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

$a|b$ implies $a|bc$ for any integer $c$;

$a|b$ and $b|c$ imply $a|c$;

$a|b$ and $a|c$ imply $a|(bx + cy)$ for any integers $x$ and $y$;

$a|b$ and $b|a$ imply $a=±b$;

$a|b$, $a>0$, $b>0$, imply $a\leq b$;

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?

@KaviRamaMurthy Corrected. Thanks. – Jonathan Komar Oct 09 '18 at 07:29 — Jonathan Komar, Oct 09 '18 at 07:29

Siong Thye Goh · Accepted Answer · 2018-11-08T14:14:16.677

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

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

edited Nov 08 '18 at 14:14

answered Oct 10 '18 at 04:30

Siong Thye Goh

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It is not obvious to me that property 3 "admits" (strange word choice by author) an extension to any finite sense. Your answer clears it up with mathematical induction ($\sum_{j=1}^{k+1}$ with the extra $1$ being the extension of the finite set). I'll accept this. By the way, it really disturbs me when authors write "I will leave $x$ as an exercise", when I clearly paid for a book that I expect would explain things. In this case, it is fine, of course. – Jonathan Komar Nov 08 '18 at 10:29
I think it would be clearer to write $al + ad$ rather than $ad+al$ after "Hence, we have", because the $al$ corresponds to the item on the left, $\sum_{j=1}^{k+1}x_jb_j$. (Of course the commutative property of the + operator makes it correct either way, but the order helps to create mental associations between the lines...) – Jonathan Komar Nov 08 '18 at 10:38
Please specify whether there is is typo in the text, because strictly, that was the original question. I can then accept this answer. – Jonathan Komar Nov 08 '18 at 10:44
There is no typo from my understanding. He's mentioned that item $3$ can be generalized (and then briefly mentioned item $2$ can be generalized too), then he gives a proof of item $3$. Also, I made a typo in the $d+l$ part, it should have been $d+x_{k+1}l$. – Siong Thye Goh Nov 08 '18 at 14:15

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

Theorem 1.1

1 Answers1