Proposition 1

Let $(a_i)_{i \in I}$ be a finite family of integers. Then the cardinals $\sum_{i \in I} a_i$ and $\prod_{i \in I} a_i$ are integers.

Let us begin by showing that if $a$ and $b$ are integers, then so is $a + b$. We proceed by induction on $b$. The assertion is true for $b = 0$ since $a + 0 = a$. If $a + b$ is an integer, then so is $(a + b) + 1$ (§ 4, no. 1, Proposition 1). But $(a + b) + 1 = a + (b + 1)$ (§ 3, no. 3, Corollary to Proposition 5); hence $a + (b + 1)$ is an integer, and consequently $a + b$ is an integer for all integers $b$.

Next we show by induction on $n = \mathrm{Card}\ (\mathrm{I})$ that $\sum\limits_{i \in \mathrm{I}} a_i$ is an integer. This is clear if $n = 0$, for then $\mathrm{I} = \emptyset$ and $\sum\limits_{i \in \mathrm{I}} a_i = 0$. If $\mathrm{Card}\ (\mathrm{I}) = n + 1$, we may write $\mathrm{I} = \mathrm{J} \cup \{k\}$, where $\mathrm{Card}\ (\mathrm{J}) = n$ and $k \notin \mathrm{J}$. Then

($\S\,3$, no. 3, Proposition 5). The inductive hypothesis is that $\sum\limits_{i \in \mathrm{J}} a_i$ is an integer; hence so is $a_k + \sum\limits_{i \in \mathrm{J}} a_i$, from the first paragraph of the proof. This shows that $\sum\limits_{i \in \mathrm{I}} a_i$ is an integer for all $n$.

Since the product $ab$ of two integers $a$ and $b$ is the sum of a finite family of integers all equal to $a$ ($\S\,3$, no. 4, Proposition 6, Corollary 2), $ab$ is an integer. We shall prove by induction on $n = \mathrm{Card}\ (\mathrm{I})$ that $\prod\limits_{i \in \mathrm{I}} a_i$ is an integer. This is true for $n = 0$, because then

If $\mathrm{Card}\ (\mathrm{I}) = n + 1$, we have (with the same notation as above)

($\S\,3$, no. 3, Proposition 5), and the inductive hypothesis therefore implies that $\prod\limits_{i \in \mathrm{I}} a_i$ is an integer. Consequently $\prod\limits_{i \in \mathrm{I}} a_i$ is an integer for all $n$.