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

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Algebra of limits: sums, scalar multiples, products and quotients

Statement

Let $(x_{k})$ and $(y_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) converging to $x$ and $y$ respectively (Limits and Cauchy sequences of reals), and let $c \in R$ . Then

$x_{k} + y_{k} \to x + y, c x_{k} \to c x, x_{k} - y_{k} \to x - y, x_{k} y_{k} \to x y,$

and if in addition $y \neq 0$ and $y_{k} \neq 0$ for every $k \in N$ , then

$\frac{1}{y_{k}} \to \frac{1}{y} and \frac{x_{k}}{y_{k}} \to \frac{x}{y} .$

The quotient case rests on an eventual lower bound for $∣ y_{k} ∣$ , proved below rather than assumed: $∣ y_{k} ∣ > ∣ y ∣ / 2$ for all sufficiently large $k$ .

Facts & Assumptions

Given: Sequences $(x_{k})$ , $(y_{k})$ of reals with $(x_{k})$ converging to $x$ and $(y_{k})$ converging to $y$ , and a real $c$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). For the last two claims we assume in addition $y \neq 0$ and $y_{k} \neq 0$ for every $k$ .

[L1]

Convergence, quantified over rational $ε > 0$ (Limits and Cauchy sequences of reals).

[L2]

Absolute value and the triangle inequality: $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u ∣ \geq 0$ , $∣ u ∣ = 0$ if and only if $u = 0$ , $∣ - u ∣ = ∣ u ∣$ , and $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ (Basic properties of the absolute value, The triangle inequality).

[L3]

Real versus rational $ε$ : for every real $η > 0$ there is a rational $ε$ with $0 < ε < η$ , by density (The rationals embed densely in the reals) or by the Archimedean property (Every complete ordered field is Archimedean) applied to $1 / η$ (Inverses of positives are positive, and reciprocation reverses order); consequently the convergence test of Limits and Cauchy sequences of reals may equivalently be run with real $ε > 0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Convergent sequences are bounded (Every convergent sequence is bounded), and a constant sequence $(c)$ is bounded by $∣ c ∣$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L5]

A null sequence times a bounded sequence is null (A null sequence times a bounded sequence is null).

[L6]

Reverse triangle inequality: $∣ ∣ u ∣ - ∣ v ∣ ∣ \leq ∣ u - v ∣$ , hence $∣ u ∣ \geq ∣ v ∣ - ∣ v - u ∣$ (The reverse triangle inequality).

[L7]

Inverses and order: $a > 0$ implies $a^{- 1} > 0$ ; $0 < a < b$ implies $0 < b^{- 1} < a^{- 1}$ ; $(u^{- 1})^{- 1} = u$ for $u \neq 0$ (Inverses of positives are positive, and reciprocation reverses order, Field).

[L8]

Order arithmetic in $R$ : adding a constant and adding inequalities preserve the order, multiplying a strict inequality by a positive factor preserves it, and $\leq$ and $<$ compose transitively; trichotomy holds, and since $u \geq 0$ means $u > 0$ or $u = 0$ , an element with $u \geq 0$ and $u \neq 0$ satisfies $u > 0$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Complete ordered field (least-upper-bound property), Ordered field). Moreover $2 = 1 + 1 > 0$ and $2$ is invertible: $0 < 1$ in any ordered field (The multiplicative identity is positive) and the positives are closed under addition, so $2 > 0$ and in particular $2 \neq 0$ (Ordered field), whence $2^{- 1}$ exists (Field).

[L9]

Rational arithmetic: $ε / 2$ is a rational $> 0$ whenever $ε$ is, and $ε / 2 + ε / 2 = ε$ (The rationals form a totally ordered field); the order on $N$ is total, so finitely many thresholds admit a common index ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1L1L2

Reduction to null sequences: for any sequence $(z_{k})$ of reals and any real $z$ , the statements " $(z_{k})$ converges to $z$ " and " $(z_{k} - z)$ converges to $0$ " are literally the same condition, because $∣ (z_{k} - z) - 0 ∣ = ∣ z_{k} - z ∣$ for every $k$ .

1.2L1L2L8L9algebra

Sum rule, in general form. Let $(u_{k}) \to u$ and $(v_{k}) \to v$ be any convergent sequences of reals and let $ε > 0$ be rational; take $K_{1}$ with $∣ u_{k} - u ∣ < ε / 2$ for $k \geq K_{1}$ and $K_{2}$ with $∣ v_{k} - v ∣ < ε / 2$ for $k \geq K_{2}$ , and let $K$ be an index at least as large as both. For $k \geq K$ , $∣ (u_{k} + v_{k}) - (u + v) ∣ = ∣ (u_{k} - u) + (v_{k} - v) ∣ \leq ∣ u_{k} - u ∣ + ∣ v_{k} - v ∣ < ε / 2 + ε / 2 = ε$ ; hence $u_{k} + v_{k} \to u + v$ , and in particular $x_{k} + y_{k} \to x + y$ .

1.3L4

Boundedness: every convergent sequence of reals is bounded, and every constant sequence $(d)$ is bounded by $∣ d ∣$ .

1.4L1L2L3L7L8choose

Quotient preparation. Assume $y \neq 0$ and $y_{k} \neq 0$ for every $k$ . Then $∣ y ∣ > 0$ by [L2], so $∣ y ∣ / 2 > 0$ ; running the convergence test of $(y_{k})$ with the real number $∣ y ∣ / 2$ as tolerance, which [L3] licenses, produces $K_{0} \in N$ with $∣ y_{k} - y ∣ < ∣ y ∣ / 2$ for all $k \geq K_{0}$ .

2.1step 1.1step 1.3L5algebra

Scalar rule, in general form. Let $(u_{k}) \to u$ and let $d \in R$ . By step 1.1 the sequence $(u_{k} - u)$ is null and by step 1.3 the constant sequence $(d)$ is bounded, so $(d (u_{k} - u)) = (d u_{k} - d u)$ is null by [L5]; by step 1.1 again, $d u_{k} \to d u$ , and in particular $c x_{k} \to c x$ .

2.2step 1.1step 1.2step 1.3L5algebra

Product rule, in general form. Let $(u_{k}) \to u$ and $(v_{k}) \to v$ , and write $u_{k} v_{k} - u v = u_{k} (v_{k} - v) + v (u_{k} - u)$ . By step 1.1 both $(v_{k} - v)$ and $(u_{k} - u)$ are null; by step 1.3 both $(u_{k})$ and the constant sequence $(v)$ are bounded; so both $(u_{k} (v_{k} - v))$ and $(v (u_{k} - u))$ are null by [L5], and their sum is null by step 1.2 applied with both limits equal to $0$ . By step 1.1, $u_{k} v_{k} \to u v$ , and in particular $x_{k} y_{k} \to x y$ .

2.3step 1.4L2L6L8

Eventual lower bound. For every $k \geq K_{0}$ , the reverse triangle inequality gives $∣ y_{k} ∣ \geq ∣ y ∣ - ∣ y - y_{k} ∣ = ∣ y ∣ - ∣ y_{k} - y ∣ > ∣ y ∣ - ∣ y ∣ / 2 = ∣ y ∣ / 2 > 0$ ; so $∣ y_{k} ∣ > ∣ y ∣ / 2$ for all $k \geq K_{0}$ , and in particular $∣ y_{k} ∣ \neq 0$ there.

3.1step 1.2step 2.1L8algebra

Difference rule. Applying step 2.1 to the sequence $(y_{k})$ with $d = - 1$ gives $- y_{k} \to - y$ ; the sum rule of step 1.2 applied to $(x_{k})$ and $(- y_{k})$ then gives $x_{k} - y_{k} = x_{k} + (- y_{k}) \to x + (- y) = x - y$ .

3.2step 2.3L2L7L8algebra

Reciprocal estimate. For $k \geq K_{0}$ we have $y_{k} \neq 0$ and $∣ y_{k} ∣ > ∣ y ∣ / 2 > 0$ , so [L7] applied to $0 < ∣ y ∣ / 2 < ∣ y_{k} ∣$ gives $0 < ∣ y_{k} ∣^{- 1} < 2 / ∣ y ∣$ , and therefore $∣ \frac{1}{y_{k}} - \frac{1}{y} ∣ = \frac{∣ y - y_{k} ∣}{∣ y_{k} ∣ ∣ y ∣} \leq \frac{2}{∣ y ∣^{2}} ∣ y_{k} - y ∣$ .

4.1step 3.2L1L3L7L8L9algebra

Reciprocal rule. Let $ε > 0$ be an arbitrary real and put $η : = ε ∣ y ∣^{2} / 2$ , a real $> 0$ ; by [L3] there is $K_{1}$ with $∣ y_{k} - y ∣ < η$ for all $k \geq K_{1}$ . For every $k$ at least as large as both $K_{0}$ and $K_{1}$ , step 3.2 gives $∣ \frac{1}{y_{k}} - \frac{1}{y} ∣ \leq \frac{2}{∣ y ∣^{2}} ∣ y_{k} - y ∣ < \frac{2}{∣ y ∣^{2}} η = ε$ ; hence $1 / y_{k} \to 1 / y$ .

5.1step 2.2step 4.1algebra

Quotient rule. By step 4.1 the sequence $(1 / y_{k})$ converges to $1 / y$ , so the product rule of step 2.2 applied to $(x_{k})$ and $(1 / y_{k})$ gives $x_{k} / y_{k} = x_{k} \cdot (1 / y_{k}) \to x \cdot (1 / y) = x / y$ .

6.1step 1.2step 2.1step 2.2step 3.1step 4.1step 5.1∎

All the claims are established: the sum rule in step 1.2, the scalar rule in step 2.1, the difference rule in step 3.1, the product rule in step 2.2, and the reciprocal and quotient rules in steps 4.1 and 5.1.

Remarks

The quotient case is where proofs usually cheat. The estimate $∣ 1 / y_{k} - 1 / y ∣ = ∣ y - y_{k} ∣ / (∣ y_{k} ∣ ∣ y ∣)$ is worthless until $∣ y_{k} ∣$ is known to stay away from $0$ : without a lower bound the denominator can be arbitrarily small and the fraction arbitrarily large, even while $∣ y - y_{k} ∣$ shrinks. Step 2.3 supplies that bound, $∣ y_{k} ∣ > ∣ y ∣ / 2$ for $k \geq K_{0}$ , and it is proved from the reverse triangle inequality, not assumed.
The hypothesis $y_{k} \neq 0$ for every $k$ is only there so that $1 / y_{k}$ is defined for every index. It is not needed for the limit: step 2.3 shows $∣ y_{k} ∣ > ∣ y ∣ / 2 > 0$ from $K_{0}$ on, so a sequence with $y \neq 0$ has at most finitely many zero terms, and by Convergence depends only on the tail one may pass to the $K_{0}$ -th tail and read the conclusion there.
The hypothesis $y \neq 0$ cannot be dropped. With $x_{k} = 1$ and $y_{k} = 1 / (k + 1)$ , both sequences converge: the first is constant (Sequences of reals: bounded, eventually, frequently, tails, subsequences) and the second is null (FALSE: limits preserve strict inequalities), so $y = 0$ . Yet $x_{k} / y_{k} = k + 1$ , and no real $M$ bounds every $k + 1$ , by the Archimedean property (Every complete ordered field is Archimedean); so the quotient sequence is unbounded, hence not convergent by Every convergent sequence is bounded.
Nothing in the proof uses completeness of $R$ beyond the Archimedean property invoked in [L3], so the same rules hold verbatim for sequences of rationals.

Algebra of limits: sums, scalar multiples, products and quotients

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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