Proving a+b=b+a in Axiomatic Construction of Naturals

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Discussion Overview

The discussion revolves around proving the commutative property of addition for natural numbers, specifically that \( a + b = b + a \), using the axiomatic construction of natural numbers based on Peano's axioms. Participants explore various approaches to this proof, including induction and informal reasoning.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes defining a set \( S \) to show that \( a + b = b + a \) holds for natural numbers, questioning the validity of using certain definitions in their proof.
  • Another participant describes their method of proving properties of natural numbers from Peano's axioms, emphasizing the need for two parts in the definition of addition.
  • A layman’s approach is suggested, using physical objects (buttons) to illustrate the commutative property, while acknowledging that examples do not constitute a formal proof.
  • One participant challenges the assumption that \( a + b = b + a \) holds universally, noting that there are number systems where addition is not commutative, but asserts that it is true for positive integers under Peano's axioms.
  • Another participant reflects on the inductive reasoning behind the commutative law, suggesting that it may not always hold in nature, which raises questions about the foundations of arithmetic.

Areas of Agreement / Disagreement

Participants express varying degrees of agreement on the validity of different approaches to proving the commutative property. Some agree on the use of induction, while others emphasize the limitations of informal reasoning. The discussion remains unresolved regarding the best method to prove the property.

Contextual Notes

Participants highlight the dependence on definitions and the assumptions made in their proofs. There are unresolved mathematical steps and varying interpretations of the axioms and definitions involved in the proof process.

Who May Find This Useful

This discussion may be of interest to those studying mathematical foundations, particularly in the context of natural numbers and axiomatic systems, as well as individuals exploring the philosophical implications of mathematical proofs.

sutupidmath
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Prove that a+b=b+a?

Hi all,

I am learning the axiomatic construction of Naturals, from the Peano axiom. Hence proving every property of naturals based solely on these axioms. I have come to some point where i am not very sure that what i am doing is okay, since there is not an explicit proof of this, it is left as an exercise to the reader.

I am trying to prove that for every a,b from naturals, a+b=b+a.

Here it is how i am going about it

[tex]S=\{b\in N | a+b=b+a\}[/tex] where a is any fixed natural numeber.

My attempt is to show that S is an inductive set.
We first need to prove that

[tex]1\in S(?)[/tex]

Here it is how i go about it, this is the part that i am not sure,

Using another theorem, where it says that a+1=a', where a' is the successor of a...this in fact is the theorem where they define the operation '+'. But to prove this, the author defines this: 1+b=b' for every b in N.------(1)

Now on my proof above, i am not using the definition of '+' but rather this other definition 1+b=b' for every b in N. ... so my question is am i allowed to do this, or there is another way of going about it.

Here is how i go about it
a+1=a' from the definition 0f +
=1+a from definition (1)
this actually means that 1 is an element of S. the rest i can easily prove if this is correct.

So?

thnx
 
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I did much the same thing, proving all the properties of the natural numbers from Peano's axioms, the going on to integers, rational numbers, and real numbers.

The definition of addition should have two parts: a+ 1 is defined as a' and, if b is NOT 1, so that b= c' for some c, a+ b is defined as (a+ c)'.

Also, I used a separate induction to prove that a+ 1= 1+ a: let U(a) be the set of all a such that a+ 1= 1+ a. It is clear that 1 is in that set. Assume a is in that set. Then a'+ 1 is, by definition (a')'. 1+ a', by the second part of the definition of addition, is (1+ a)' and the rest follows from the fact that a+ 1= 1+ a.
 


Knowing that you mathematicians must have a mathematical proof for everything, let's try a layman’s test of a + b = b + a.

If I put 3 buttons in a box, then put 4 more buttons in the box, I have 7 buttons. If I take them out and put 4 buttons in, followed up by 3 buttons, I again have 7.

Understanding that one example, or even Z examples, does not make a proof, can we derive a layman’s proof from this?

Maybe: If we put some portion ( X ) of N buttons in a box, then put in the remainder of N buttons, ( Y ), then we have N buttons in the box. If we do this in reverse, we always have N buttons in the box.

I could add some more verbiage to ensure preciseness, but the concept is clear.
 


First off, a+b is not always equal to b+a. Mathematicians have invented several number systems in which addition is not commutative.

You are talking about the positive integers, which are described by the Peano postulates. Addition is commutative in the positive integers. One way to prove this is by using induction in conjunction with the associativity of addition.

First, prove that a+1=1+a for all positive integers a
  1. Assume the hypothesis is true for some a.
  2. (a+1)+1 = (1+a)+1 by assumption.
  3. (1+a)+1 = 1+(a+1) by associativity of addition.
  4. (a+1)+1 = 1+(a+1) by transitivity of equality.
  5. a+1=1+a for a=1 by reflexivity of equality (all natural numbers are equal to themselves).
  6. i]a[/i]+1=1+a for all positive integers a by induction.

Now prove a+b=b+a:
  1. Assume the hypothesis is true for some positive integers a, b.
  2. (b+1)+a = (1+b)+a by 1+k=k+1.
  3. (1+b)+a = 1+(b)+a) by associativity of addition.
  4. 1+(b)+a) = (b)+a)+1 by 1+k=k+1.
  5. (b)+a)+1 = (a)+b)+1 by assumption.
  6. (a)+b)+1 = a+(b+1) by associativity of addition.
  7. (b+1)+a = a+(b+1) by transitivity of equality.
  8. b+a = a+b for b=1 by 1+k=k+1.
  9. b+a = a+b for all positive integers a, b by induction.
 


bkelly said:
Knowing that you mathematicians must have a mathematical proof for everything, let's try a layman’s test of a + b = b + a.

If I put 3 buttons in a box, then put 4 more buttons in the box, I have 7 buttons. If I take them out and put 4 buttons in, followed up by 3 buttons, I again have 7.

Understanding that one example, or even Z examples, does not make a proof, can we derive a layman’s proof from this?

Maybe: If we put some portion ( X ) of N buttons in a box, then put in the remainder of N buttons, ( Y ), then we have N buttons in the box. If we do this in reverse, we always have N buttons in the box.

I could add some more verbiage to ensure preciseness, but the concept is clear.

It's interesting that you bring this up. Arithmetic is axiomized from the everyday experience with quantities of whole objects. The commutative law is obtained by inductive reasoning; then we proceed to list it as an axiom for the deductive system we wish to build (arithmetic). Because a result obtained by inductive reasoning can never be certain, it may be that the commutative law sometimes does not hold in nature (as much as this is against our intuition). If this is the case, arithmetic is not a theory which fits the behavior of whole objects.
 


HallsofIvy said:
I did much the same thing, proving all the properties of the natural numbers from Peano's axioms, the going on to integers, rational numbers, and real numbers.

The definition of addition should have two parts: a+ 1 is defined as a' and, if b is NOT 1, so that b= c' for some c, a+ b is defined as (a+ c)'.

Also, I used a separate induction to prove that a+ 1= 1+ a: let U(a) be the set of all a such that a+ 1= 1+ a. It is clear that 1 is in that set. Assume a is in that set. Then a'+ 1 is, by definition (a')'. 1+ a', by the second part of the definition of addition, is (1+ a)' and the rest follows from the fact that a+ 1= 1+ a.

Well, i guess this is what i was looking for. I did not look at it carefuly, i am in a rush right now, but i am going to come to it later, so if i have anything unclear, i will get back to u guys.

Because i was only looking how to prove that a+1=1+a, because i can do the rest.

THank you guys.
 

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