# Euclidean Algorithm

## Objective

The **Euclidean Algorithm** efficiently computes the **Greatest Common Divisor (GCD)** of two integers $$A$$ and $$B$$.

The core idea is based on the following identity:

$$
\gcd(A, B) = \gcd(B, A \bmod B)
$$

## Code

```python
def gcd(a, b):
    if b == 0:
        return a
    else:
        return gcd(b, a % b)
```

## Proof

Let $$G = \gcd(A, B)$$. Then, we can write $$A=aG$$, $$B=bG$$, where $$a$$ and $$b$$ are coprime integers.

Dividing $$A$$ by $$B$$ gives:

$$
A = Bq + R
$$

Then:

$$
R = A - Bq = (a - bq)G = rG
$$

So $$R = rG$$, and we want to prove that $$\gcd(B, R) = G$$ is true if and only if $$b$$ and $$r$$ are coprime.

Suppose, for contradiction, that $$b$$ and $$r$$ are **not** coprime. Then there exists a common divisor $$m > 1$$ such that:

$$
b = mb', \quad r = mr'
$$

where $$b'$$ and $$r'$$ are coprime. Then:

$$
\begin{align\*}
a &= bq + r \\
&= mb'q + mr' \\
&= m(b'q + r')
\end{align\*}
$$

This means $$a$$ is also divisible by $$m$$, implying $$a$$ and $$b$$ share a common factor $$m$$, contradicting the assumption that $$a$$ and $$b$$ are coprime. Thus, $$b$$ and $$r$$ must be coprime.

Proving the reverse is trivial, so we don't prove it here.

> Written by [Ryan Kim](mailto:undefined) of Fractalyze


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://fractalyze.gitbook.io/intro/primitives/euclidean-algorithm.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.