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In the construction of natural number system, I'm not sure how the ordering of elements of N is defined. It seems that almost every approach to that is quite abstract without mentioning an actual number except 1. Then, how do you determine whether 2 is bigger than 1? It would be great if you recommend a book that will be perfect for my question—traditional textbooks for set theory doesn't seem to cover this topic.

Asaf Karagila
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eca2ed291a2f572f66f4a5fcf57511
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    I think that it is a definition. $1+1>1$, and it is defined $1+1=2$. Hence we have $2>1$. – SchrodingersCat Nov 14 '15 at 14:57
  • $1+1 > 1\Leftrightarrow 1+1+(-1) > 1+(-1)\Leftrightarrow 1>0$. Based on the axioms of real numbers. What we call $1+1$ is just a notation thing. – AlvinL Nov 14 '15 at 14:59
  • @Aniket Where can I find the definition of 2? That's the point of my question. – eca2ed291a2f572f66f4a5fcf57511 Nov 14 '15 at 15:07
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    It is a conventional definition. If you want, you can call $1+1=6$, but the only problem is that others won't understand this until and unless, you tell them about your unique definition. It has been unanimously accepted that $1+1=2$. And please don't ask why everybody accepted that definition. Then you have to check history books. – SchrodingersCat Nov 14 '15 at 15:12

2 Answers2

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Standard construction is:

  • $0:=\varnothing$

  • $n+1:=n\cup\{n\}$

Order $<$ is actually the same as order $\in$.

Then $1=\{0\}$ and $2=\{0,1\}$ so that $1\in2$ or equivalently $1<2$.

drhab
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  • It seems you're assuming that $1+1:=2$, which is basically not a proof. The ordering is conventional and its ambiguity never be resolved? – eca2ed291a2f572f66f4a5fcf57511 Nov 14 '15 at 15:29
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    In my answer I could also have said: $n^+:=n\cup{n}$ in stead of $n+1:=n\cup{n}$. After defining that you can define on the naturals the operation $+$ by stating that $n+0:=n$ and $n+m^+:=(n+m)^+$. The sets $\varnothing$, $\varnothing^+$ and $(\varnothing^+)^+$ are denoted/abbreviated by the symbols $0,1,2$ respectively. In that context $1+1=(1+0)^+=1^+=2$. In that sense $1+1=2$ by definition, so no proof is needed. The precedence of $1$ w.r.t $2$ (title of your question) is on base of $1\in2$. – drhab Nov 14 '15 at 17:52
  • I still don't understand $1^{+}=2$. The natural numbers are basically nothing but drawings. Who determines 2 should be next to 1? All your reasoning seems to be based on that assumption. I'm just wondering if there has been efforts to address this ambiguity—something like making a international standard. If not, the fact that all those fields in mathematics, which is the most seemingly logical thing in the world, were built up upon the conventional consensus looks somewhat paradox to me. – eca2ed291a2f572f66f4a5fcf57511 Nov 16 '15 at 18:11
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    The natural numbers are basically sets (not drawings). Only to denote these sets drawings are needed: let's take $0$ as drawing for the empty set, $1$ for the singleton that contains set $0$, et cetera. We chose drawing $2$ for set $1^+={0,1}$ (why not?). Then let us define operation $+$ as described in my former comment. Consequence of this all: $1+1=2$. The drawings are international standard. Nobody is served by changing them. – drhab Nov 16 '15 at 20:37
  • Your comments helped me a lot. Thanks again. – eca2ed291a2f572f66f4a5fcf57511 Dec 06 '15 at 08:06
  • You are very welcome. – drhab Dec 06 '15 at 12:54
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As with almost every concept in mathematics there is not "the definition" of $2$. There are multiple ones. The most common ones are:

  • $2$ is the successor of $1$
  • $2 = 1 + 1$

But nothing stops you from using more complicated definitions, like:

  • $2$ is the smallest prime (consider that primes have to be greater than $1$)

All these definitions are equivalent and the all imply that $1<2$. For example in the case $2 = 1 + 1$ we have $1<2$ because $a<b \Leftrightarrow \exists n>0 : a + n = b$ (take $a=n=1$ and $b=2$)

Stefan Perko
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