Proof of If S(m)=S(n), then m=n

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SUMMARY

The discussion centers on the proof of the statement "If S(m)=S(n), then m=n," where S(m) represents the successor of the natural number m. Two distinct proofs are presented: the first, a concise proof from Enderton's book, utilizes the property of transitive sets to establish equality, while the second, drawn from Halmos' and Hrbacek and Jech's works, involves two lemmas regarding natural numbers and their subsets. Participants debate the merits of brevity versus thoroughness in mathematical proofs, concluding that both approaches offer valuable insights.

PREREQUISITES
  • Understanding of successor functions in set theory
  • Familiarity with transitive sets and their properties
  • Knowledge of basic set theory concepts, including subsets
  • Awareness of mathematical proof techniques and their equivalences
NEXT STEPS
  • Study Enderton's "Elements of Set Theory" for concise proof techniques
  • Explore Halmos' "Naive Set Theory" for deeper insights into natural number properties
  • Investigate the concept of transitive sets and their implications in set theory
  • Examine different proof strategies in mathematics to understand their unique contributions
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Mathematicians, students of set theory, and educators seeking to deepen their understanding of proof techniques and the nature of mathematical reasoning.

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Proof of "If S(m)=S(n), then m=n"

Hello all. I have a question regarding the statement

If S(m)=S(n), then m=n, where S(m)=m\cup{m}, the successor of the natural number m.

I have come across two proofs for the above.

1. This one is as simple as observing the fact that
if S(m)=S(n), then \bigcupS(m)=\bigcupS(n) and since m and n are transitive (a set A is transitive if x\ina\inA\Rightarrowx\inA), it implies that m=n (for transitive sets like A, it can be proved that \bigcupS(A)=A).

This cheeky and cute proof is from Enderton's book.

2. This proof is a longer one which is given in Halmos' book and Hrbacek and Jech's book
follows a similar line of thought.

It involves proving two Lemmas. To quote him,

Naive Set Theory, P. R. Halmos, pp.47
(i) no natural number is a subset of any of its elements, and (ii)
every element of a natural number is a subset of it.

This is basically proving the ideas related to order defined on the set of natural numbers.

I wish to know the reason, if there is any, as to why one proof might be better than the other. Proper perspective always helps in understanding abstract mathematical concepts. Your input is greatly appreciated.
 
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Hey StatOnTheSide.

I just wanted to make a general comment about one proof being better than the other: I don't think that one proof is necessarily better than the other since all ways will provide some kind of insight that another may not (assuming all proofs don't drag on too much).

They both say the same thing anyway (in terms of the statement of the proof) but one is not more correct than the other (since it is a proof and all proper proofs in mathematics are equivalent in terms of the two statements being equal under the appropriate logical system): it's just that one establishes it in a different way.

Mathematics is like this: there are many ways to do something and when things are consistent and in a system that makes sense, then all paths from A to B take you from A to B regardless of what has happened in-between.
 


Thanks Chiro. I thought of asking because the length of the two proofs are very different. Proof 1. is like a one line statement in Enderton's book while proof 2. runs to like a page or two in Halmos' book and even in Hrbacek's book. I was just curious to know why one would write a one page proof when there is a one line proof for the same statement.

Thanks very much for your input. :)
 

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