## MaBloWriMo 24: Bezout’s identity

A few days ago we made use of Bézout’s Identity, which states that if $a$ and $b$ have a greatest common divisor $d$, then there exist integers $x$ and $y$ such that $ax + by = d$. For completeness, let’s prove it.

Consider the set of all linear combinations of $a$ and $b$, that is, $\{ax + by \mid x, y \in \mathbb{Z} \}$,

and suppose $d = as + bt$ is the smallest positive integer in this set. For example, if $a = 10$ and $b = 6$, then you can check that, for example, $16 = a + b$, and $8 = 2a - 2b$, and $4 = a - b$ are all in this set, as is, for example, $-4 = -a + b$, but the smallest positive integer you can get is $2 = 2a - 3b$. We will prove that in fact, $d$ is the greatest common divisor of $a$ and $b$.

Consider dividing $a$ by $d$. This will result in some remainder $r$ such that $0 \leq r < d$. I claim the remainder is also of the form $ax + by$ for some integers $x$ and $y$: note that $a = a1 + b0$ is of this form, and $d$ is of this form by definition, and we get the remainder by subtracting some number of copies of $d$ from $a$. Subtracting two numbers of the form $ax + by$ works by subtracting coefficients, yielding another number of the same form again. But $d$ is supposed to be the smallest positive number of this form, and $r$ is less than $d$—which means $r$ has to be zero, that is, $d$ evenly divides $a$. The same argument shows that $d$ evenly divides $b$ as well. So $d$ is a common divisor of $a$ and $b$.

To see that $d$ is the greatest common divisor, suppose $c$ also divides $a$ and $b$. Then since $d = ax + by$, we can see that $c$ must divide $d$ as well—so it must be less than or equal to $d$.

Voila! This proof doesn’t show us how to actually compute some appropriate $x$ and $y$ given $a$ and $b$—that can be done using the extended Euclidean algorithm; perhaps I’ll write about that some other time. But this proof will do for today.

For the remainder of the month, as suggested by commented janhrcek, we’ll prove the thing I hinted at in an earlier post: namely, that the order of any group element is always a divisor of the order of the group. This is a really cool proof that hints at some much deeper group theory. Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 2 Responses to MaBloWriMo 24: Bezout’s identity

1. Logan says:

When showing that $d$ is a common divisor, you state “…that is, $d$ evenly divides $a$.” The next sentence states “…shows that $d$ evenly divided $a$ as well.” Shouldn’t the second be $b$?

• Brent says:

Yup, good catch, thanks! Fixed now.