5 ms·

>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 ell

by a1a 13y ago

>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

rntz 13y ago
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]
- a1a 13y ago
  Can you please translate that to mathematics? My request was regarding a formal mathematical proof.
  - rntz 13y ago
    The code I gave translates easily into Agda, a computerized proof checker, and as such more formal than most mathematics. However, here's the same thing in a modernized version of Peano arithmetic. We assume all the usual properties of equality: reflexivity, symmetry, transitivity, and substitution. -- Axioms (only 1 and 2 are relevant) 1. 0 ∈ N 2. ∀ x∈N. S(x) ∈ N 3. ∀ x∈N. 0 ≠ S(x) 4. ∀ x∈N, y∈N. S(x) = S(y) ⊃ x = y 5. P(0) ∧ (∀ x∈N. P(x) ⊃ P(S(x))) ⊃ ∀ x∈N. P(x) -- Definition of addition 6. ∀ a∈N. a + 0 = a 7. ∀ a,b ∈ N. a + S(b) = S(a+b) -- Proof that S(0) + S(0) = S(S(0)) 9. S(0) ∈ N [from 2 and 1] 10. S(0) + S(0) = S(S(0) + 0) [from 7 and 9] 11. S(0) + 0 = S(0) [from 6 and 9] 12. S(S(0) + 0) = S(S(0)) [substitution of equals, from 11] 13. S(0) + S(0) = S(S(0)) [transitivity from 10 and 12] Edit: Ah, I see I was beaten to it by pavelrub.
pavelrub 13y ago
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=umhistmath&cc=umhistmath&idno=aat3201.0001.001&frm=frameset&view=image&seq=401 http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhist...
- a1a 13y ago
  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/PeanosAxioms.html http://mathworld.wolfram.com/Successor.html http://mathworld.wolfram.com/Successor.html
  - pavelrub 13y ago
    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."

by a1a 13y ago

rntz 13y ago

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]

a1a 13y ago
Can you please translate that to mathematics? My request was regarding a formal mathematical proof.
- rntz 13y ago
  The code I gave translates easily into Agda, a computerized proof checker, and as such more formal than most mathematics. However, here's the same thing in a modernized version of Peano arithmetic. We assume all the usual properties of equality: reflexivity, symmetry, transitivity, and substitution. -- Axioms (only 1 and 2 are relevant) 1. 0 ∈ N 2. ∀ x∈N. S(x) ∈ N 3. ∀ x∈N. 0 ≠ S(x) 4. ∀ x∈N, y∈N. S(x) = S(y) ⊃ x = y 5. P(0) ∧ (∀ x∈N. P(x) ⊃ P(S(x))) ⊃ ∀ x∈N. P(x) -- Definition of addition 6. ∀ a∈N. a + 0 = a 7. ∀ a,b ∈ N. a + S(b) = S(a+b) -- Proof that S(0) + S(0) = S(S(0)) 9. S(0) ∈ N [from 2 and 1] 10. S(0) + S(0) = S(S(0) + 0) [from 7 and 9] 11. S(0) + 0 = S(0) [from 6 and 9] 12. S(S(0) + 0) = S(S(0)) [substitution of equals, from 11] 13. S(0) + S(0) = S(S(0)) [transitivity from 10 and 12] Edit: Ah, I see I was beaten to it by pavelrub.

pavelrub 13y ago

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=umhistmath&cc=umhistmath&idno=aat3201.0001.001&frm=frameset&view=image&seq=401 http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhist...

a1a 13y ago
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/PeanosAxioms.html http://mathworld.wolfram.com/Successor.html http://mathworld.wolfram.com/Successor.html
- pavelrub 13y ago
  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."