Science Student Posted March 6, 2014 Share Posted March 6, 2014 In other words, what axiom implies 1 < 2? 1 Link to comment Share on other sites More sharing options...
John Posted March 7, 2014 Share Posted March 7, 2014 (edited) One way we might (begin to) construct the natural numbers is to define 0 as the empty set Ø and define the successor function S(x) = x ∪ {x} for any set x. It follows that each natural number n ≠ 0 equals {0, 1, 2, ..., n - 1}. So we have 0 = Ø 1 = S(0) = {0} = {Ø} 2 = S(1) = {0, 1} = {Ø, {Ø}} 3 = S(2) = {0, 1, 2} = {Ø, {Ø}, {Ø, {Ø}}} etc. Of course, only heretics take zero to be a natural number, and this construction still works if we begin with 1 = Ø instead of 0 = Ø, but I think the latter is slightly easier to conceptualize. In any case, we can then define the relation < in terms of elementhood such that a < b ↔ a ∈ b. Thus, from the construction given above, 1 < 2. Of course, there are other possible constructions (though as far as I know, they all follow the same general process). A nice example, along with a more detailed definition and discussion of <, can be found here. Edited March 7, 2014 by John 5 Link to comment Share on other sites More sharing options...
Science Student Posted March 7, 2014 Author Share Posted March 7, 2014 One way we might (begin to) construct the natural numbers is to define 0 as the empty set Ø and define the successor function S(x) = x ∪ {x} for any set x. It follows that each natural number n ≠ 0 equals {0, 1, 2, ..., n - 1}. So we have 0 = Ø 1 = S(0) = {0} = {Ø} 2 = S(1) = {0, 1} = {Ø, {Ø}} 3 = S(2) = {0, 1, 2} = {Ø, {Ø}, {Ø, {Ø}}} etc. Of course, only heretics take zero to be a natural number, and this construction still works if we begin with 1 = Ø instead of 0 = Ø, but I think the latter is slightly easier to conceptualize. In any case, we can then define the relation < in terms of elementhood such that a < b ↔ a ∈ b. Thus, from the construction given above, 1 < 2. Of course, there are other possible constructions (though as far as I know, they all follow the same general process). A nice example, along with a more detailed definition and discussion of <, can be found here. Oh god, this is well beyond what I was expecting - thanks a lot! 1 Link to comment Share on other sites More sharing options...
Olinguito Posted March 13, 2014 Share Posted March 13, 2014 In other words, what axiom implies 1 < 2? It depends of what your set of axioms is for. For example, your axioms might be for the real numbers as an ordered field. An field [latex]\langle F,+,\cdot\rangle[/latex] is said to be (totally) ordered by [latex]\leq[/latex] iff the following holds: [latex]\forall\,a,b,c\in F[/latex]: [latex]\text{either}\ a\leq b\ \text{or}\ b\leq a[/latex] [latex]a\leq b\ \text{and}\ b\leq a\ \Rightarrow\ a=b[/latex] [latex]a\leq b\ \text{and}\ b\leq c\ \Rightarrow\ a\leq c[/latex] [latex]a\leq b\ \Rightarrow\ a+c\leq b+c[/latex] [latex]0\leq a\ \text{and}\ 0\leq b\ \Rightarrow\ 0\leq ab[/latex] We first show that [latex]0\leq 1[/latex]. By axiom (1) either [latex]0\leq 1[/latex] or [latex]1\leq 0[/latex]. If [latex]1\leq 0[/latex] then [latex]1+(-1)\leq 0+(-1)[/latex] (axiom (4)) i.e. [latex]0\leq -1[/latex]. Then [latex]0\leq -1[/latex] and [latex]0\leq -1[/latex] imply (axiom (5)) [latex]0\leq (-1)(-1)=1[/latex]. (This is actually a contradiction because [latex]1\leq0[/latex] and [latex]0\leq 1[/latex] imply (axiom (2)) [latex]0=1[/latex].) So we can’t have [latex]1\leq 0[/latex]; hence we must have [latex]0\leq 1[/latex]. And now we are done, for [latex]0\leq 1[/latex] [latex]\implies[/latex] (axiom (4)) [latex]1=0+1\leq 1+1=2[/latex]. http://en.wikipedia.org/wiki/Ordered_field 2 Link to comment Share on other sites More sharing options...
ajb Posted March 15, 2014 Share Posted March 15, 2014 As a comment, it too Russell and Whitehead 378 pages to even begin how to discuss how 1+1=2. Of course, they were trying to do it all using logic. So, my thoughts are 1<2 is obvious but trying to prove it could be difficult, depending on where you start. 1 Link to comment Share on other sites More sharing options...
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