TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$

Statement

For all $x, y \in R$ and every $n \in N$ ,

$(x + y)^{n} = \sum_{k < n + 1} ι (\binom{n}{k}) x^{k} y^{n - k},$

where the powers are the integer powers of Integer powers $a^{m}$ , the sum is the real finite sum of Finite sums and finite products, by recursion over $k = 0, 1, \dots, n$ , the difference $n - k$ is a genuine one because $k \leq n$ throughout the range, and $ι$ is the canonical natural of The canonical natural $ι (n) = n \cdot 1_{F}$ of a field.

The coefficient is $ι (\binom{n}{k})$ and not $(\binom{n}{k})$ . A binomial coefficient is a natural number, that is a von Neumann natural, that is a set; it is not an element of $R$ , and it enters the field through $ι$ .

The identity is stated in $R$ and only in $R$ . The same proof uses nothing but commutativity, associativity, distributivity and natural-number multiples of a ring element, so a commutative-ring version is available wherever rings are; rings are not available at this point in the reading order, and the ring statement is a separate statement, to be made where they are. See the Remarks below.

Statement references

Integer powers $a^{m}$

Finite sums and finite products, by recursion

The canonical natural $ι (n) = n \cdot 1_{F}$ of a field

Facts & Assumptions

Given: Reals $x, y$ ; a natural $n$ ; and the abbreviation $c_{k} : = ι (\binom{n}{k})$ for every $k \in N$ , so that $c_{k} = ι (0) = 0$ whenever $k > n$ .

[L1]

Induction (The principle of mathematical induction).

[L2]

Integer powers (Integer powers $a^{m}$ ): $a^{0} = 1$ for every real $a$ , including $a = 0$ , and $a^{σ (m)} = a^{m} \cdot a$ . An immediate induction gives $1^{m} = 1$ .

[L3]

Real finite sums (Finite sums and finite products, by recursion): $\sum_{k < 0} u_{k} = 0$ and $\sum_{k < σ (N)} u_{k} = \sum_{k < N} u_{k} + u_{N}$ ; additivity $\sum (u_{k} + v_{k}) = \sum u_{k} + \sum v_{k}$ , scaling $\sum λ u_{k} = λ \sum u_{k}$ , and splitting $\sum_{k < N} u_{k} = \sum_{k < p} u_{k} + \sum_{j < N - p} u_{p + j}$ for $p \leq N$ (Laws of finite sums and finite products, clauses 1, 2 and 3).

[L4]

$ι$ is additive and multiplicative with $ι (0) = 0$ and $ι (1) = 1$ (clause 0 of Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L5]

Binomial coefficients (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ): $(\binom{m}{0}) = 1$ and $(\binom{m}{j}) = 0$ for $j > m$ ; Pascal's rule $(\binom{σ (m)}{σ (j)}) = (\binom{m}{j}) + (\binom{m}{σ (j)})$ for all $m, j$ (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ , clause 1).

[L6]

Field arithmetic of $R$ : associativity, commutativity, distributivity, $0 \cdot a = 0$ (Field, Ordered field, Multiplication by zero: $0 \cdot a = 0$ ).

[L7]

Arithmetic of $N$ : for $k \leq n$ , $k + (n - k) = n$ , and hence $(n - k) + 1 = σ (n) - k$ and $σ (n) - σ (k) = n - k$ ; every nonzero natural is a successor (Order on the natural numbers, Addition is cancellative, Every nonzero natural number is a successor, Finite sums and finite products of natural numbers, $\sum_{k < n} a_{k}$ and $\prod_{k < n} a_{k}$ in $N$ for the truncated difference).

Proof

technique · induction

1.1givenL4L5

Both sides are functions of $n$ with $x, y$ fixed, and the induction is on $n$ . Note first that $c_{σ (n)} = ι (\binom{n}{σ (n)}) = ι (0) = 0$ , since $σ (n) > n$ .

1.2baseL2L3L4L5

Base case $n = 0$ . The left-hand side is $(x + y)^{0} = 1$ by [L2]. The right-hand side is $\sum_{k < 1} ι (\binom{0}{k}) x^{k} y^{0 - k} = ι (\binom{0}{0}) x^{0} y^{0} = 1 \cdot 1 \cdot 1 = 1$ , using [L3], $(\binom{0}{0}) = 1$ , $ι (1) = 1$ and [L2]. This is correct at $x = 0$ and at $y = 0$ as well, because $a^{0} = 1$ for every real $a$ .

1.3ih

Inductive hypothesis: fix $n$ and assume $(x + y)^{n} = \sum_{k < σ (n)} c_{k} x^{k} y^{n - k}$ for all $x, y \in R$ .

2.1step 1.3L2L3L6

Expanding one factor. By [L2] and distributivity, $(x + y)^{σ (n)} = (x + y)^{n} (x + y) = (\sum_{k < σ (n)} c_{k} x^{k} y^{n - k}) x + (\sum_{k < σ (n)} c_{k} x^{k} y^{n - k}) y$ , using the hypothesis of step 1.3; and by the scaling clause of [L3] together with $x^{k} x = x^{σ (k)}$ and $y^{n - k} y = y^{(n - k) + 1}$ this equals $Σ_{1} + Σ_{2}$ with $Σ_{1} : = \sum_{k < σ (n)} c_{k} x^{σ (k)} y^{n - k}$ and $Σ_{2} : = \sum_{k < σ (n)} c_{k} x^{k} y^{(n - k) + 1}$ .

3.1step 1.1step 2.1L3L6L7

Rewriting $Σ_{2}$ . For $k \leq n$ one has $(n - k) + 1 = σ (n) - k$ by [L7], so $Σ_{2} = \sum_{k < σ (n)} c_{k} x^{k} y^{σ (n) - k}$ . Extending the range by one term costs nothing: by the recursion clause of [L3], $\sum_{k < σ (σ (n))} c_{k} x^{k} y^{σ (n) - k} = Σ_{2} + c_{σ (n)} x^{σ (n)} y^{0}$ , and $c_{σ (n)} = 0$ by step 1.1, so the added term is $0$ by [L6] and $Σ_{2} = \sum_{k < σ (σ (n))} c_{k} x^{k} y^{σ (n) - k}$ .

3.2step 2.1L3L6L7

Rewriting $Σ_{1}$ . Define a list $b$ of length $σ (σ (n))$ by $b_{0} : = 0$ and $b_{σ (i)} : = c_{i} x^{σ (i)} y^{n - i}$ for $i < σ (n)$ ; every index below $σ (σ (n))$ is $0$ or a successor $σ (i)$ with $i < σ (n)$ , by [L7], so $b$ is well defined. Splitting at $p = 1$ by [L3] and using $σ (σ (n)) - 1 = σ (n)$ and $1 + i = σ (i)$ , $\sum_{j < σ (σ (n))} b_{j} = b_{0} + \sum_{i < σ (n)} b_{σ (i)} = 0 + Σ_{1} = Σ_{1}$ .

4.1step 3.1step 3.2L3L4L5L6L7

Adding the two. By step 3.1, step 3.2 and the additivity clause of [L3], $(x + y)^{σ (n)} = \sum_{k < σ (σ (n))} (b_{k} + c_{k} x^{k} y^{σ (n) - k})$ . Evaluate the general term. At $k = 0$ it is $0 + ι (\binom{n}{0}) x^{0} y^{σ (n)} = y^{σ (n)} = ι (\binom{σ (n)}{0}) x^{0} y^{σ (n) - 0}$ , both coefficients being $ι (1) = 1$ . At $k = σ (i)$ with $i < σ (n)$ it is $c_{i} x^{σ (i)} y^{n - i} + c_{σ (i)} x^{σ (i)} y^{σ (n) - σ (i)}$ , and $σ (n) - σ (i) = n - i$ by [L7], so the term equals $(ι (\binom{n}{i}) + ι (\binom{n}{σ (i)})) x^{σ (i)} y^{σ (n) - σ (i)} = ι (\binom{σ (n)}{σ (i)}) x^{σ (i)} y^{σ (n) - σ (i)}$ by the additivity of $ι$ and Pascal's rule. Hence $(x + y)^{σ (n)} = \sum_{k < σ (σ (n))} ι (\binom{σ (n)}{k}) x^{k} y^{σ (n) - k}$ , which is the claim at $σ (n)$ .

5.1step 1.2step 4.1L1L2discharge-induction∎

By step 1.2, step 4.1 and induction the identity holds for every $n \in N$ and all reals $x$ , $y$ ; in particular at $x = 0$ or $y = 0$ , where the convention $a^{0} = 1$ of Integer powers $a^{m}$ is what makes the extreme terms come out right and no exceptional case is needed.

Remarks

Two index traps, both checked. The sum runs over $k < n + 1$ , that is $k \leq n$ , so the exponent $n - k$ is never a truncated difference in disguise; and the inductive step needs the coefficient $(\binom{n}{σ (n)})$ , which is $0$ by the boundary values of The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ rather than undefined. Step 1.1 records that once and both rewritings use it.
Where $0^{0}$ matters. At $x = 0$ the term with $k = 0$ is $ι (\binom{n}{0}) 0^{0} y^{n} = y^{n}$ , and the identity reads $y^{n} = y^{n}$ . A treatment leaving $0^{0}$ undefined would have to state the theorem with exceptions; Integer powers $a^{m}$ fixes $a^{0} = 1$ for every real $a$ , so there are none.
The ring version is a different statement. It says the same thing about $x, y$ in a commutative ring, with $ι (\binom{n}{k})$ replaced by the $(\binom{n}{k})$ -fold multiple of the ring element. Making it requires rings, which come later in the reading order; the pointer to Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides ↗ is orientation only and nothing above rests on it.

The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The binomial theorem in RR: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k(x+y)n=∑k<n+1​ι(kn​)xkyn−k

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$