# Intuition for $\,c\mid a,b \iff c\mid \gcd(a,b)$ and the Bezout GCD identity

Context: I was looking at http://math.stackexchange.com/questions/62496/show-that-gcda-1-dots-a-n-gcda-1-dots-a-n-2-gcda-n-1-a-n. I didn't know the result referred to in Brian M. Scott's answer is hard enough to require a separate proof ― http://www.proofwiki.org/wiki/Common_Divisor_Divides_GCD. ― http://math.stackexchange.com/questions/362975/concise-proof-that-every-common-divisor-divides-gcd-without-bezouts-identity?rq=1.

(1) ― What's the intuition? Is Bezout Identity supposed to be intuitive?

(2) — Why doesn't this try work? Let $$c$$ be any common divisor of $$a, b$$.
Therefore $$c\mid a, c\mid b \iff cj_1 = a, cj_2 = b$$ for some integers $$j_1, j_2$$.

Call $$g = \gcd$$. By definition $$\gcd(a,b)\mid a$$ and $$\gcd(a,b)\mid b$$ and $$c \le \gcd(a,b)$$.
Therefore $$g\mid a, g\mid b \iff gk_1 = a, gk_2 = b$$ for some integers $$k_1, k_2$$.

The two paragraphs together precipitate $$cj_1 = a = gk_1, cj_2 = b = gk_2$$
but this only proves $$c\mid gk_1, c\mid gk_2$$?

Origin - Elementary Number Theory, Jones, p9, Exercise 1.8

• Note that here $c$ and $k_1$ are coprime, so you do get that $c$ divides $g$. Jan 1, 2014 at 18:10
• @fkraiem How are $gcd(c, k_i) = 1$ for $i = 1, 2$ ? Jan 5, 2014 at 14:14
• Suppose $c$ and $k_1$ have a common factor $p$. Then, since $p$ is a factor of $c$, it is a common factor of $a$ and $b$. This contradicts the fact that $g$ is greatest, because then $gp$ is a common factor of $a$ and $b$. Jan 5, 2014 at 14:30

The proof of the linked GCD associativity property follows immediately from this basic property:

$$\rm\qquad\quad\ \ d\mid (a_1,a_2) \iff d\mid a_1,a_2\quad\ [\color{#c00}{UP} =$$ Universal Property of GCD]  (proof below)

Hence $$\rm\ \ d\mid (a_1,a_2),a_3,a_4,\ldots\!\!\overset{\color{#c00}{UP}}\iff d\mid a_1,a_2,\,\ d\mid a_3,a_4,\ldots\!\!\iff d\mid a_1,a_2,a_3,a_4,\ldots$$

Therefore $$\rm\ (a_1,a_2),a_3,a_4,\ldots\,$$ and $$\rm\,\ a_1,a_2,a_3,a_4\ldots\,$$ have the same set $$\rm\,S\,$$ of common divisors $$\,\rm d,\,$$ hence they have the same greatest common divisor $$\rm (= max\ S).\quad$$ QED

Hence the associativity of GCD boils down to the associativity of AND (which is implicit in above, i.e. $$\rm\ d\mid a,\,b,\ldots\,$$ means $$\rm\,d\mid a\,$$ AND $$\rm\,d\mid b,\ldots\,).$$

Lemma $$\ \ d\mid a,b,c\iff d\mid (a,b,c)\ \ \$$ [GCD Universal Property]

$${\bf Proof}\quad\ d\mid a,b,c\,\Rightarrow\, d\mid (a,b,c) = ia\!+\!jb\!+\!kc,\$$ for some $$\, i,j,k\in\Bbb Z,\,$$ by Bezout.

$$\qquad\qquad\, d\mid (a,b,c)\Rightarrow d\mid a,b,c\$$ by $$\ d\mid (a,b,c)\mid a,b,c\,$$ and transitivity of  "divides".

Remark  See my other answer here for a conceptual proof of the Bezout identity. In more general domains the Bezout identity may not hold (e.g. $$\,\Bbb Z[x]\,$$ and $$\,\Bbb Q[x,y])\,$$ so the above proof does not work. But both are UFDs, where we can use from prime factorizations to prove the GCD universal property. The proof essentially boils down to the universal property of $$\,\min,\,$$ occurring in the exponents of prime powers un the unique prime factorizations of the integers, i.e.

$$p^{\large i}\mid p^{\large j},p^{\large k}\iff i\le j,k\iff i\le \min(j,k)\iff p^{\large i}\mid p^{\large\min(j,k)}\!=\gcd(p^{\large j},p^{\large k})$$

More generally, in gcd-domains the universal property is actually taken as the definition of a GCD (and dually for LCM) - follow the above link for further discussion.

Key Idea $$\$$ Sets of positive integers closed under subtraction $$(>0)$$ have a special form, namely

Lemma $$\$$ Suppose a set $$\,\rm S\,$$ of positive integers satisfies $$\rm\ n, m\: \in\, S \, \Rightarrow\, n-m\, \in\, S\,$$ for all $$\rm \,n>m.\,$$ Then every element of $$\rm\:S\:$$ is a multiple of the least element $$\rm\:\ell\in S.$$

Proof $$\bf\, 1$$ $$\,$$ If not there is a least nonmultiple $$\rm\:n\in S,\,$$ contra $$\rm\:n-\ell \in S\:$$ is a nonmultiple of $$\rm\:\ell.$$

Proof $$\bf\, 2$$ $$\rm\ \ S\,$$ closed under subtraction $$\rm\:\Rightarrow\:S\,$$ closed under remainder (mod), when it's $$\ne 0,$$ since mod may be computed by repeated subtraction, i.e. $$\rm\: a\ mod\ b\, =\, a - k b = a\!-\!b\!-\!b\!-\cdots -\!b.\,$$ Hence $$\rm\:n\in S\:$$ $$\Rightarrow$$ $$\rm\: n\ mod\ \ell = 0,\:$$ else it is $$\rm\,\in S\,$$ and smaller than $$\rm\:\ell,\:$$ contra mimimality of $$\rm\:\ell.$$

This above property yields a simple conceptual proof of Bezout's identity for the gcd.

The set $$\rm\,S\,$$ of all integers of the form $$\rm\,a\,x + b\,y,\,\ x,y\in \mathbb Z,\,$$ is closed under subtraction so, by the above Lemma, all positive $$\rm\,n\in S\,$$ are divisible by $$\rm\,d =$$ least positive $$\rm\in S,\,$$ so $$\rm\,a,b\in S\,$$ $$\Rightarrow$$ $$\rm\,d\mid a,b.\,$$ So $$\rm\,d\,$$ is a common divisor of $$\rm\,a,b,\,$$ necessarily greatest,  by $$\rm\,c\,|\,a,b\,$$ $$\Rightarrow$$ $$\rm\,c\mid d\! =\! a\,x\!+\! b\,y\,$$ $$\Rightarrow$$ $$\rm\,c\le d.$$ Thus any common divisor of $$\rm\,a,b\,$$ that is of linear form $$\rm\,ax+by\,$$ is always a greatest one. The goal of the (extended) Euclidean algorithm is to search $$\,\rm S\,$$ for such a linear common divisor.

Remark $$\$$ In a nutshell, two applications of induction yield the following inferences

$$\rm S\ closed\ under\ {\bf subtraction}$$
$$\Rightarrow\:\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction$$
$$\Rightarrow\:\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm)$$

Interpreted constructively, this yields the extended Euclidean algorithm for the gcd. Namely,  starting from the two elements of $$\rm\,S\,$$ that we know: $$\rm\ a \,=\, 1\cdot a + 0\cdot b,\ \ b \,=\, 0\cdot a + 1\cdot b,\$$ we search for the least element of $$\rm\,S\,$$ by repeatedly subtracting elements to produce smaller elements of $$\rm\,S\,$$ (while keeping track of every elements linear representation in terms of $$\rm\,a\,$$ and $$\rm\,b).\:$$ This is essentially the subtractive form of the Euclidean algorithm (vs. the mod/remainder form).

Note: in more general numbers systems enjoying Division with Remainder (i.e. Euclidean domains) it is not true that $$\!\bmod\!$$ is equivalent to repeated subtraction, so in such rings the above descent is achieved by $$\!\bmod\!$$ (vs. subtraction), as in Proof $$2,\,$$ e.g. this is true for polynomial rings over a field.

The set of nonnegative integers is partially ordered by divisibility: indeed the relation denoted by $a\mid b$ is reflexive, antisymmetric and transitive.

The maximum element is $0$, the minimum element is $1$. The greatest common divisor between $a$ and $b$, provided it exists, is a number $d$ such that

1. $d\mid a$ and $d\mid b$;
2. for all $c$, if $c\mid a$ and $c\mid b$, then $c\mid d$.

This definition is slightly different than saying that the greatest common divisor is the “larges divisor” (in the sense of the $\le$ relation). But, for nonzero $a,b$, the fact that $a\mid b$ implies $a\le b$. Also $0$ can't be the gcd of two numbers under the above definition, unless they are both $0$.

So, if a gcd of two nonzero numbers exists, it is also the largest divisor of both, because of condition 2 (and the fact that it divides both, condition 1).

Note that the gcd, in this definition, is just the greater lower bound in the sense of order relations, so, in particular, it is unique if it exists.

How to prove the gcd exists? With the Euclidean algorithm, of course. If one of the numbers $a$ or $b$ is $0$, then their greatest common divisor is the other one, because $0$ is the maximum element. If the numbers are equal, then they are their gcd. So we can assume $a>b>0$. Then we can write $b=r_0$ and $a=r_0q+r_1$, where $r_1<r_0$; then

$$\gcd(a,b)=\gcd(a,r_0)=\gcd(r_0,r_1)$$

in the sense that the first number exists if and only if the last number exists and, in this case, they are equal: just compare the set of common divisors of $a$ and $b$ and the set of divisors of $r_0$ and $r_1$, finding out these sets are the same.

We have started a recursion:

$$\gcd(r_0,r_1)=\gcd(r_1,r_2)=\dots=\gcd(r_{n-1},r_n)=\gcd(r_n,0)$$

where all the equalities are “conditional” as before. But the sequence of the remainders $r_k$ is strictly decreasing, so it must stop at $0$. In the above formula, $r_n$ denotes the last nonzero remainder. Since $\gcd(r_n,0)$ exists, all “conditional” equalities become true equalities.

A proof by induction on the maximum number of steps, which avoids the “conditional equality” can be easily written.

If you define the gcd as the largest (in the sense of $\le$) common divisor, then you must prove the two properties above and one way is using the Bézout identity, which can be derived from the Euclidean algorithm.

I prefer the other method, that basically takes what you want to prove as the definition, because in this way we can show that $(\mathbb{N},|)$ is not only a partially ordered set, but also a lattice: any two elements also have a smallest upper bound (the lowest common multiple, of course). The minimal elements in the subset $\mathbb{N}\setminus\{1\}$ are exactly the prime numbers.

Moreover, the sublattice $\mathbb{N}\setminus\{0\}$ is (order) isomorphic to the lattice of sequences of natural numbers $$(s_0,s_1,s_2,\dotsc)$$ which are eventually zero, ordered by the product order: $$(s_0,s_1,s_2,\dotsc)\le(t_0,t_1,t_2,\dotsc)$$ if and only if $s_k\le t_k$ for all $k$. This isomorphism can be used to derive the middle-school definitions of gcd and lcm as well as the formula $$\gcd(a,b)\operatorname{lcm}(a,b)=ab$$ valid for all natural numbers $a$ and $b$.