[proofplan] Both identities are derived by sandwiching between the "obvious" lower bound and an upper bound that collapses via the previous theorem $\aleph_\beta \cdot \aleph_\beta = \aleph_\beta$ ([theorem 1503](/theorems/1503)). For multiplication, we bound $\aleph_\alpha \cdot \aleph_\beta$ below by $\aleph_\beta$ (using $\aleph_\alpha \geq 1$) and above by $\aleph_\beta \cdot \aleph_\beta = \aleph_\beta$ (using $\aleph_\alpha \leq \aleph_\beta$); the [Cantor–Schröder–Bernstein theorem](/theorems/1479) then forces equality. For addition, we bound $\aleph_\alpha + \aleph_\beta$ below by $\aleph_\beta$ and above by $\aleph_\beta + \aleph_\beta = 2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta$, again sandwiching to force equality. [/proofplan]

[step:Recall the cardinal operations and set up the notation] For cardinals $\kappa$ and $\lambda$ represented by disjoint sets $K$ and $L$ of those cardinalities, cardinal addition and multiplication are defined by \begin{align*} \kappa + \lambda &:= \operatorname{card}(K \sqcup L), \\ \kappa \cdot \lambda &:= \operatorname{card}(K \times L). \end{align*} Both are well-defined, commutative, and associative operations on cardinals. Cardinal inequalities are defined by injections: $\kappa \leq \lambda$ iff there exists an injection from a set of cardinality $\kappa$ into a set of cardinality $\lambda$. The [Cantor–Schröder–Bernstein theorem](/theorems/1479) asserts $\kappa \leq \lambda$ and $\lambda \leq \kappa$ together imply $\kappa = \lambda$. Throughout this proof, $\alpha \leq \beta$ are fixed ordinals, so $\aleph_\alpha \leq \aleph_\beta$ (the aleph function is monotone: $\aleph_\alpha \leq \aleph_\beta$ iff $\alpha \leq \beta$). [/step]

[step:Prove $\aleph_\alpha \cdot \aleph_\beta = \aleph_\beta$ by a two-sided cardinality bound]We establish \begin{align*} \aleph_\beta \leq \aleph_\alpha \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta, \end{align*} and then apply Cantor–Schröder–Bernstein. **Lower bound $\aleph_\beta \leq \aleph_\alpha \cdot \aleph_\beta$.** Let $A$ and $B$ be disjoint sets of cardinalities $\aleph_\alpha$ and $\aleph_\beta$ respectively. Fix any $a_0 \in A$ (possible since $\aleph_\alpha \geq \aleph_0 \geq 1$, so $A \neq \varnothing$). The map \begin{align*} \iota: B &\to A \times B \\ b &\mapsto (a_0, b) \end{align*} is injective: if $\iota(b) = \iota(b')$ then $(a_0, b) = (a_0, b')$, hence $b = b'$. Therefore $\aleph_\beta = \operatorname{card}(B) \leq \operatorname{card}(A \times B) = \aleph_\alpha \cdot \aleph_\beta$. **Upper bound $\aleph_\alpha \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta$.** Since $\aleph_\alpha \leq \aleph_\beta$, there exists an injection $f: A \hookrightarrow B'$ where $B'$ is a set of cardinality $\aleph_\beta$. Then \begin{align*} f \times \operatorname{id}_B: A \times B &\to B' \times B \\ (a, b) &\mapsto (f(a), b) \end{align*} is injective: if $(f(a), b) = (f(a'), b')$, then $b = b'$ and $f(a) = f(a')$; the latter forces $a = a'$ since $f$ is injective. Hence $\operatorname{card}(A \times B) \leq \operatorname{card}(B' \times B) = \aleph_\beta \cdot \aleph_\beta$. **The collapse $\aleph_\beta \cdot \aleph_\beta = \aleph_\beta$.** This is exactly the content of [theorem 1503](/theorems/1503) (Cardinal Multiplication for Alephs), which asserts $\aleph_\gamma \cdot \aleph_\gamma = \aleph_\gamma$ for every ordinal $\gamma$; applied with $\gamma = \beta$. **Conclusion.** Combining the three estimates, \begin{align*} \aleph_\beta \leq \aleph_\alpha \cdot \aleph_\beta \leq \aleph_\beta, \end{align*} so by the [Cantor–Schröder–Bernstein theorem](/theorems/1479), $\aleph_\alpha \cdot \aleph_\beta = \aleph_\beta$.[/step]

[guided]To prove $\aleph_\alpha \cdot \aleph_\beta = \aleph_\beta$ we build a two-sided cardinality bound and apply [Cantor–Schröder–Bernstein](/theorems/1479). The strategy is \begin{align*} \aleph_\beta \leq \aleph_\alpha \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta, \end{align*} with equality forced by the sandwich. **Lower bound.** To embed $B$ (of cardinality $\aleph_\beta$) into $A \times B$ (of cardinality $\aleph_\alpha \cdot \aleph_\beta$), we just fix a single element $a_0 \in A$ — this uses only that $A$ is non-empty, which is certainly true since $\operatorname{card}(A) = \aleph_\alpha \geq \aleph_0$. Then the "injection into the fibre over $a_0$" \begin{align*} \iota(b) = (a_0, b) \end{align*} is visibly an injection: the second coordinate is preserved by $\iota$, so if $\iota(b) = \iota(b')$ then $b = b'$. The existence of an injection $B \hookrightarrow A \times B$ gives the cardinality bound $\aleph_\beta \leq \aleph_\alpha \cdot \aleph_\beta$. **Upper bound.** We use the hypothesis $\alpha \leq \beta$, which gives $\aleph_\alpha \leq \aleph_\beta$. Why does the aleph function preserve order? Because $\aleph$ is a class function $\operatorname{Ord} \to \operatorname{Card}$ defined recursively so that $\aleph_\alpha$ is strictly increasing in $\alpha$. Hence there is an injection from a set of cardinality $\aleph_\alpha$ into a set of cardinality $\aleph_\beta$. Fix such sets: $A$ of cardinality $\aleph_\alpha$, $B'$ of cardinality $\aleph_\beta$, and an injection $f: A \hookrightarrow B'$. Take also a set $B$ of cardinality $\aleph_\beta$. Then $(f, \operatorname{id}_B): A \times B \to B' \times B$, given by $(a, b) \mapsto (f(a), b)$, is an injection. Injectivity: if $(f(a), b) = (f(a'), b')$, then the two coordinates separately agree, giving $b = b'$ and $f(a) = f(a')$; the latter, since $f$ is injective, forces $a = a'$. Hence $A \times B \hookrightarrow B' \times B$, so $\aleph_\alpha \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta$. **The middle collapse.** By [theorem 1503](/theorems/1503) applied with index $\beta$, $\aleph_\beta \cdot \aleph_\beta = \aleph_\beta$. This is the deep part — the previous theorem's work absorbs all the difficulty. **Sandwich.** We now have \begin{align*} \aleph_\beta \leq \aleph_\alpha \cdot \aleph_\beta \leq \aleph_\beta. \end{align*} Cantor–Schröder–Bernstein converts this two-sided inequality of cardinals into equality: any two cardinals with injections going each way are equal. Hence $\aleph_\alpha \cdot \aleph_\beta = \aleph_\beta$.[/guided]

[step:Prove $\aleph_\alpha + \aleph_\beta = \aleph_\beta$ by the same sandwich, using the multiplication result]We establish \begin{align*} \aleph_\beta \leq \aleph_\alpha + \aleph_\beta \leq 2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta, \end{align*} and again apply Cantor–Schröder–Bernstein. **Lower bound $\aleph_\beta \leq \aleph_\alpha + \aleph_\beta$.** Let $A, B$ be disjoint sets of cardinalities $\aleph_\alpha$ and $\aleph_\beta$. The inclusion \begin{align*} B \hookrightarrow A \sqcup B \end{align*} is an injection (just the subset inclusion), so $\aleph_\beta = \operatorname{card}(B) \leq \operatorname{card}(A \sqcup B) = \aleph_\alpha + \aleph_\beta$. **Upper bound $\aleph_\alpha + \aleph_\beta \leq \aleph_\beta + \aleph_\beta$.** Since $\aleph_\alpha \leq \aleph_\beta$ (from $\alpha \leq \beta$), there is an injection $f: A \hookrightarrow B'$ with $\operatorname{card}(B') = \aleph_\beta$ and $B' \cap B = \varnothing$ (we may take $B'$ disjoint from $B$ by relabeling). Then the map \begin{align*} f \sqcup \operatorname{id}_B: A \sqcup B &\to B' \sqcup B \\ x &\mapsto \begin{cases} f(x) & \text{if } x \in A, \\ x & \text{if } x \in B, \end{cases} \end{align*} is injective (it is injective on each piece of the disjoint union, and the images in $B'$ and $B$ are disjoint). Hence $\aleph_\alpha + \aleph_\beta = \operatorname{card}(A \sqcup B) \leq \operatorname{card}(B' \sqcup B) = \aleph_\beta + \aleph_\beta$. **Identifying $\aleph_\beta + \aleph_\beta$ with $2 \cdot \aleph_\beta$.** Let $B_1, B_2$ be disjoint copies of a set of cardinality $\aleph_\beta$, and let $\{0, 1\}$ be a two-element set. The map \begin{align*} B_1 \sqcup B_2 &\to \{0, 1\} \times B_1 \\ x &\mapsto \begin{cases} (0, x) & \text{if } x \in B_1, \\ (1, \phi(x)) & \text{if } x \in B_2, \end{cases} \end{align*} where $\phi: B_2 \to B_1$ is any fixed bijection (exists since $\operatorname{card}(B_1) = \operatorname{card}(B_2)$), is a bijection. Hence $\aleph_\beta + \aleph_\beta = 2 \cdot \aleph_\beta$. **The chain $2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta$.** The inequality $2 \leq \aleph_\beta$ holds since $\aleph_\beta \geq \aleph_0 > 2$. Multiplying both sides by $\aleph_\beta$ (monotonicity of cardinal multiplication: if $\kappa \leq \lambda$ then $\kappa \cdot \mu \leq \lambda \cdot \mu$, proved by composing an injection $\kappa \hookrightarrow \lambda$ with the identity on $\mu$), \begin{align*} 2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta. \end{align*} By [theorem 1503](/theorems/1503), $\aleph_\beta \cdot \aleph_\beta = \aleph_\beta$. **Conclusion.** Combining, \begin{align*} \aleph_\beta \leq \aleph_\alpha + \aleph_\beta \leq 2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta, \end{align*} so by [Cantor–Schröder–Bernstein](/theorems/1479), $\aleph_\alpha + \aleph_\beta = \aleph_\beta$.[/step]

[guided]The addition identity reduces to the multiplication identity via the chain \begin{align*} \aleph_\beta \leq \aleph_\alpha + \aleph_\beta \leq 2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta. \end{align*} **Lower bound.** With $A, B$ disjoint of cardinalities $\aleph_\alpha, \aleph_\beta$, the inclusion $B \subseteq A \sqcup B$ gives an injection $B \hookrightarrow A \sqcup B$. Hence $\aleph_\beta \leq \aleph_\alpha + \aleph_\beta$. **Upper bound $\aleph_\alpha + \aleph_\beta \leq \aleph_\beta + \aleph_\beta$.** Use $\aleph_\alpha \leq \aleph_\beta$ to obtain an injection of $A$ into a disjoint copy $B'$ of a set of cardinality $\aleph_\beta$. Combining with the identity on $B$ gives an injection of $A \sqcup B$ into $B' \sqcup B$ — each half is injectively embedded, and the two halves land in disjoint targets. Hence $\aleph_\alpha + \aleph_\beta \leq \aleph_\beta + \aleph_\beta$. **Identifying $\aleph_\beta + \aleph_\beta$ with $2 \cdot \aleph_\beta$.** For any cardinal $\kappa$, $\kappa + \kappa = 2 \cdot \kappa$ by the canonical bijection $B_1 \sqcup B_2 \leftrightarrow \{0, 1\} \times B$, where we tag elements of $B_1$ with $0$ and elements of $B_2$ (after bijecting to $B_1$) with $1$. Applied to $\kappa = \aleph_\beta$ this gives $2 \cdot \aleph_\beta$. **Closing with the multiplication result.** We use $2 \leq \aleph_\beta$ (which holds since $\aleph_\beta \geq \aleph_0$, an infinite cardinal) and monotonicity of multiplication to get $2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta$. Finally [theorem 1503](/theorems/1503) gives $\aleph_\beta \cdot \aleph_\beta = \aleph_\beta$. **Sandwich.** Every inequality in \begin{align*} \aleph_\beta \leq \aleph_\alpha + \aleph_\beta \leq 2 \cdot \aleph_\beta \leq \aleph_\beta \cdot \aleph_\beta = \aleph_\beta \end{align*} is non-strict, and the two ends match. By [Cantor–Schröder–Bernstein](/theorems/1479), $\aleph_\alpha + \aleph_\beta = \aleph_\beta$. This completes both statements of the theorem. The moral is that cardinal addition and multiplication among alephs are "boring" in the sense that they simply pick the larger of the two operands — all interesting behaviour of cardinal arithmetic is concentrated in **exponentiation**, which is highly non-trivial even for $\aleph_0$ (e.g., $2^{\aleph_0} = \aleph_1$ is the Continuum Hypothesis, independent of ZFC).[/guided]