LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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Laws of integer exponents

Statement

Let $a, b$ be elements of a field (Field) and let integer powers be as in Integer powers $a^{m}$ .

For all $m, n \in N$ : $a^{m + n} = a^{m} a^{n}$ , $(a^{m})^{n} = a^{m n}$ and $(a b)^{n} = a^{n} b^{n}$ .
If $a \neq 0$ then $a^{n} \neq 0$ for every $n \in N$ , and $a^{- m} = (a^{m})^{- 1}$ for every $m \in Z$ (Arithmetic on the integers).
If $a \neq 0$ and $b \neq 0$ then all three identities of claim 1 hold for all $m, n \in Z$ .

Facts & Assumptions

Given: Elements $a, b$ of a field $F$ , naturals $m, n, p, q, k$ and integers ranged over by $m, n$ in claims 2 and 3.

[L1]

Definition of powers (Integer powers $a^{m}$ ): $a^{0} = 1$ and $a^{n + 1} = a^{n} a$ for $n \in N$ ; and $a^{- n} : = (a^{n})^{- 1}$ for $a \neq 0$ and $n \geq 1$ , the two clauses agreeing at $n = 0$ .

[L2]

Induction principle (The principle of mathematical induction).

[L3]

Field arithmetic: multiplication is associative and commutative with identity $1$ , and every nonzero element has an inverse (Field); inverses are unique (Identities and inverses in a field are unique, which states uniqueness and nothing further), and HENCE, for $x, y \neq 0$ , $(x y)^{- 1} = x^{- 1} y^{- 1}$ and $(x^{- 1})^{- 1} = x$ , since $(x y) (x^{- 1} y^{- 1}) = 1$ and $x^{- 1} x = 1$ exhibit inverses that uniqueness then identifies.

[L4]

A field has no zero divisors: $x y = 0$ implies $x = 0$ or $y = 0$ (A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ ).

[L5]

$Z$ is a commutative ring in which every element is $k$ or $- k$ for a unique natural $k$ (The integers form a commutative ring, The naturals embed in the integers, Arithmetic on the integers); here $k$ in an integer expression denotes the image of the natural $k$ under the embedding $ι : N \to Z$ .

Proof

technique · induction

1.1baseL1L3

Base cases at $n = 0$ for the addition law, the product law and nonvanishing: $a^{m + 0} = a^{m} = a^{m} \cdot 1 = a^{m} a^{0}$ for every $m \in N$ ; $(a b)^{0} = 1 = 1 \cdot 1 = a^{0} b^{0}$ ; and if $a \neq 0$ then $a^{0} = 1 \neq 0$ .

1.2ih

Inductive hypothesis: fix $n \in N$ and assume $a^{m + n} = a^{m} a^{n}$ for all $m \in N$ , $(a b)^{n} = a^{n} b^{n}$ , and $a^{n} \neq 0$ whenever $a \neq 0$ . The iterated-power law $(a^{m})^{n} = a^{m n}$ is deliberately NOT carried in this hypothesis: its successor step needs the addition law at the exponent pair $(m n, m)$ , whose second entry is not the current stage, so that law must be finished first and the iterated law proved afterwards.

1.3L1L2L3L4L5

For $a \neq 0$ and every integer $j$ , $a^{- j} = (a^{j})^{- 1}$ : for $j \geq 0$ this is the definition together with the agreement of the two clauses at $j = 0$ , and for $j = - k$ with $k \geq 1$ it reads $a^{k} = (a^{- k})^{- 1}$ , which holds because $a^{- k} = (a^{k})^{- 1}$ and $(x^{- 1})^{- 1} = x$ at $x = a^{k}$ . That last substitution needs $a^{k} \neq 0$ , which is NOT free here and must not be read off the definition, since the definition of the negative clause is what is being justified; it is instead a self-contained induction on $k$ , from $a^{0} = 1 \neq 0$ and the fact that $a^{k + 1} = a^{k} a$ is a product of two nonzero elements of a field, hence nonzero.

2.1step 1.2L1L3L4

Successor step for the addition law, the product law and nonvanishing: $a^{m + (n + 1)} = a^{(m + n) + 1} = a^{m + n} a = (a^{m} a^{n}) a = a^{m} (a^{n} a) = a^{m} a^{n + 1}$ for every $m \in N$ ; $(a b)^{n + 1} = (a b)^{n} (a b) = (a^{n} b^{n}) (a b) = (a^{n} a) (b^{n} b) = a^{n + 1} b^{n + 1}$ ; and if $a \neq 0$ then $a^{n + 1} = a^{n} a$ is a product of two nonzero elements, hence nonzero.

3.1step 1.1step 2.1L2

By the induction principle, for all $m, n \in N$ : $a^{m + n} = a^{m} a^{n}$ and $(a b)^{n} = a^{n} b^{n}$ , and $a^{n} \neq 0$ whenever $a \neq 0$ . The addition law is thereby available at EVERY pair of natural exponents, which is exactly what the iterated-power law needs.

4.1step 3.1L1L2L3

The iterated-power law for natural exponents, $(a^{m})^{n} = a^{m n}$ , by a second induction on $n$ with $m \in N$ fixed: at $n = 0$ both sides are $1$ , since $(a^{m})^{0} = 1 = a^{0} = a^{m \cdot 0}$ ; and if $(a^{m})^{n} = a^{m n}$ then $(a^{m})^{n + 1} = (a^{m})^{n} a^{m} = a^{m n} a^{m} = a^{m n + m} = a^{m (n + 1)}$ , where the third equality is the addition law of step 3.1 at the pair $(m n, m)$ , legitimate precisely because that law is by now proved for all pairs of naturals. This completes claim 1.

4.2step 3.1step 1.3L1L3

For $a \neq 0$ and every integer $j$ , $a^{j + 1} = a^{j} a$ : for $j \geq 0$ this is the recursion clause, and for $j = - k$ with $k \geq 1$ we compute $a^{- k} a = (a^{k})^{- 1} a = (a^{k - 1} a)^{- 1} a = (a^{k - 1})^{- 1} a^{- 1} a = (a^{k - 1})^{- 1} = a^{- (k - 1)} = a^{- k + 1}$ .

4.3step 3.1step 1.3L3L5

For $a, b \neq 0$ the product law holds for all integers $n$ : for $n \geq 0$ it is step 3.1, and for $n = - k$ with $k \geq 1$ we get $(a b)^{- k} = ((a b)^{k})^{- 1} = (a^{k} b^{k})^{- 1} = (a^{k})^{- 1} (b^{k})^{- 1} = a^{- k} b^{- k}$ .

5.1step 4.2L1L2L3

For $a \neq 0$ , every integer $j$ and every natural $k$ , $a^{j + k} = a^{j} a^{k}$ , by induction on $k$ : the case $k = 0$ is $a^{j} = a^{j} \cdot 1$ , and if $a^{j + k} = a^{j} a^{k}$ then $a^{j + k + 1} = a^{j + k} a = a^{j} a^{k} a = a^{j} a^{k + 1}$ by step 4.2 applied to the integer $j + k$ and by the recursion clause.

6.1step 5.1step 1.3L3L5

For $a \neq 0$ the addition law holds for all integers $m, n$ : writing $n = k$ or $n = - k$ with $k \in N$ , the case $n = k$ is step 5.1, while for $n = - k$ step 5.1 applied to the integer $m - k$ gives $a^{m} = a^{(m - k) + k} = a^{m - k} a^{k}$ , hence $a^{m - k} = a^{m} (a^{k})^{- 1} = a^{m} a^{- k}$ .

7.1step 6.1step 3.1step 1.3L2L3L5

For $a \neq 0$ the iterated-power law holds for all integers $m, n$ : for $n \geq 0$ induction on $n$ gives $(a^{m})^{n + 1} = (a^{m})^{n} a^{m} = a^{m n} a^{m} = a^{m n + m} = a^{m (n + 1)}$ , the third equality by the integer addition law of step 6.1 at the pair $(m n, m)$ , with base $(a^{m})^{0} = 1 = a^{0}$ ; and for $n = - q$ with $q \geq 1$ , $(a^{m})^{- q} = ((a^{m})^{q})^{- 1} = (a^{m q})^{- 1} = a^{- m q} = a^{m (- q)}$ , using that $a^{m} \neq 0$ by step 3.1 and step 1.3.

8.1step 3.1step 4.1step 1.3step 6.1step 4.3step 7.1discharge-induction∎

Claims 1, 2 and 3 are therefore established: the addition, product and iterated-power laws for natural exponents together with nonvanishing by steps 3.1 and 4.1, the identity $a^{- m} = (a^{m})^{- 1}$ by step 1.3, and the three integer-exponent laws by steps 6.1, 4.3 and 7.1.

Laws of integer exponents

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources