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How do I prove that 1 < 2?

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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. :P

 

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 by John

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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. :P

 

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!

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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

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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.

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