4 ms·

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

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 2 is not defined as 1+1. Conceptually this reflects the fact that knowing what 2 means is different from knowing what 1 means and what addition means. 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.

tomp 13y ago
Maybe your parent meant that "1 + 1 = 2" is the definition of "2", in the sense that we write the symbol "2" to denote the Peano successor of "1", the Peano successor of "0", which is not a successor.
- pavelrub 13y ago
  Maybe. Formally this is inaccurate as the successor function is not the addition function, and one has to prove that S(S(0))=S(0)+S(0). Edit: it should also be noted that this equality shouldn't seem "obvious": the name "successor" here is misleading, as it automatically links to our everyday understanding of a successor as being "the next number", in which case it is indeed obvious. But formally it's just a meaningless symbol that can be changed to other meaningless symbols by certain meaningless rules. Our everyday life association of (apple) -> S(0), (apple,apple)->S(S(0)) is then a kind of a model for those meaningless things, but from within the formal system - they have no meaning. In general I also do not like the invocation of formal arguments here, I think it misses the point. The bigger issue is that for us humans there is an obvious difference in kind between 1+1=2 and 0^0=1 - this should be obvious. Trying to hide this difference by appealing to different kinds of language trickery and formalism is an indication of doing something wrong, not of discovering anything new. The difference is still there, and we should understand why it's there, not force it to disappear (compare to the claim that every human act is egoistic in nature - what information does it really give us about the human nature?).
  - rntz 13y ago
    I did not in fact mean that 2 is defined as 1+1; you (and several others) are correct in saying that 1+1=2 requires proof for all reasonable definitions of +. 0^0 = 1 follows much more directly from the definitions in which it is true. My overall point has been articulated by arketyp in a sibling post to tomp's; like ^, + is "just a definition" which we arranged so as to line up with our intuitions. I disagree with your assertion that 1+1=2 and 0^0=1 are of completely diferent kinds. That depends on what intuitive angle you approach them from; my intuition for ^ is that it is the iteration of multiplication, and in this context the only sensible definition is that x^0 = 1 (the unit of multiplication), for any x. The generalization to continuous bases and exponents should preserve this property. But this is certainly a matter of taste, and certainly you'll find more disagreement here than in the matter of 1+1=2. What this tells us about human nature, I think, is not much: humans will disagree more when things get more complicated (^ is more complicated/abstract than +) and there is more room for disagreement.
thomasahle 13y ago
What do you think the definition of ^ is then? The ^ operator is defined as: 0^0 = 1 x^y = exp(ylog(x)) if x != 0 0^y = 0 if y != 0 Or similarly set theoretically.
- pavelrub 13y ago
  Whatever it is, 2^1=2 is not part of it - it follows from it.
  - thomasahle 13y ago
    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.
    - pavelrub 13y ago
      "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.
- Aardwolf 13y ago
  Isn't that a circular definition, since, exp is e^x?
  - taejo 13y ago
    If one defines x^y this way, one usually defines exp by a power series, and then defines e as the number such that e^x = exp(x).
    - 13y ago
arketyp 13y ago
This is a good point. Your grandparent's original distinction was that 1+1=2, just as 2^1=2, follows from a definition while 0^0=1 is immediately defined. However, the parent's still valid point was clearly that even the plus operator is in the larger sense just as arbitrary as the definition of 0^0, concurring with the overall theme of this thread that mathematics ultimately gives no credibility to intuition.
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.

tomp 13y ago

Maybe your parent meant that "1 + 1 = 2" is the definition of "2", in the sense that we write the symbol "2" to denote the Peano successor of "1", the Peano successor of "0", which is not a successor.

pavelrub 13y ago
Maybe. Formally this is inaccurate as the successor function is not the addition function, and one has to prove that S(S(0))=S(0)+S(0). Edit: it should also be noted that this equality shouldn't seem "obvious": the name "successor" here is misleading, as it automatically links to our everyday understanding of a successor as being "the next number", in which case it is indeed obvious. But formally it's just a meaningless symbol that can be changed to other meaningless symbols by certain meaningless rules. Our everyday life association of (apple) -> S(0), (apple,apple)->S(S(0)) is then a kind of a model for those meaningless things, but from within the formal system - they have no meaning. In general I also do not like the invocation of formal arguments here, I think it misses the point. The bigger issue is that for us humans there is an obvious difference in kind between 1+1=2 and 0^0=1 - this should be obvious. Trying to hide this difference by appealing to different kinds of language trickery and formalism is an indication of doing something wrong, not of discovering anything new. The difference is still there, and we should understand why it's there, not force it to disappear (compare to the claim that every human act is egoistic in nature - what information does it really give us about the human nature?).
- rntz 13y ago
  I did not in fact mean that 2 is defined as 1+1; you (and several others) are correct in saying that 1+1=2 requires proof for all reasonable definitions of +. 0^0 = 1 follows much more directly from the definitions in which it is true. My overall point has been articulated by arketyp in a sibling post to tomp's; like ^, + is "just a definition" which we arranged so as to line up with our intuitions. I disagree with your assertion that 1+1=2 and 0^0=1 are of completely diferent kinds. That depends on what intuitive angle you approach them from; my intuition for ^ is that it is the iteration of multiplication, and in this context the only sensible definition is that x^0 = 1 (the unit of multiplication), for any x. The generalization to continuous bases and exponents should preserve this property. But this is certainly a matter of taste, and certainly you'll find more disagreement here than in the matter of 1+1=2. What this tells us about human nature, I think, is not much: humans will disagree more when things get more complicated (^ is more complicated/abstract than +) and there is more room for disagreement.

thomasahle 13y ago

What do you think the definition of ^ is then? The ^ operator is defined as: 0^0 = 1 x^y = exp(ylog(x)) if x != 0 0^y = 0 if y != 0 Or similarly set theoretically.

pavelrub 13y ago
Whatever it is, 2^1=2 is not part of it - it follows from it.
- thomasahle 13y ago
  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.
  - pavelrub 13y ago
    "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.
Aardwolf 13y ago
Isn't that a circular definition, since, exp is e^x?
- taejo 13y ago
  If one defines x^y this way, one usually defines exp by a power series, and then defines e as the number such that e^x = exp(x).
  - 13y ago

arketyp 13y ago

This is a good point. Your grandparent's original distinction was that 1+1=2, just as 2^1=2, follows from a definition while 0^0=1 is immediately defined. However, the parent's still valid point was clearly that even the plus operator is in the larger sense just as arbitrary as the definition of 0^0, concurring with the overall theme of this thread that mathematics ultimately gives no credibility to intuition.

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.