Question

Commutative is true on natural numbers under addition

Answer 1

Natural Number Addition is Commutative

Contents

 [hide] 

1Theorem2Proof 13Proof 23.1Basis for the Induction3.2Induction Hypothesis3.3Induction Step4Proof 34.1Basis for the Induction4.2Induction Hypothesis4.3Induction Step5Sources

Theorem

The operation of addition on the set of natural numbers N is commutative:

∀m,n∈N:m+n=n+m


Proof 1

Consider the natural numbers defined as a naturally ordered semigroup.


By definition, the operation in a naturally ordered semigroup is commutative.

Hence the result.

◼


Proof 2

Proof by induction.

Consider the natural numbers N defined as the elements of the minimal infinite successor set ω.


From the definition of addition in ω‎, we have that:

∀m,n∈N:  m+0=mm+n+=(m+n)+


For all n∈N, let P(n) be the proposition:

∀m∈N:m+n=n+m


Basis for the Induction

From Natural Number Addition Commutes with Zero, we have:

∀m∈N:m+0=m=0+m

Thus P(0) is seen to be true.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if P(k) is true, where k≥0, then it logically follows that P(k+) is true.


So this is our induction hypothesis P(k):

∀m∈N:m+k=k+m


Then we need to show that P(k+) follows directly from P(k):

∀m∈N:m+k+=k++m


Induction Step

This is our induction step:


k++m=(k+m)+ Natural Number Addition Commutativity with Successor=(m+k)+ from the induction hypothesis=m+k+ by definition

So P(k)⟹P(k+) and the result follows by the Principle of Mathematical Induction.


Therefore:

∀m,n∈N:m+n=n+m

◼


Proof 3

In the axiomatisation of 1-based natural numbers, this is rendered:

∀x,y∈N>0:x+y=y+x


Using the following axioms:

(A)  :    ∃11∈N>0:a×1=a=1×a            (B)  :    ∀a,b∈N>0:a×(b+1)=(a×b)+a            (C)  :    ∀a,b∈N>0:a+(b+1)=(a+b)+1            (D)  :    ∀a∈N>0,a≠1:∃1b∈N>0:a=b+1            (E)  :    ∀a,b∈N>0:Exactly one of these three holds:            a=b∨(∃x∈N>0:a+x=b)∨(∃y∈N>0:a=b+y)            (F)  :    ∀A⊆N>0:(1∈A∧(z∈A⟹z+1∈A))⟹A=N>0            


Let x∈N>0 be arbitrary.

For all n∈N>0, let P(n) be the proposition:

x+n=n+x


Basis for the Induction

From Natural Number Commutes with 1 under Addition, we have that:

∀x∈N>0:x+1=1+x

and so P(1) holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if P(k) is true, where k≥1, then it logically follows that P(k+1)is true.


So this is our induction hypothesis:

x+k=k+x


Then we need to show:

x+(k+1)=(k+1)+x


Induction Step

This is our induction step:


x+(k+1)=(x+k)+1 Natural Number Addition is Associative=(k+x)+1 Induction Hypothesis=k+(x+1) Natural Number Addition is Associative=k+(1+x) Basis for the Induction=(k+1)+x Natural Number Addition is Associative

So P(k)⟹P(k+1) and the result follows by the Principle of Mathematical Induction.

◼


Sources

1951: Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts ... (previous) ... (next): Introduction §4: The natural numbers: A21972: A.G. Howson: A Handbook of Terms used in Algebra and Analysis ... (previous) ... (next): §4: Number systems I: Peano's Axioms

Categories: Proven ResultsNatural NumbersNatural Number Addition is Commutative

Navigation menu

Log inRequest account

PageDiscussion

ReadView sourceView history

Search

Main PageCommunity discussionCommunity portalRecent changesRandom proofHelpFAQ

ProofWiki.org

Proof IndexDefinition IndexSymbol IndexAxiom IndexMathematiciansBooksSandboxAll CategoriesJokes

To Do

Proofread ArticlesWanted ProofsStub ArticlesTidy ArticlesHelp NeededQuestionable ContentImprovements InvitedRefactoringMissing LinksProposed MergersProposed DeletionsMaintenance Needed

Tools

What links hereRelated changesSpecial pagesPrintable versionPermanent linkPage information

Advertisements

This page was last modified on 10 March 2018, at 11:52 and is 1,391 bytesContent is available under Creative Commons Attribution-ShareAlike License unless otherwise noted.Privacy policyAbout ProofWikiDisclaimers