Is there a typo about the theorem of divisibility in The Theory of Numbers book by Ivan Niven; Herbert Zuckerman; Hugh Montgomery?

Question

Theorem 1.1

1. $a|b$ implies $a|bc$ for any integer $c$;
2. $a|b$ and $b|c$ imply $a|c$;
3. $a|b$ and $a|c$ imply $a|(bx + cy)$ for any integers $x$ and $y$;
4. $a|b$ and $b|a$ imply $a=±b$;
5. $a|b$, $a>0$, $b>0$, imply $a\leq b$;
6. if $m ≠ 0$, $a|b$ implies and is implied by $ma|mb$.

Proof: The proofs of these results follow at once from the definition of divisibility. Property 3 admids an obvious extension to an finite set, thus:

$a|b_1$, $a|b_2$,...$a|b_n$ imply $a|\sum_{j=1}^{n}{b_jx_j}$ for any integers $x_j$

p.5

Is "Property 3" a mistake? Shouldn't it be Property 2?

Property 2 can be extended similarly.

To give a sample proof, consider item 3. Since $a|b$ and $a|c$ are given, this implies that there are integers $r$ and $s$ such that $a|b$ and $b = ar$ and $c = as$. Hence, $bx + cy$ can be written as $a(rx +sy)$, and this proves that $a$ is a divisor of $bx + cy$.

p.5

Is "Property 2" a mistake? Shouldn't it be Property 3?

I hope to completely understand this (it is not initially obvious as the book often proclaims i.e. it would be nice to know why something is obvious). Could somebody elaborate on the proofs please?

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Ivan Niven; Herbert Zuckerman; Hugh Montgomery. An Introduction to The Theory of Numbers. Copyright 1991. Fifth Edition. 2017 reprint.

Answer 1

I think what the book meant is property $3$ can be extended to a finite set. (Similarly for property $2$. Just mentioning without diving into details.)

Then it gives a proof of property $3$.

Mathematical induction is a common tool to prove extension to finite set.

We want to prove that

$$a|b_1, a|b_2 \ldots a|b_n \implies a | \sum_{j=1}^n x_jb_j, \forall x_j \in \mathbb{Z}$$

We have proven the base case. Also, we assume that we have

$$a|b_1, a|b_2 \ldots a|b_k \implies a | \sum_{j=1}^k x_jb_j, \forall x_j \in \mathbb{Z}$$

and we have to check that

$$a|b_1, a|b_2 \ldots a|b_k , a|b_{k+1}\implies a | \sum_{j=1}^{k+1} x_jb_j, \forall x_j \in \mathbb{Z}$$

We can write $\sum_{j=1}^k x_jb_j = ad$ and $b_{k+1}=al$ for some $d, l \in \mathbb{Z}$.

Hence, we have \begin{align} \sum_{j=1}^{k+1} x_j b_j &= \sum_{j=1}^k x_j b_j + x_{k+1}b_{k+1} \\ &=ad + x_{k+1}(al) \\ &= a(d+x_{k+1}l) \end{align}

Since $d+x_{k+1}l \in \mathbb{Z}$, we conclude that $a | \sum_{j=1}^{k+1} x_jb_j$.

I will leave the exercise of the extension of property $2$ as an exercise, prove that

$$a|a_1, a_1 | a_2 \ldots, a_{n-1}|a_n \implies a|a_n$$