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I have the following question for homework.

Fix positive integers $a$ and $b$ . Here’s an inductive definition of a set $S$ :

Foundation rule: $a,b ∈ S$.

Constructor rule: If $m,n ∈ S$ , then $m − n ∈ S$.

(a) Suppose $h$ is a common factor of $a$ and $b$ . Use the exclusion rule to prove that for every $n ∈ S$ , $h$ divides $n$ .

(b) Suppose $k ∈ S$ is a positive integer which is not a factor of $a$ . Prove that there is some $l ∈ S $ such that $0 < l < k$ . (Hint: Consider the sequence $a,a − k,a − 2 k,...$ and use the fact that $\mathbb{N}$ is well-ordered.)

(c) In the same way that you proved (b), we may also prove the following fact: if $k ∈ S$ is a positive integer which is not a factor of $b$ , then there is some $l ∈ S$ such that $0 < l < k$ . Use (b) and the above fact to prove that there is some positive integer in $S$ which is a common factor of $a$ and $b$ . (Hint: Use the fact that $\mathbb{N}$ is well-ordered.)

(d) Use (a) and (c) to conclude that S contains gcd($a,b$).

I'm unsure about how to even start (a) and (b).

For (a) I thought something along the lines of "because $h$ is a common factor of $a$ and $b$, as $n ∈ S$, $h$ must be a divisor of $n$" yet apparently this is quite far off what we are supposed to do.

I have literally no idea how to even start (b).

Any help would be appreciated.

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zhl44304
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    What course are you studying? What is the exclusion rule? Also, please ask one question at a time. – Shaun Mar 13 '20 at 00:11
  • Can you use induction on a)? Base cases: $h|a$ and $hb$. Inductive step if $h|m$ and $h|n$ then $h|m-n$. So $h$ divides all constructable elements. That works, doesn't it? – fleablood Mar 13 '20 at 00:29
  • @Shaun It's a discrete math course, although because of the recent coronavirus outbreak he hasn't been able to teach or give office hours for the past week or 2, however the homework is still due. I wasn't sure if it was useful to have the information about the rest of the homework problem, as it is given in steps. – zhl44304 Mar 13 '20 at 00:52
  • @zhl44304 Do you know about the thing called Ideal? I think it is in https://en.wikipedia.org/wiki/Ring_(mathematics), you can try reading about some of its results. I think your (a) and (b) are direct results of this – Nikola Tolzsek Mar 13 '20 at 03:26
  • @Shaun, it is not necessary in this case. For once, this question can be asked 4 at a times. They are direct results of each other. – Nikola Tolzsek Mar 13 '20 at 04:00
  • See the Lemma and following in the linked dupe. – Bill Dubuque Jun 03 '21 at 20:21

2 Answers2

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(a)
Bass case. As h divides a and b, show h divides a - b.
Induction step. If h divides r,s, show h divides r - s.

There's such a thing as the exclusion rule?

William Elliot
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Some greetings

Hello @zhl44304, welcome to MSE. Concerning your problem, I think you should first read about Ideal and Ring theory.

My proof of a direct result down here will help you solve all four questions in one note.

Restate the question

Let there be a set $\mathbb{I}$ with 2 foundation numbers $a$ and $b$, and if $x$ and $y$ $\in \mathbb{I}$ then $x+y \in \mathbb{I}$

We now prove that $\mathbb{I} = k\mathbb{Z}$, which is all the integers that is divisible by $k$, for some $k \in \mathbb{Z}$

Prove the properties

  • If $\mathbb{I} = \left\{ 0 \right\}$, then $k=0$
  • If $\mathbb{I} \neq \left\{ 0 \right\}$, then $k=0$

Consider the set $\mathbb{I^*}$ that consists of positive numbers.

It is easy to show that $\mathbb{I^*}$ has a lower bound, therefore, $\exists$ $m \in \mathbb{I^*}, m= min \left\{ \mathbb{I^*} \right\}$

We now prove that $k=m$.

Indeed, take a random $e \in \mathbb{I^*}$, write $e=qm +r $. If r is not 0, then it is easy to show that $r \in \mathbb{I^*}$, which is a contradiction since $r<m$ and $m= min \left\{ \mathbb{I^*} \right\}$

Thus, all number of $\mathbb {I^*}$ is divisible by m.

Reversely, it can also be conclude that because the smallest number is already $1\times m$, then every numbers divisible by $m$ is also belongs to $\mathbb{I^*}$

Back to your question...

So now, it is clear that your set $\mathbb{S}$ is an ideal on $\mathbb{R}$, thus it has the form of $ \mathbb{S}=k\mathbb{Z}$

Therefore, $k \vert gcd(a,b) $. But it is also apparent that because a and b are foundation numbers, which is, the set $\mathbb{S}$ is constructed upon these two numbers, then all numbers must be divisible by k, and their common divisor must be no more than k (otherwise, due to the proof above, either $a$ or $b$ does not belong to $\mathbb{S}$

Thus, the following conclusions are made:

  1. $\mathbb{S}$ consists of and only of all the integers divisible by k.
  2. $k=gcd(a,b)$

And hence we have a, b, c, d proven. Q.E.D

P/s: If you like this solution, vote up. If you don't like it, comment below so I can improve it. Thank you for your very interesting question!

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Nikola Tolzsek
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