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Question
The well-ordering principle can be used to show that there is a unique gcd of two positive integers. Let $a$ and $b$ be positive integers, and let $S$ be the set of positive integers of the form $as+bt$, where $s$ and $t$ are integers.

a) Show that $S$ is non-empty.
b) Use the well-ordering property to show that $S$ has a smallest element $c$.
c) Show that if $d$ is a common divisor of $a$ and $b$, then $d$ is a divisor $c$.
d) Show that $c|a$ and $c|b$.
e) Conclude from (c) and (d) that the greatest common divisor of $a$ and $b$ exists. Finish the proof by showing that this GCD of two positive integers is unique.

This is my first use of the well-ordering principle so I need confirmation. More about my problem of this question in the section Problem below.

My Attempt
a) $S$ is non-empty since $s$ and $t$ in $as+bt$ can be any non-negative integers.

b) From (a), since $S$ is a non-empty set of positive integers, by the Well-Ordering principle there is a least element $c$.

c) Since $d|a$ and $d|b$, $$a = dk_1, k_1 \in \mathbb{Z}$$ $$b = dk_2, k_2 \in \mathbb{Z}$$ Thus, $$c = as + bt = (dk_1)s + (dk_2)t = d(k_1s) + d(k_2s) \equiv 0 (\bmod d)$$

d) Suppose $c \nmid a$, then $a = qc + r, 0 < r < c$. Suppose now $r \in S$, this is a contradiction since we've already established that $c$ is the least element in set $S$.

e)

  • Existence: We have shown a least element $c$ in a non-empty set $S$, such that $c\mid a$ and $c \mid b$. Other divisors such as $d$, divides $a, b, c$, thus $d \leq c$, hence gcd $c$ exist.

  • Uniqueness: Suppose $\exists e \mid \gcd(a, b) = e$ then $$e \mid a, e \mid b \rightarrow e \mid c,$$ which is true if and only if $e \leq c$, considering $\gcd(a, b) = e$, thus $e = c$.

Problem:
(a) and (b) was my first use of the well-ordering principle (at least consciously) thus I need confirmation that it is being used correctly. (c) I'm quite confident so it's unlikely that it's wrong. (d) I'm a little shaky on that, is that also a correct use of the well-ordering principle? (e) I can't find anything wrong with it, but if it's wrong point it out.

Bill Dubuque
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JoeyAndres
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    If I could give $+10$ to this post, I would. This is a perfect example of how to show work when asking a question. :) – apnorton Jul 15 '14 at 02:25
  • @bof I guess what I meant is any non-negative integers would do since $a$ and $b$ are already positive integers. – JoeyAndres Jul 15 '14 at 02:29
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    "Any non-negative integers" won't do, because $s=t=0$ is no good. Anyway, to show that $S$ is nonempty, all you have to do is exhibit one element of $S$. Why not just say "$S$ is nonempty because $a+b\in S$, as we can see by taking $s=t=1$"? – bof Jul 15 '14 at 02:41
  • In "(b)", you should say "has a least element, $c$", so that we know what $c$ refers to in part "(c)". In part "d", best to say "...then $a = qc + r$ for some $q \in \mathbb Z$ and some $r$ with $0 < r < c$. Your part-d argument isn't clear to me .. needs further detail, I believe. But the parts a/b stuff is fine (i.e., the use or well-ordering), subject to bof's comment. (And I agree with anorton completely -- congrats!) – John Hughes Jul 15 '14 at 02:47
  • @bof I don't get why "Any non-negative integers" won't do. Isn't Wel-Ordering Property is "Every non-empty set of non-negative...", if $s=t=0$, then $c = 0$, which is non-negative, enough for $c$ to be in set $S$. – JoeyAndres Jul 15 '14 at 02:51
  • A bit of a quibble about logical wording, but in part (d) we do not "suppose" that $r \in S$, as since $c \in S$ we can write $c = ta + sb$ we know we can re-arrange that equation to $$0 < r = qc + a = [q(t+1)]a + qsb \in S$$. Suppose should be restricted to the single assumption used for the contradiction. – jxnh Jul 15 '14 at 02:52
  • $0\notin S$ because $S$ is defined as "the set of positive integers of the form . . .". Good thing, too. If $0$ were allowed as an element of $S$, then of course the least element of $S$ would always be $0$, but that won't work as a greatest common divisor. – bof Jul 15 '14 at 03:35
  • Actually, the best way to do part (a) is to take $s=a$ and $t=b$ and show that $a^2+b^2\in S$. This way we can see that we don't really need to assume that $a$ and $b$ are both positive, all we need is that they are not both zero. – bof Jul 15 '14 at 03:38
  • @bof I see. I'll take that into account. Cheers. – JoeyAndres Jul 15 '14 at 03:56
  • A search for squares arithmetic progression found this paper with a parameterization of primitive sets of squares. – Ross Millikan Jun 23 '21 at 14:15
  • "d) Suppose c∤a, then a=qc+r,0<r<c. Suppose now r∈S," .... and what if $r$ isn't in $S$? How would that be a contradiction? – fleablood Jul 06 '24 at 21:31

1 Answers1

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For a, you didn't use the well-ordering principle. You showed there was at least one element of $S$ by exhibiting one. It would be a bit better to say "take $s=1, t=0$, then $as+bt=a$, so $S$ is not empty." but you have the idea. b is correct as stated-you used the well ordering principle correctly. c is fine as well. For d you don't justify that $r \in S$, but you can. This is the critical point. You have $r=a-qc$ Now substitute in $c=as+bt$ and you justify what you were assuming.

Ross Millikan
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  • Hello! The OP says "We have shown a least element c in a non-empty set S, such that c∣a and c∣b. Other divisors such as d, divides a,b,c, thus d≤c, hence gcd c exists." The OP initially assumed c to be the smallest element of the set. Isnt it then contradictory to the line "Other divisors such as d, divides a,b,c, thus d≤c". Shouldnt d always be equal to c, since a number less than d doesnt exist? Im a mere beginner in number theory, please bear with me! – newbie105 Jun 23 '21 at 05:57
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    @newbie105: $c$ is the smallest element of the set of numbers you can form as $as+bt$. Other divisors $d$ of $a,b$ are not members of $S$, but they are $\le c$ – Ross Millikan Jun 23 '21 at 13:31