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Re: 0^0

#231

Post by a1a »

rntz wrote: post #228
a1a wrote: post #220

Earlier quoted context omitted.

>There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. I would love to see a proof of 1+1=2. For example in the case of elliptic curves, as far as I know addition is simply an axiom. E.g. A+A'=0= inf , where inf is the point at infinity and A' is the reflection of A. Math is not related to nature. Try to reason your way to matrix multiplication with apples: A * B != B * A

Here is a simple definition of the natural numbers and addition, in Haskell: data Nat = Zero | Suc Nat plus :: (Nat, Nat) -> Nat plus (Zero, y) = y -- axiom 1 plus (Suc x, y) = Suc (plus (x, y)) -- axiom 2 one = Suc Zero two = Suc one Here is a proof that plus (one, one) = two: plus (Suc Zero, Suc Zero) = Suc (plus (Zero, Suc Zero)) [by axiom 2] = Suc (Suc Zero) [by axiom 1]

Can you please translate that to mathematics? My request was regarding a formal mathematical proof.

Re: 0^0

#232

Post by pavelrub »

thomasahle wrote: post #216
pavelrub wrote: post #204

Earlier quoted context omitted.

Whatever it is, 2^1=2 is not part of it - it follows from it.

Uhm, anything that is part of a definition follows from that definition. `A and B implies A' and so on. In the end a function, in the mathematical sense, is just a long (possibly infinite) table of `from' and `to' values. We could argue that one definition of ^ is more or less complex than another, for some definition of complexity, but I'm not sure which would win, and if it would be useful.

"Follows" in the sense that it is not an axiom - it has a proof of more than one line in the sense of the Hilbert system. Seeing functions as collections of ordered pairs tells us nothing here.

Your definition of ^ doesn't directly tell me that 2^1=2. It does tell me however that 0^0=1.

Re: 0^0

#233

Post by pavelrub »

a1a wrote: post #220
pavelrub wrote: post #193

Earlier quoted context omitted.

This is simply wrong. 1+1=2 is the same kind of statement as 2^1=2 - it follows from the definition of the operation and from a general rule (it would be absurd here to say that 2^1 is the definition of 2, or that 2^1=2 is part of the definition of ^). 0^0 is just a meaningless symbol unless given explicit definition. There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. The symbol…

>There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. I would love to see a proof of 1+1=2. For example in the case of elliptic curves, as far as I know addition is simply an axiom. E.g. A+A'=0= inf , where inf is the point at infinity and A' is the reflection of A. Math is not related to nature. Try to reason your way to matrix multiplication with apples: A * B != B * A

Using the Peano axioms, we need to prove that S(0)+S(0)=S(S(0)).

From the definition of +: S(0)+S(0)=S(S(0)+0) Again using the definition of +: S(0)+0=S(0). And we get: S(0)+S(0)=S(S(0))

Q.E.D.

There is also a famous proof by Whitehead and Russell on page 379 of Principia Mathematica: http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhist...

Re: 0^0

#234

Post by a1a »

pavelrub wrote: post #233
a1a wrote: post #220

Earlier quoted context omitted.

>There is no popular set of axioms where 1+1=2 is taken as an axiom - it is always proven. I would love to see a proof of 1+1=2. For example in the case of elliptic curves, as far as I know addition is simply an axiom. E.g. A+A'=0= inf , where inf is the point at infinity and A' is the reflection of A. Math is not related to nature. Try to reason your way to matrix multiplication with apples: A * B != B * A

Using the Peano axioms, we need to prove that S(0)+S(0)=S(S(0)). From the definition of +: S(0)+S(0)=S(S(0)+0) Again using the definition of +: S(0)+0=S(0). And we get: S(0)+S(0)=S(S(0)) Q.E.D. There is also a famous proof by Whitehead and Russell on page 379 of Principia Mathematica: http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhist...

Thank you, I didn't know you could prove that!

I believe my disagreement was specifically about this (quoted):

> In principle a computer can correctly recognize 2 apples as

> being 2 apples without knowing anything about addition and

> without being able to recognize 1 apple.

I have a hard time seeing this. If I am not mistaken again, there is no definition of the symbol 2 that does not include addition. The proof above doesn't really prove this either as Peano's axiom rely on the definition of "successor". Successor definition: "a+1 is the successor of a".

http://mathworld.wolfram.com/PeanosAxioms.html

http://mathworld.wolfram.com/Successor.html

Re: 0^0

#235

Post by pavelrub »

a1a wrote: post #234
pavelrub wrote: post #233

Earlier quoted context omitted.

Using the Peano axioms, we need to prove that S(0)+S(0)=S(S(0)). From the definition of +: S(0)+S(0)=S(S(0)+0) Again using the definition of +: S(0)+0=S(0). And we get: S(0)+S(0)=S(S(0)) Q.E.D. There is also a famous proof by Whitehead and Russell on page 379 of Principia Mathematica: http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhist...

Thank you, I didn't know you could prove that! I believe my disagreement was specifically about this (quoted): > In principle a computer can correctly recognize 2 apples as > being 2 apples without knowing anything about addition and > without being able to recognize 1 apple. I have a hard time seeing this. If I am not mistaken again, there is no definition of the symbol 2 that does not include addition. The proof ab…

See the edit in my reply to tomp. In the Peano axioms, 2 is defined as S(S(0)), not as S(0)+1. The correct way to think about this is to ignore any inclination to give those symbols any "real world" meaning. From the point of view of the formal system - they are just strings, and the only thing we know about them is how to manipulate them to form other strings. Under this perspective, the connection between S(S(0)) and S(0)+1, or in general between S(a) and a+1, is again something which requires proof: S(a) = S(a+0) = a + S(0).

A computer can be given an explicit map between the symbols S(0), S(S(0)), s(S(S(0))), ... and 1,2,3..., so to identify S(S(S(S(S(S(0)))))) with 6 - it wouldn't need to know anything at all about + or about 1.

Edit: also consider the following quote from Wittgenstein's Philosophical Investigations and whether the person described needs to have any concept of 'addition' in order to correctly use numbers: "Now think of the following use of language: I send someone shopping. I give him a slip marked 'five red apples'. He takes the slip to the shopkeeper, who opens the drawer marked 'apples', then he looks up the word 'red' in a table and finds a colour sample opposite it; then he says the series of cardinal numbers--I assume that he knows them by heart--up to the word 'five' and for each number he takes an apple of the same colour as the sample out of the drawer.--It is in this and simlar ways that one operates with words--"But how does he know where and how he is to look up the word 'red' and what he is to do with the word 'five'?" ---Well, I assume that he 'acts' as I have described. Explanations come to an end somewhere.--But what is the meaning of the word 'five'? --No such thing was in question here, only how the word 'five' is used."

Re: 0^0

#236

Post by ubercow13 »

einhverfr wrote: post #225
ubercow13 wrote: post #218

Earlier quoted context omitted.

y=x^2/x and y=x aren't the same thing though. The first is undefined at x=0, the second is not. How does this break algebra?

Because if you can't simplify, you can't treat them as the same, which is a fundamental premise of algebra. I.e. the promise/premise of algebra is that if you take an equation, it remains equally valid when you add, subtract, multiply, or divide both sides by the same expression. If you treat these as distinct, then every division operation where you divide by a variable expression reduces the domain of possible answ…

I don't really understand what you mean. You can never divide by zero. You can't treat the functions as the same because they are not the same function. It's not a fundamental premise of algebra that you should be able to.

The corner cases aren't nasty, you have to take care of them every time. Like with your example, the solution is f(x)=x+2, except where x=-2 where f(-2) can be any number. There are a continuum of solutions to the equation each with a different value of f at x=-2.

Re: 0^0

#237

Post by lotharbot »

einhverfr wrote: post #189
lotharbot wrote: post #181

Earlier quoted context omitted.

> "x * f(x) = x cannot be equivalent to f(x) = x/x" Sure it's equivalent, over a domain not including x=0. This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. We work in restricted domains in mathematics all the time. > " f'(x) = 2x/x" f'(x) = lim (h->0) [2(x+h)-2x]/h. Since h is approaching (and therefore not equal to) zero, t…

> This does not break algebra any more than, say, restricting the domain of the square root (when working in the reals) to non-negative numbers. Sure it does, because if that is the case, you restrict your domain when you divide by a variable expression. If you divide both sides by x-1, then you effectively rule out 1 from the domain. That's the problem. Now this is not the same as 0/0. The point is that 0/0 is only…

> "you restrict your domain when you divide by a variable expression"

Why is this a problem?

Whenever you perform an operation that has a restricted domain, you restrict your domain. This may result in an actual "not defined at x=1" result, or simply "the value at x=1 is found through an alternative method" result.

> "0/0 is only undefined when it persists after simplification"

When you're working in the context of limits, it wasn't an actual 0/0 to begin with; it was near-0/near-0, which is perfectly OK to simplify. The limit defines a function that already has a restricted domain -- h->0 means h is not actually zero. The expression naively evaluating to 0/0 simply tells you that you need to do more work to properly evaluate it -- 0/0 is not the actual result.

Note that using the limit to find the derivative gives you a function that you'd like to be continuous in x, but the divide-by-zero is in h. Consider f(x)=x^2. The derivative is

lim h->0 [(x+h)^2 - x^2 ] /h

lim h->0 [ x^2 + 2xh + h^2 - x^2 ] /h

lim h->0 [ 2xh + h^2 ] / h

lim h->0 [2x + h] * h/h

since h does NOT equal zero, we can treat h/h=1, and the limit trivially collapses to 2x. Note that we never had the variable x in the denominator of our fraction; we never placed a restriction on x or suggested anything about a discontinuity relative to x. We only restricted h, which was already restricted by the limit itself.

Re: 0^0

#238

Post by krick »

philbarr wrote: post #217
rickhanlonii wrote: post #86

It's very important to note here that 0^0=1 is a shorthand and not a truth . Mathematicians are absolutely not stating that they have proven, or that it is true, that 0^0=1. It is a definition, not a claim of equality. They're not saying "0^0 is 1" in the sense that they say "1+1 is 2" or "0.999... is 1". They're saying "we define 0^0 to be 1". The difference is more than just pedantry, it strikes at the core of why…

Can anyone recommend a really good book on mathematics that demonstrates the "beauty" of it? I'd love to really learn and understand about Fast Fourier Transforms and the like, but any book I have is from my old college days and is just so bloody tedious. I used to really love maths when I was younger but it got beaten out of me by endless repetition and "now do Questions 1 - 50" tedium. Now I get scared whenever I s…

> book on mathematics that demonstrates the "beauty" of it

I'd say that you're asking wrong question. Yes, there're quite a few books intended to demonstrate the very thing you're asking for: beauty of mathematics. You are even likely to find some showcases interesting, might be somewhat beautiful. But essentially it won't be the same "beauty" that mathematicians see. It will be more like a thousand more-or-less scientific examples of how number φ appears in different places in our life, which isn't the essence of it obviously.

So if you're scared by "summation" symbol you're better to reconsider or forget about understanding "beauty of mathematics" forever. Because "beauty of math" is math itself.

About FFT: are you sure that you mean FFT? Because it's merely an algorithm to compute Discrete Fourier Transform. If you want to understand the Fourier Transformation itself, here's basic explanation: http://betterexplained.com/articles/an-interactive-guide-to-...

Anyway, here're some easy-to-read books about math considered good by various people:

  Hardy, "Mathematician's Apology"
  Hofstadter, "Gödel, Escher, Bach"
  Klein, "Mathematics and the Search for Knowledge" (he wrote many sci-pop books which I expect to be good, but I read only few of them, so I'll recommend only this one)
  Сонин, "Постижение совершенства" — a good one, but I don't know if it was ever translated to english, sorry

Re: 0^0

#239

Post by a1a »

pavelrub wrote: post #235
a1a wrote: post #234

Earlier quoted context omitted.

Thank you, I didn't know you could prove that! I believe my disagreement was specifically about this (quoted): > In principle a computer can correctly recognize 2 apples as > being 2 apples without knowing anything about addition and > without being able to recognize 1 apple. I have a hard time seeing this. If I am not mistaken again, there is no definition of the symbol 2 that does not include addition. The proof ab…

See the edit in my reply to tomp. In the Peano axioms, 2 is defined as S(S(0)), not as S(0)+1. The correct way to think about this is to ignore any inclination to give those symbols any "real world" meaning. From the point of view of the formal system - they are just strings, and the only thing we know about them is how to manipulate them to form other strings. Under this perspective, the connection between S(S(0)) a…

This is quite beyond my knowledge, I like the Wittgenstein philosophy, but I would still argue:

You cannot calculate S^(x+1)(0) before the result of S^x(0) is known.

To calculate S^6(0) you start by calculating S(0) ----> It's not possible to "identify S^6(0) with 6" if you haven't calculated S(0) first, because you cannot know S^6(0) at this point.

Re: 0^0

#240

Post by pavelrub »

a1a wrote: post #239
pavelrub wrote: post #235

Earlier quoted context omitted.

See the edit in my reply to tomp. In the Peano axioms, 2 is defined as S(S(0)), not as S(0)+1. The correct way to think about this is to ignore any inclination to give those symbols any "real world" meaning. From the point of view of the formal system - they are just strings, and the only thing we know about them is how to manipulate them to form other strings. Under this perspective, the connection between S(S(0)) a…

This is quite beyond my knowledge, I like the Wittgenstein philosophy, but I would still argue: You cannot calculate S^(x+1)(0) before the result of S^x(0) is known. To calculate S^6(0) you start by calculating S(0) ----> It's not possible to "identify S^6(0) with 6" if you haven't calculated S(0) first, because you cannot know S^6(0) at this point.

The computer has a table in memory that tells it S^6(0) = 6 (or equivalently O O O O O O = 6 apples). When it sees a symbol - for example S^6(0) - it searches this table for the same symbol, and then outputs the corresponding number. In no stage of this process does the computer "calculate" anything - it doesn't even need to know that S^5(0)=5 in order to find that S^6(0)=6. In fact S^5 might not even be in the table.
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