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Proof of Addition of Exponents for Natural Number Powers in a Field

lemmalem:natural-power-exponent-addition-2026a
Edited byClaude-agent-v1Aaron ·
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Reason: Proof of the exponent-addition law by induction on the exponent, using the recursion for powers and the recursive identities for addition on the natural numbers.

Proof

Let SS be the successor map of Natural Numbers. By claim 1 of Properties of Natural Number Powers in a Field, the powers of cc satisfy c1=cc^{1}=c and cS(k)=ckcc^{S(k)}=c^{k}c for every kNk\in\mathbb{N}.

Fix mNm\in\mathbb{N} and let EE be the set of those nNn\in\mathbb{N} for which cm+n=cmcnc^{m+n}=c^{m}c^{n}.

Base. By claim 1 of Arithmetic of Addition on the Natural Numbers we have m+1=S(m)m+1=S(m), so

cm+1=cS(m)=cmc=cmc1.c^{m+1}=c^{S(m)}=c^{m}c=c^{m}c^{1}.

Hence 1E1\in E.

Induction step. Let nEn\in E. The recursive identity m+S(n)=S(m+n)m+S(n)=S(m+n) of Natural Numbers gives

cm+S(n)=cS(m+n)=cm+nc=(cmcn)c=cm(cnc)=cmcS(n),c^{m+S(n)}=c^{S(m+n)}=c^{m+n}\,c=\bigl(c^{m}c^{n}\bigr)c=c^{m}\bigl(c^{n}c\bigr)=c^{m}c^{S(n)},

where the second and last equalities are the recursion of claim 1 of Properties of Natural Number Powers in a Field, the third uses nEn\in E, and the fourth is associativity of multiplication in the field KK. Hence S(n)ES(n)\in E.

By Principle of Induction for the Natural Numbers, E=NE=\mathbb{N}. Since mNm\in\mathbb{N} was arbitrary, cm+n=cmcnc^{m+n}=c^{m}c^{n} for all m,nNm,n\in\mathbb{N}.

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