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Let a > b be positive integers. In applying the Euclidean algorithm, we have $a = b q_0$ + $r_0$, $b = r_0 q_1 + r_1$, and $r_{n-1} = r_n q_{n+1} + r_{n+1}$, for all $n > 0$. Prove by induction that $r_n$ is in the set $\{ka + lb\}$ such that l and k are integers every $n>-1$

This i find very frustrating but i am horrible at induction :),

i started with my base case's $s=0,1$

$r_0 = a - b q_0$

$b = r_0 q_1 r_1$ thus $r_1 = b - r_0 q_1$

from there i assume ( cause i have no idea what else to do) $r_1$ must divide b and $r_0$ which it should but may not be useful to me. Now i assume that its true for $r_d$ and $r_{d-1}$ and want to show it for $r_s+1$

well $r_{d+1} = r_{d-1} - r_d q_{d+1}$ and now im lost, though im fairly certain i was lost at prove by induction :)

im certain if i can prove divisibility here with this step i can merely write a and b as multiples of $r_n$ and show that it divides the set.

That being said i could very well be on the wrong track altogether any advice much appreciated.

Faust
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  • I have somewhat reformatted your LaTeX code. Please take a look at the editing I have done (when it shows up). – Andreas Caranti Feb 07 '13 at 13:58
  • In your second paragraph, the part beginning "in my opinion" seems confused, for two reasons. First, it describes $r_n$ as something that doesn't mention $n$, whereas in fact, as $n$ varies, the algorithm produces different values of $r_n$. Second, you wrote that $r_n$ is defined in a certain way, whereas you've already quoted, in your first paragraph, the definition of $r_n$ and it looks entirely different. – Andreas Blass Feb 07 '13 at 14:05
  • i removed it, i was merely implying that i do not like the question, the gcd (a,b)= gcd (b,$r_0$)= gcd($r_0$,$r_1$) etc the gcd is the last $r$ that does not equal 0 but each $r_n$ must divide that set i just have no idea how to prove anything with induction :) I will change my assumption case to d instead of n so its clearer to understand. i was running out of alphabet :) – Faust Feb 07 '13 at 14:23

3 Answers3

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Here are steps to show the question you pose in the last sentence of your first paragraph, which I think is your question:

(1) Let $\mathbb{Z} m := \{ x m | x \in \mathbb{Z} \}$. Show: every subgroup of the additive group of the integers is of form $\mathbb{Z} m$, with $m$ being the smallest positive integer in the subgroup. (proof outline: (a) Let $H$ be any subgroup. If it is trivial ($H = \{0 \}$), $H = \mathbb{Z} 0$; if not, it contains an integer $x$, and so, as it is a subgroup, also $-x$, so it contains a positive integer. Call the smallest such integer $m$; (b) show, using $H$ is a subgroup, that $\mathbb{Z} m \subset H$: $mk = m + \dots m \in H, -mk \in H, 0 = 0m$ as well (immediate as $H$ is a subgroup); (c) using Euclid, write any $n \in H$ as $n = qm + r, q, r \in \mathbb{Z}, 1 \leq r < m$. $qm \in H$ by (b), and $n$ by choice, so also $r = n - qm$ as $H$ is a subgroup. But by our definition of $m$ as the smallest positive integer in $H$, and $0 \leq r < m$, then $r= 0$, and so $n = qm$, which was to be shown.). 

(b) by (a), $\{ka + lb, k, l \in \mathbb{Z} \} = \mathbb{Z} m$ for $m$ the smallest positive integer in this subgroup. Now it is easy to show: (1) there are integers $r, s$ such that $m = ra + sb$ (Bezout's identity: as $m = 1 m \in \mathbb{Z} m$); (2) $m$ divides $a$ and $b$ (directly from (1)); and (3) if an integer $d$ divides $a$ and $b$, it also divides $m$ (also directly from (1)). But (2) and (3) are the definition of the greatest common divisor, so you are done, as euclid's algorithm has $r_n = \operatorname{gcd}(a, b)$.

gnometorule
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  • Exactly what i want to prove but i have to prove it using induction :( – Faust Feb 07 '13 at 14:29
  • @Faust7: oy...:) So you essentially need a direct proof by induction of Euclid (as it typically includes the proof Bezout, which you really need). This can also be done. Sitting at 28F while sipping coffee an smoking and typing this on phone, so too cold to add that. When I'm back home and not posted yet, can do that too. – gnometorule Feb 07 '13 at 14:35
  • I just realized that when i read Bezouts identity, i will look aa a proof by induction of that as i am sure i have seen it before i just can't recall it. Thanks for your help :) – Faust Feb 07 '13 at 14:39
  • @Faust7: You're welcome! With that, and the other answer now posted, I'm sure you'll get there after sufficient thinking about it. – gnometorule Feb 07 '13 at 14:50
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All the $q_n$ are integers.

Base case n=0: $$r_0 = 1 * a - q_0 b,$$ so $r_0$ is in the set with $k = 1$ and $l = q_0$.

For n=1: $$r_1 = b - q_1 r+0 \\ = b - q_1 (a - q_0 b) \\ = -q_1 a + (1-q_0)b,$$ so $r_1$ is in the set with $k = -q_1$ and $l = 1 - q_0$.

Rearrange your formula for $r_{n+1}.$ Assuming that $r_{n-1}, r_n$ are integer multiples of a plus integer multiples of b, then you can show that $r_{n+1}$ is too.

oks
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Hint $\, $ The set $\rm\,I = a\,\Bbb Z + b\,\Bbb Z\subset \Bbb Z\,$ is closed under $\rm\color{#C00}{addition}$ and $\rm\color{#0A0}{scalings}$ by $\rm\,\color{#0A0}n\in \Bbb Z.\:$ Indeed, $\rm\,ai\!+\!bj \color{#C00}{\bf +} (ai'\!+\!bj') = a(i\!+\!i')+b(j\!+\!j')\in S,\ $ and $\rm\,\ \color{#0A0}{\bf n}\,(ai+bj) = a(ni) + b(nj)\in S.$

Therefore, by scaling closure $\rm\,r_n \in I\,\Rightarrow\, \color{#0A0}{q_{n+1}} r_n\in I.\,$ Adding $\rm\,r_{n-1}\in I\,$ to this yields $\rm\,r_{n+1}\in I,$ by addition closure, i.e. $$\rm r_n,r_{n-1}\in I\ \Rightarrow\ r_{n+1} = \color{#0A0}{q_{n+1}} r_n \color{#C00}{\bf +} r_{n-1} \in I\quad $$

Thus the induction step follows simply because successive remainders are computed from prior remainders $\rm\in I$ using operations that remain in $\rm\,I.$

Remark $\ $ We can simplify in the ring $\rm\,\Bbb Z.\,$ Integer scalings are repeated additions or subtractions, so if $\rm\,I\,$ is closed under subtraction then it is also closed under addition and scalings. In group/ring theory language: $\rm\ I\,$ is a ideal of the ring $\rm\,\Bbb Z$ $\!\iff\!$ $\rm I\,$ is an additive subgroup of the additive group $\rm\,\Bbb Z\ $ (since, by the subgroup test, subgroups are precisely the subsets closed under subtraction).

The essence of the Bezout identity for the gcd is: subgroups of $\rm\,\Bbb Z\,$ are cyclic (or ideals are principal), generated by their least positive element $\rm\,d,\,$ since closed under subtraction implies closed under mod or remainder (obtained by repeated subtraction, i.e. the Division Algorithm). So we conclude $\rm\,d\,$ divides all $\rm\,i\in I\,$ (else $\rm\,i\ mod\ d \in I\,$ is nonzero and smaller than $\rm\,d).\,$ That is essentially the inductive step of the Euclidean algorithm. It generalizes to "Euclidean" rings which enjoy division with "smaller" remainder, e.g. polynomials over a field, where smaller means smaller degree.

Nonempty subsets of a ring closed under addition and scaling by ring elements are known as ideals. If you study university algebra you will learn that ideals play a fundamental role in number theory and algebra. Ideals abstract the innate structure that governs many proofs in elementary number theory, e.g. denominator ideals and order ideals.

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