. We have Lim (x --> 0): x^0 = 1. But we also have Lim (y --> 0): 0^y = 0. There is an obvious discontinuity in the function z = x^y at (0, 0). (See "0^0 with Continuous Exponents" at https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero#Continuous_exponents)
Your algebra professor, on the other hand, may tell you that you can assume that 0^0 = 1 on the natural numbers--for convenience mostly. They may justify it with analogies to various conventions, e.g. usually the convention of so-called empty products--the product of no numbers? Many simply define it to be 1. It’s apparently not something you can actually prove. As for 0^0 being undefined on the real numbers, and exponentiation being entirely consistent in both domains, that is a mere coincidence. We are talking about entirely different functions here, they will say. If that strikes you as being just a bit too, well, “hand wavy” for your liking, read on!
Here, I will develop the exponentiation function on the natural numbers with 0^0 undefined, given only the operations of addition and multiplication on N. I use what I believe to be a novel approach that looks at all possible functions that satisfy the usual requirements for an exponentiation function on N. In so doing, we can justify leaving 0^0 undefined, as it is on the set of real numbers R. I will also look at some implications for the usual laws of exponents on N for undefined 0^0.
Exponentiation Defined as Repeated Multiplication on N
When you were first introduced to exponents in elementary or high school, you probably started with the exponents greater than or equal to two. After all, you need at least 2 numbers to multiply. For all a ε N, we have:
a^2 = a.a
a^3 = a.a.a = a^2. a
a^4 = a.a.a.a = a^3. a
a^5 = a.a.a.a.a = a^4. a
and so on.
This infinite sequence of equations can be recursively summarized in just two equations for all a, b ε N as follows:
1. a^2 = a.a
2. a^(b+1) = a^b . a
These two equations, by themselves, do not, however, tell us anything about exponents 0 or 1. It turns out that there are infinitely many such exponent-like functions on N that satisfy these equations. Proof (21 lines) at http://dcproof.com/PowFunctionInfinitelyManyBase2.htm
Fortunately, these infinitely many functions differ only in the value assigned to 0^0. Proof (194 lines) at http://dcproof.com/PowFunctionEquivBases2.htm
This suggests that, in our definition of exponentiation on N, we should simply leave 0^0 undefined. To this end, we can construct (i.e. prove the existence of) a unique partial function for exponentiation on N. Proof (618 lines) at http://dcproof.com/PowPartialFunction.htm
Thus we can define exponentiation on N as follows:
1. a^b ε N (for a or b ≠ 0)
2. 0^1 = 0
3. a^0 = 1 (for a ≠ 0)
4. a^b+1 = a^b . a (for a or b ≠ 0)
The Laws of Exponents on N for undefined 0^0
Using the above definition, we can derive the 3 Laws of Exponents on N:
1. The Product of Powers Rule: a^b. a^c = a^(b+c) (for a ≠ 0 OR both b, c ≠ 0) Proof at http://dcproof.com/ProductOfPowersV3.htm
2. The Power of a Power Rule: (a^b)^c = a^(b.c) (for a ≠ 0 OR both b, c ≠ 0) Proof at http://dcproof.com/PowerOfAPowerV2.htm
3. The Power of a Product Rule: (a.b)^c = a^c . b^c (for c ≠ 0 OR both a, b ≠ 0) Proof at http://dcproof.com/PowerOfAProductV4.htm
Interestingly, these restrictions would not apply if 0^0 was defined to be either 0 or 1. So, adding these laws to the requirements for exponentiation would narrow down the infinitely many possibilities to only two. But we would still be left with some ambiguity—is 0^0 equal to 0 or 1? Oh, the ambiguity!
My Recommendation
If, as in most well used, centuries-old textbook applications where an informal proof will usually suffice, it is convenient and probably safe to assume 0^0=1 on the natural numbers, if not on the reals. If, however, you are pushing the limits of number theory and only a formal proof will do, you should take the extra care and avoid assuming any particular value for 0^0 in both the real numbers and the natural numbers. There are easy work-arounds in many cases. Applications of the binomial theorem to obtain the usual expansions of (x+y)^n, for example, will result in terms including 0^0 when either x or y are zero. In such cases, however, there is a simple work-around, e.g. (x+0)^n = x^n for x or n ≠ 0.
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
bassam karzeddin
Nov 20, 2019, 9:08:36 AM11/20/19
to
You have hardly understood anything from the lesson I gave you so far
1) You didn't specify clearly what numbers are your exponent but followed the common wrong beliefs among your mainstreams and masters as well
2) You didn't realize the emptiness of the case so-called zero in your so-called modern maths that implies nothing at all, but a tendency of short vision mathematickers to generalize anything as they like where finally they get stuck before a very thick wall from their own inventions where their skulls are truly much thicker
3) You didn't explain the power of unity when you scatter it to parts that seems even worse than your surficial thinking
4)..., 5)..., ..., n) You better leave this task for others, for sure
You are certainly so funny Dan, (thinking deeply inside a circle)
BKK
Dan Christensen
Nov 20, 2019, 9:35:25 AM11/20/19
to
From psycho troll BKK who wrote here:
“Those many challenges of mine (in my posts) weren't actually designed for human beings, but for the future artificial beings that would certainly replace them not far away from now, for sure.”
-- BKK, Dec. 6, 2017
“You know certainly that I'm the man, and more specially the KING who is going to upside down most of your current false mathematics for all future generations.”
-- BKK, Nov. 22, 2018
Math failure, BKK, doesn't believe in negative numbers, zero or numbers like pi and root 2. He doesn't even believe in 40 degree angles or circles. Really! Needless to say his own goofy system is getting nowhere and never will. As such he is insanely jealous of wildly successful mainstream mathematics. He seems to believe these super-intelligent artificial beings of his will somehow be enlisting his aid to "reform" mathematics worldwide when they take over the planet in the near future. He is truly delusional.
Mostowski Collapse
Dec 4, 2021, 11:06:22 PM12/4/21
to
Your 0^0 nonsense only shows that your notion of function
is broken. Try this function F : R \ {0} -> R, where
F(x,y) <=> x*y = 1
Also known as reciprocal. Do you claim that we
don't know whether ~F(0,1) holds or not?
is broken. Try this function F : R \ {0} -> R, where
F(x,y) <=> x*y = 1
Also known as reciprocal. Do you claim that we
don't know whether ~F(0,1) holds or not?
Mostowski Collapse
Dec 4, 2021, 11:12:05 PM12/4/21
to
Now assume the mathematician Bert talks to the
mathematician Carlo, look I have a function
F : R \ {0} -> R, but Bert does not reveal:
F(x,y) <=> x*y = 1
Shouldn't be Carlo already be able to deduce ~F(0,1) ?
According to Wikipedia yes. According to Dan-O-Matiks
chimpansee definition no.
LMAO!
mathematician Carlo, look I have a function
F : R \ {0} -> R, but Bert does not reveal:
F(x,y) <=> x*y = 1
According to Wikipedia yes. According to Dan-O-Matiks
chimpansee definition no.
LMAO!
Dan Christensen
Dec 5, 2021, 12:37:08 AM12/5/21
to
On Saturday, December 4, 2021 at 11:06:22 PM UTC-5, Mostowski Collapse wrote:
> Your 0^0 nonsense only shows that your notion of function
> is broken.
Sadly for cranks like you, it works all too well, Jan Burse!
> Your 0^0 nonsense only shows that your notion of function
> is broken.
>Try this function F : R \ {0} -> R, where
>
> F(x,y) <=> x*y = 1
>
> Also known as reciprocal. Do you claim that we
> don't know whether ~F(0,1) holds or not?
Mostowski Collapse
Dec 5, 2021, 7:33:34 AM12/5/21
to
There is no "if". The algebraic geometry definition is:
F(x,y) <=> x*y = 1
Algebraic geometry is a branch of mathematics,
classically studying zeros of multivariate polynomials.
https://en.wikipedia.org/wiki/Algebraic_geometry
Its easy to prove ~F(0,1), just substitute:
~F(0,1) <=>
~0*1 = 1 <=>
~0 = 1 <=>
true
F(x,y) <=> x*y = 1
classically studying zeros of multivariate polynomials.
https://en.wikipedia.org/wiki/Algebraic_geometry
Its easy to prove ~F(0,1), just substitute:
~F(0,1) <=>
~0*1 = 1 <=>
~0 = 1 <=>
true
Dan Christensen
Dec 5, 2021, 8:47:11 AM12/5/21
to
On Sunday, December 5, 2021 at 7:33:34 AM UTC-5, Mostowski Collapse wrote:
> Dan Christensen schrieb am Sonntag, 5. Dezember 2021 um 06:37:08 UTC+1:
> > On Saturday, December 4, 2021 at 11:06:22 PM UTC-5, Mostowski Collapse wrote:
> > > Your 0^0 nonsense only shows that your notion of function
> > > is broken.
> > Sadly for cranks like you, it works all too well, Jan Burse!
> > >Try this function F : R \ {0} -> R, where
> > >
> > > F(x,y) <=> x*y = 1
> > >
> > > Also known as reciprocal. Do you claim that we
> > > don't know whether ~F(0,1) holds or not?
> > Just try to formally prove it. That was YOUR assignment, Jan Burse. We are STILL waiting for your formal proof that f(0)=1 is false given: For all real x and y, if x=/=0 then we have f(x)=y <=> x*y=1. Do your homework!
> Dan Christensen schrieb am Sonntag, 5. Dezember 2021 um 06:37:08 UTC+1:
> > On Saturday, December 4, 2021 at 11:06:22 PM UTC-5, Mostowski Collapse wrote:
> > > Your 0^0 nonsense only shows that your notion of function
> > > is broken.
> > Sadly for cranks like you, it works all too well, Jan Burse!
> > >Try this function F : R \ {0} -> R, where
> > >
> > > F(x,y) <=> x*y = 1
> > >
> > > Also known as reciprocal. Do you claim that we
> > > don't know whether ~F(0,1) holds or not?
> > Just try to formally prove it. That was YOUR assignment, Jan Burse. We are STILL waiting for your formal proof that f(0)=1 is false given: For all real x and y, if x=/=0 then we have f(x)=y <=> x*y=1. Do your homework!
> There is no "if". The algebraic geometry definition is:
> F(x,y) <=> x*y = 1...
Giving up already??? Oh, well...
Deny it if you don't mind looking foolish, but there is no solution in the reals for 0y=1.
Mostowski Collapse
Dec 5, 2021, 9:32:11 AM12/5/21
to
Your replies don't make any sense.
Mostowski Collapse
Dec 5, 2021, 9:35:30 AM12/5/21
to
Maybe have a look at a hyperbola:
https://www.wolframalpha.com/input/?i=x*y+%3D+1
Did you just invent dark boolean values? To
realize WM with extra steps?
Do you see white or black at the point (0,1) in the
graph of the hyperbola. If its black it would be F(0,1),
if it is white it would be ~F(0,1).
Do you distribute DC poop with tin foil hat
and a cupper braclet? What does DC mean?
DC = double chimpanzee?
https://www.wolframalpha.com/input/?i=x*y+%3D+1
Did you just invent dark boolean values? To
realize WM with extra steps?
Do you see white or black at the point (0,1) in the
graph of the hyperbola. If its black it would be F(0,1),
if it is white it would be ~F(0,1).
Do you distribute DC poop with tin foil hat
and a cupper braclet? What does DC mean?
DC = double chimpanzee?
Dan Christensen
Dec 5, 2021, 9:46:29 AM12/5/21
to
On Sunday, December 5, 2021 at 9:35:30 AM UTC-5, Mostowski Collapse wrote:
> Maybe have a look at a hyperbola:
> https://www.wolframalpha.com/input/?i=x*y+%3D+1
>
Follow your own advice, Jan Burse! What value, if any, is given at x=0?
> Maybe have a look at a hyperbola:
> https://www.wolframalpha.com/input/?i=x*y+%3D+1
>
Dan Christensen
Dec 5, 2021, 10:12:48 AM12/5/21
to
On Sunday, December 5, 2021 at 9:35:30 AM UTC-5, Mostowski Collapse wrote:
> Maybe have a look at a hyperbola:
> https://www.wolframalpha.com/input/?i=x*y+%3D+1
>
The solution given there is "x=/=0, y=1/x".
> Maybe have a look at a hyperbola:
> https://www.wolframalpha.com/input/?i=x*y+%3D+1
>
So, 1/x is UNDEFINED for x=0. How then can you determine that 1/0=1 (F(1, 0) in your notation) is false if the LHS is undefined? Hint: You can't. Deal with it, Jan Burse!
Mostowski Collapse
Dec 5, 2021, 10:57:45 AM12/5/21
to
Can we buy these double chimpazee copper bracelets
already on Amazon? Here is your fallacy explained.
We can prove:
/* provable */
F is hyperbola => ALL(x):[x≠0 => x*F(x)=1]
But we cannot prove:
/* not provable */
ALL(x):[x≠0 => x*F(x)=1] => F is hyperbola
Got it?
already on Amazon? Here is your fallacy explained.
We can prove:
/* provable */
F is hyperbola => ALL(x):[x≠0 => x*F(x)=1]
But we cannot prove:
/* not provable */
ALL(x):[x≠0 => x*F(x)=1] => F is hyperbola
Got it?
Mostowski Collapse
Dec 5, 2021, 11:09:38 AM12/5/21
to
Even a blind mole can show both claims:
> /* provable */
> F is hyperbola => ALL(x):[x≠0 => x*F(x)=1]
/* serial */
∀a∃bFab ∧
/* functional */
(∀a∀b∀c((Fab ∧ Fac) → b=c) ∧
/* hyperbola */
∀a∀b(Fab ↔ m(a,b)=z))
entails
/* double chimpanzee */
∀a(¬a=z → ∃b(Fab ∧ m(a,b)=z)).
https://www.umsu.de/trees/#~6a~7bFab~1~6a~6b~6c%28Fab~1Fac~5b=c%29~1~6a~6b%28Fab~4m%28a,b%29=z%29|=~6a%28~3a=z~5~7b%28Fab~1m%28a,b%29=z%29%29
> /* not provable */
> ALL(x):[x≠0 => x*F(x)=1] => F is hyperbola
/* serial */
∀a∃bFab ∧
/* functional */
(∀a∀b∀c((Fab ∧ Fac) → b=c) ∧
/* double chimpanzee */
∀a(¬a=z → ∃b(Fab ∧ m(a,b)=z)))
does not entail
/* hyperbola */
∀a∀b(Fab ↔ m(a,b)=z).
https://www.umsu.de/trees/#~6a~7bFab~1~6a~6b~6c%28Fab~1Fac~5b=c%29~1~6a%28~3a=z~5~7b%28Fab~1m%28a,b%29=z%29%29|=~6a~6b%28Fab~4m%28a,b%29=z%29
> /* provable */
> F is hyperbola => ALL(x):[x≠0 => x*F(x)=1]
∀a∃bFab ∧
/* functional */
(∀a∀b∀c((Fab ∧ Fac) → b=c) ∧
/* hyperbola */
∀a∀b(Fab ↔ m(a,b)=z))
entails
/* double chimpanzee */
∀a(¬a=z → ∃b(Fab ∧ m(a,b)=z)).
https://www.umsu.de/trees/#~6a~7bFab~1~6a~6b~6c%28Fab~1Fac~5b=c%29~1~6a~6b%28Fab~4m%28a,b%29=z%29|=~6a%28~3a=z~5~7b%28Fab~1m%28a,b%29=z%29%29
> /* not provable */
> ALL(x):[x≠0 => x*F(x)=1] => F is hyperbola
∀a∃bFab ∧
/* functional */
(∀a∀b∀c((Fab ∧ Fac) → b=c) ∧
/* double chimpanzee */
∀a(¬a=z → ∃b(Fab ∧ m(a,b)=z)))
does not entail
/* hyperbola */
∀a∀b(Fab ↔ m(a,b)=z).
https://www.umsu.de/trees/#~6a~7bFab~1~6a~6b~6c%28Fab~1Fac~5b=c%29~1~6a%28~3a=z~5~7b%28Fab~1m%28a,b%29=z%29%29|=~6a~6b%28Fab~4m%28a,b%29=z%29
Dan Christensen
Dec 5, 2021, 11:12:48 AM12/5/21
to
On Sunday, December 5, 2021 at 12:37:08 AM UTC-5, Dan Christensen wrote:
> On Saturday, December 4, 2021 at 11:06:22 PM UTC-5, Mostowski Collapse wrote:
> > Your 0^0 nonsense only shows that your notion of function
> > is broken.
> Sadly for cranks like you, it works all too well, Jan Burse!
> >Try this function F : R \ {0} -> R, where
> >
> > F(x,y) <=> x*y = 1
> >
> > Also known as reciprocal. Do you claim that we
> > don't know whether ~F(0,1) holds or not?
> Just try to formally prove it. That was YOUR assignment, Jan Burse. We are STILL waiting for your formal proof that f(0)=1 is false given: For all real x and y, if x=/=0 then we have f(x)=y <=> x*y=1. Do your homework!
STILL waiting!
> On Saturday, December 4, 2021 at 11:06:22 PM UTC-5, Mostowski Collapse wrote:
> > Your 0^0 nonsense only shows that your notion of function
> > is broken.
> Sadly for cranks like you, it works all too well, Jan Burse!
> >Try this function F : R \ {0} -> R, where
> >
> > F(x,y) <=> x*y = 1
> >
> > Also known as reciprocal. Do you claim that we
> > don't know whether ~F(0,1) holds or not?
> Just try to formally prove it. That was YOUR assignment, Jan Burse. We are STILL waiting for your formal proof that f(0)=1 is false given: For all real x and y, if x=/=0 then we have f(x)=y <=> x*y=1. Do your homework!
Mostowski Collapse
Dec 5, 2021, 11:20:24 AM12/5/21
to
Corr.:
Need to fix my blind mole proof, serial shouldn't
be there. Don't know how much is needed to capture
"hyperbola" axiomatically.
But I already proved it, what Dan-O-Matik thinks I
didn't prove. For F is hyperbola we have ~F(0,1).
For F is double chimpanzee we cannot decide ~F(0,1).
Thats the fallacy of the double chimpanzee:
Mostowski Collapse schrieb am Sonntag, 5. Dezember 2021 um 13:33:34 UTC+1:
> There is no "if". The algebraic geometry definition is:
Need to fix my blind mole proof, serial shouldn't
be there. Don't know how much is needed to capture
"hyperbola" axiomatically.
But I already proved it, what Dan-O-Matik thinks I
didn't prove. For F is hyperbola we have ~F(0,1).
For F is double chimpanzee we cannot decide ~F(0,1).
Thats the fallacy of the double chimpanzee:
Mostowski Collapse schrieb am Sonntag, 5. Dezember 2021 um 13:33:34 UTC+1:
> There is no "if". The algebraic geometry definition is:
> F(x,y) <=> x*y = 1
> Algebraic geometry is a branch of mathematics,
> classically studying zeros of multivariate polynomials.
> https://en.wikipedia.org/wiki/Algebraic_geometry
>
> Its easy to prove ~F(0,1), just substitute:
>
> ~F(0,1) <=>
> ~0*1 = 1 <=>
> ~0 = 1 <=>
> true
https://groups.google.com/g/sci.logic/c/DBQJkCTGO_I/m/ny6lfI6jBQAJ
> classically studying zeros of multivariate polynomials.
> https://en.wikipedia.org/wiki/Algebraic_geometry
>
> Its easy to prove ~F(0,1), just substitute:
>
> ~F(0,1) <=>
> ~0*1 = 1 <=>
> ~0 = 1 <=>
> true
Mostowski Collapse
Dec 5, 2021, 11:45:00 AM12/5/21
to
Ok here is a corrected proof, now saying for the reciprocal
hyperbola, that the reciprocal hyperbola is defined for x≠0
and otherwise undefined for x=0.
Also fixed a typo concerning the one (i.e. 1=o):
> /* provable */
> F is hyperbola => ALL(x):[x≠0 => x*F(x)=1]
∀a∀b(Fab ↔ m(a,b)=o))
entails ∀a(¬a=z → ∃b(Fab ∧ m(a,b)=o)).
https://www.umsu.de/trees
> /* not provable */
> ALL(x):[x≠0 => x*F(x)=1] => F is hyperbola
does not entail
∀a(¬a=z ↔ ∃bFab) ∧
∀a∀b(Fab ↔ m(a,b)=o).
https://www.umsu.de/trees
hyperbola, that the reciprocal hyperbola is defined for x≠0
and otherwise undefined for x=0.
Also fixed a typo concerning the one (i.e. 1=o):
> /* provable */
> F is hyperbola => ALL(x):[x≠0 => x*F(x)=1]
∀a∀b∀c((Fab ∧ Fac) → b=c) ∧
(∀a(¬a=z ↔ ∃bFab) ∧
∀a∀b(Fab ↔ m(a,b)=o))
entails ∀a(¬a=z → ∃b(Fab ∧ m(a,b)=o)).
https://www.umsu.de/trees
> /* not provable */
> ALL(x):[x≠0 => x*F(x)=1] => F is hyperbola
∀a∀b∀c((Fab ∧ Fac) → b=c) ∧
∀a(¬a=z → ∃b(Fab ∧ m(a,b)=o))
does not entail
∀a(¬a=z ↔ ∃bFab) ∧
∀a∀b(Fab ↔ m(a,b)=o).
https://www.umsu.de/trees
Mostowski Collapse
Dec 13, 2021, 4:02:22 PM12/13/21
to
Donnie Darko whispered:
> 1/0 is simply undefined
This is already an inference. You have an expression
made of 1 and 0, and combined by /. How do you
derive its undefined in DC Proof?
You can't. You would need to lookup the domain
of /, and check whether (1,0) is in the domain.
But when you for example only state as an axiom:
ALL(a):ALL(b):[a in R & b in R & b≠0 => /(a,b) in R]
How would your tool derive:
undefined(1/0) ?
How would you define undefined(_)?
> 1/0 is simply undefined
This is already an inference. You have an expression
made of 1 and 0, and combined by /. How do you
derive its undefined in DC Proof?
You can't. You would need to lookup the domain
of /, and check whether (1,0) is in the domain.
But when you for example only state as an axiom:
ALL(a):ALL(b):[a in R & b in R & b≠0 => /(a,b) in R]
How would your tool derive:
undefined(1/0) ?
How would you define undefined(_)?
Mostowski Collapse
Dec 13, 2021, 4:04:01 PM12/13/21
to
You always state DC Proof can be used for mathematics textbooks.
In a mathematics textbook you find a statement. The same is easy
peasy in ZFC, for the negation there is even the operator ↓ for it:
1/0 is undefined
How do you formalize that in DC Proof. I don't think this is possible,
to defined undefined(_) such that you can plug in 1/0, and there
would be derivable then in DC Proof:
undefined(1/0)
Can you show such a proof? Or do you admit that your claim DC
proof can be used for textbook mathematics is nonsense?
For DC Proof and Dan-O-Matik f:A->B undefined is undefinable,
hence DC Proof cannot be used for textbook mathematics.
The operator ↓ seems to be found for example in:
Feferman, Solomon, 1995, “Definedness,” Erkenntnis, 43 (3): 295–320.
In a mathematics textbook you find a statement. The same is easy
peasy in ZFC, for the negation there is even the operator ↓ for it:
1/0 is undefined
How do you formalize that in DC Proof. I don't think this is possible,
to defined undefined(_) such that you can plug in 1/0, and there
would be derivable then in DC Proof:
undefined(1/0)
Can you show such a proof? Or do you admit that your claim DC
proof can be used for textbook mathematics is nonsense?
For DC Proof and Dan-O-Matik f:A->B undefined is undefinable,
hence DC Proof cannot be used for textbook mathematics.
The operator ↓ seems to be found for example in:
Feferman, Solomon, 1995, “Definedness,” Erkenntnis, 43 (3): 295–320.
Mostowski Collapse
Dec 13, 2021, 4:09:08 PM12/13/21
to
I didn't read the Feferman paper yet, but the operator is exactly
given with the 1/0 example, see yourself:
In applications of free logic involving partial functions, the e
xistence predicate ‘E!’ is often replaced by the postfix definedness
predicate ‘↓’. For any singular term t,t↓ is true if and only if t
has some definite value in D. Thus, for example, the formula
‘(1/0)↓’ is false. While some writers (e.g., Feferman (1995))
distinguish ‘↓’ from ‘E!’, the literature as a whole does not,
and ‘↓’ is often merely a syntactic variant of ‘E!’.
https://plato.stanford.edu/entries/logic-free/#Partial
You say we say something is undefined, like 1/0. By
"we say" I take that mathematics textbooks say it also.
You claim DC proof can model mathematics textbooks,
can you demonstrate the definition of the operator ↓ ?
given with the 1/0 example, see yourself:
In applications of free logic involving partial functions, the e
xistence predicate ‘E!’ is often replaced by the postfix definedness
predicate ‘↓’. For any singular term t,t↓ is true if and only if t
has some definite value in D. Thus, for example, the formula
‘(1/0)↓’ is false. While some writers (e.g., Feferman (1995))
distinguish ‘↓’ from ‘E!’, the literature as a whole does not,
and ‘↓’ is often merely a syntactic variant of ‘E!’.
https://plato.stanford.edu/entries/logic-free/#Partial
You say we say something is undefined, like 1/0. By
"we say" I take that mathematics textbooks say it also.
You claim DC proof can model mathematics textbooks,
can you demonstrate the definition of the operator ↓ ?
Dan Christensen
Dec 13, 2021, 9:51:00 PM12/13/21
to
On Monday, December 13, 2021 at 4:02:22 PM UTC-5, Mostowski Collapse wrote:
> Donnie Darko whispered:
> > 1/0 is simply undefined
>
> This is already an inference. You have an expression
> made of 1 and 0, and combined by /. How do you
> derive its undefined in DC Proof?
>
Learn some logic, Jan Burse. In this context, f(x) being undefined just means that x is not in the domain of function f. I think they teach that in elementary schools these days. It's REALLY basic stuff.
> Donnie Darko whispered:
> > 1/0 is simply undefined
>
> This is already an inference. You have an expression
> made of 1 and 0, and combined by /. How do you
> derive its undefined in DC Proof?
>
Dan
Mostowski Collapse
Feb 10, 2022, 12:24:20 PM2/10/22
to
Dan-O-Matik responded:
> To formally construct the real numbers and the required arithmetic operations
is a massive project of literally tens of thousands of lines of formal proof.
I didn't ask for that. I gave g(x) as follows:
g(x) = x / |x|
Can you prove:
g(x) is undefined at x = 0 ?
> To formally construct the real numbers and the required arithmetic operations
is a massive project of literally tens of thousands of lines of formal proof.
I didn't ask for that. I gave g(x) as follows:
g(x) = x / |x|
Can you prove:
g(x) is undefined at x = 0 ?
Mostowski Collapse
Feb 10, 2022, 12:28:58 PM2/10/22
to
I think you have THE Dan-O-Matik solution for 0^0,
I guess you have also THE Dan-Matik solution for 0/0,
so it should work smoothly to show the desired:
g(x) is undefined at x = 0 ?
Or is 0/0 also subject to he Burse Paradox?
LoL
I guess you have also THE Dan-Matik solution for 0/0,
so it should work smoothly to show the desired:
g(x) is undefined at x = 0 ?
LoL
Ross A. Finlayson
Feb 10, 2022, 12:33:15 PM2/10/22
to
Mostowski Collapse
Feb 10, 2022, 12:36:53 PM2/10/22
to
You need only to define:
(x,y,z) e / <=> [EXISTUNIQUE(t):[x = y * t] & x = y * z]
And have:
ALL(x):[0 = 0 * x]
The rest should follow from Highschool math.
(x,y,z) e / <=> [EXISTUNIQUE(t):[x = y * t] & x = y * z]
And have:
ALL(x):[0 = 0 * x]
The rest should follow from Highschool math.
Daniel Pehoushek
Feb 10, 2022, 2:29:52 PM2/10/22
to
the letter x is banned from the alphabet and from any word or label or symbol
and
the thirty third letter of russian, the tvordy znak, is dropped so that
five bits encodes the thirty two symbols used for reason
and
the universal mom day alpha bet is
0123456789 the ten solids
0abcdefghijklmnopqrstuvw yz
daniel
Mostowski Collapse
Feb 10, 2022, 3:14:57 PM2/10/22
to
I am pretty shure DC poop cannot prove:
g(0) ≠ 1.
It already cannot prove:
g(0) - g(0) ≠ 0.
g(0) ≠ 1.
It already cannot prove:
g(0) - g(0) ≠ 0.
Mostowski Collapse
May 15, 2022, 3:45:39 AM5/15/22
to
You are promoting non-well defined definitions with your
UNDEFINED nonsense. Lets take a closer look at
even natural numbers, how do they look like?
According to Dan Christensen Case 1 type definition:
{ ... dark elements ..., 0, 2, 4, ..., ... dark elements ... }
For example Even(1/2) might or might not hold.
The Euclid understanding Case 2 type definition:
{0, 2, 4, ...}
Evan(1/2) is definitively false.
What is well defined, see here:
In mathematics, a well-defined expression or unambiguous
expression is an expression whose definition assigns it a
unique interpretation or value.
https://en.wikipedia.org/wiki/Well-defined_expression
UNDEFINED nonsense. Lets take a closer look at
even natural numbers, how do they look like?
According to Dan Christensen Case 1 type definition:
{ ... dark elements ..., 0, 2, 4, ..., ... dark elements ... }
For example Even(1/2) might or might not hold.
The Euclid understanding Case 2 type definition:
{0, 2, 4, ...}
Evan(1/2) is definitively false.
What is well defined, see here:
In mathematics, a well-defined expression or unambiguous
expression is an expression whose definition assigns it a
unique interpretation or value.
https://en.wikipedia.org/wiki/Well-defined_expression
Mostowski Collapse
May 15, 2022, 3:48:17 AM5/15/22
to
Its even provable for the Case 2 type definition:
125 ALL(a):[~a ε n => ~Even(a)]
Conclusion, 114
Here is the full proof:
See here:
https://groups.google.com/g/sci.logic/c/gV50jEU6fvQ/m/ncKE65R2AgAJ
125 ALL(a):[~a ε n => ~Even(a)]
Conclusion, 114
Here is the full proof:
See here:
https://groups.google.com/g/sci.logic/c/gV50jEU6fvQ/m/ncKE65R2AgAJ
Dan Christensen
May 15, 2022, 10:57:38 AM5/15/22
to
On Sunday, May 15, 2022 at 3:45:39 AM UTC-4, Mostowski Collapse wrote:
> You are promoting non-well defined definitions with your
> UNDEFINED nonsense.
In your example here, the standard definition of evenness on the natural numbers is applicable ONLY to the natural numbers, not, as you would have it, for everything in the universe:
> You are promoting non-well defined definitions with your
> UNDEFINED nonsense.
ALL(a):[a in N => [Even(a) <=> EXIST(b):[b in n & a=2*b]]]
In words: For every natural number, we say that it is even if and only if it is a multiple of 2.
This definition cannot be applied for -1 or 1/2. In mathematics, you need to distinguish between the predicate Even(x) being false, and Even(x) being undefined.
Mostowski Collapse
May 15, 2022, 11:18:13 AM5/15/22
to
Whats the use case of your nonsense keeping Even open? You proved:
ALL(a):[a e n => [Even(a) => Odd(a+1)]]
You can only extend it to EvenZ and OddZ:
ALL(a):[a e n => [EvenZ(a) => OddZ(a+1)]]
But this works already not anymore:
ALL(a):[a e n => [EvenQ(a) => OddQ(a+1)]]
The problem is unlike the Peano axioms, which are ∀-Formulas,
they can indeed be satisfied by succN, succZ, succQ, succR, etc..
The even definition is a ∀ ∃ - Formula:
ALL(a):[a in N => [Even(a) <=> EXIST(b):[b in n & a=2*b]]]
And you have hard wired N in the EXIST subformula,
by hard wiring (b in n). So keeping even open is nonsense,
evenN, evenZ still work, but evenQ, evenR, etc.. dont work.
In Q this here flips:
~EXIST(a):[a ε n & 2*a=1]
A proper way to do an open even, Case 1 and Case 2, would be:
Case 1: With Parameter D to solve ∀ ∃ - Formula Problem
ALL(a):[a in D => [Even(D,a) <=> EXIST(b):[b in D & a=2*b]]]
Case 2: With Parameter D to solve ∀ ∃ - Formula Problem
ALL(a):[Even(D,a) <=> a e D & EXIST(b):[b in D & a=2*b]]]
But with parameter case 2 is still better, since it is well defined.
ALL(a):[a e n => [Even(a) => Odd(a+1)]]
You can only extend it to EvenZ and OddZ:
ALL(a):[a e n => [EvenZ(a) => OddZ(a+1)]]
But this works already not anymore:
ALL(a):[a e n => [EvenQ(a) => OddQ(a+1)]]
The problem is unlike the Peano axioms, which are ∀-Formulas,
they can indeed be satisfied by succN, succZ, succQ, succR, etc..
The even definition is a ∀ ∃ - Formula:
ALL(a):[a in N => [Even(a) <=> EXIST(b):[b in n & a=2*b]]]
by hard wiring (b in n). So keeping even open is nonsense,
evenN, evenZ still work, but evenQ, evenR, etc.. dont work.
In Q this here flips:
~EXIST(a):[a ε n & 2*a=1]
A proper way to do an open even, Case 1 and Case 2, would be:
Case 1: With Parameter D to solve ∀ ∃ - Formula Problem
ALL(a):[a in D => [Even(D,a) <=> EXIST(b):[b in D & a=2*b]]]
Case 2: With Parameter D to solve ∀ ∃ - Formula Problem
ALL(a):[Even(D,a) <=> a e D & EXIST(b):[b in D & a=2*b]]]
But with parameter case 2 is still better, since it is well defined.
Earle Jones
Nov 26, 2022, 2:44:03 PM11/26/22
to
On Thu Feb 10 11:29:50 2022 Daniel Pehoushek wrote:
> On Thursday, February 10, 2022 at 12:36:53 PM UTC-5, Mostowski Collapse wrote:
> > You need only to define:
> >
> > (x,y,z) e / <=> [EXISTUNIQUE(t):[x = y * t] & x = y * z]
> >
> > And have:
> >
> > ALL(x):[0 = 0 * x]
> >
> > The rest should follow from Highschool math.
> > Mostowski Collapse schrieb am Donnerstag, 10. Februar 2022 um 18:28:58 UTC+1:
> > > I think you have THE Dan-O-Matik solution for 0^0,
> > > I guess you have also THE Dan-Matik solution for 0/0,
> > > so it should work smoothly to show the desired:
> > > g(x) is undefined at x = 0 ?
> > > Or is 0/0 also subject to he Burse Paradox?
> > >
> > > LoL
> > > Mostowski Collapse schrieb am Donnerstag, 10. Februar 2022 um 18:24:20 UTC+1:
> > > > Dan-O-Matik responded:
> > > > > To formally construct the real numbers and the required arithmetic operations
> > > > is a massive project of literally tens of thousands of lines of formal proof.
> > > >
> > > > I didn't ask for that. I gave g(x) as follows:
> > > >
> > > > g(x) = x / |x|
> > > >
> > > > Can you prove:
> > > >
> > > > g(x) is undefined at x = 0 ?
>
[xxxxx]
> On Thursday, February 10, 2022 at 12:36:53 PM UTC-5, Mostowski Collapse wrote:
> > You need only to define:
> >
> > (x,y,z) e / <=> [EXISTUNIQUE(t):[x = y * t] & x = y * z]
> >
> > And have:
> >
> > ALL(x):[0 = 0 * x]
> >
> > The rest should follow from Highschool math.
> > Mostowski Collapse schrieb am Donnerstag, 10. Februar 2022 um 18:28:58 UTC+1:
> > > I think you have THE Dan-O-Matik solution for 0^0,
> > > I guess you have also THE Dan-Matik solution for 0/0,
> > > so it should work smoothly to show the desired:
> > > g(x) is undefined at x = 0 ?
> > > Or is 0/0 also subject to he Burse Paradox?
> > >
> > > LoL
> > > Mostowski Collapse schrieb am Donnerstag, 10. Februar 2022 um 18:24:20 UTC+1:
> > > > Dan-O-Matik responded:
> > > > > To formally construct the real numbers and the required arithmetic operations
> > > > is a massive project of literally tens of thousands of lines of formal proof.
> > > >
> > > > I didn't ask for that. I gave g(x) as follows:
> > > >
> > > > g(x) = x / |x|
> > > >
> > > > Can you prove:
> > > >
> > > > g(x) is undefined at x = 0 ?
>
*
To begin to understand the meaning of the expression x^x we need to examine this:
Consider the relationship: y = x^x.
First: for all positive x (x > 0)
The curve is well-behaved for positive x. For any positive x (rational or irrational) there is a real value for y.
x y
1 1
2 4
3 27
pi about 36.5
10 10^10
etc.
Aside: Interestingly enough, try some x values between 0 and 1:
Try x = 0.1 and 0.3 and 0.6 and 0.9
You will notice that there is a minimum y between x = 0 and x = 1.
Can you find that minimum y in the 0 to 1 interval?
Now let s look at negative x (x < 0)
First we consider rational x values (x = m/n for integer m and n.)
If n is odd, y will be real; if n is even, y will be imaginary.
x y
- about + 0.6934.. (real)
- about - 0.707.. i (imaginary)
Negative x results in a strange distribution of y values.
The limit of x^x as x approaches zero (FROM THE RIGHT -- POSITIVE X) = 1.
What is the limit of x^x as x approaches zero (FROM THE LEFT -- NEGATIVE X) = ??
An exercise for the reader.
earle
*
Sergi o
Nov 26, 2022, 3:29:41 PM11/26/22
to
> First we consider rational x values (x = m/n for integer m and n.)
> If n is odd, y will be real; if n is even, y will be imaginary.
> x y
> - about + 0.6934.. (real)
> - about - 0.707.. i (imaginary)
>
> Negative x results in a strange distribution of y values.
>
> The limit of x^x as x approaches zero (FROM THE RIGHT -- POSITIVE X) = 1.
> What is the limit of x^x as x approaches zero (FROM THE LEFT -- NEGATIVE X) = ??
>
> An exercise for the reader.
>
> earle
> *
1
> If n is odd, y will be real; if n is even, y will be imaginary.
> x y
> - about + 0.6934.. (real)
> - about - 0.707.. i (imaginary)
>
> Negative x results in a strange distribution of y values.
>
> The limit of x^x as x approaches zero (FROM THE RIGHT -- POSITIVE X) = 1.
> What is the limit of x^x as x approaches zero (FROM THE LEFT -- NEGATIVE X) = ??
>
> An exercise for the reader.
>
> earle
> *
where is the maximum on the negative side, between 0 and -1 ?
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