Skip to content
Back to exploration
Discrete mathematics / English

The Euclidean Algorithm: How and Why, Visually

Proof of Concept · YouTube · 13:28

Open original
READ & KEEP

The explanation, unpacked.

Reviewed learning material · Video analysis · English
Read the full overview

Explain and justify the Euclidean algorithm using subtraction, remainders and an invariant common divisor.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

3:00Reduce the larger pile7:30Why the gcd is preserved10:30Replace many subtractions by division

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

Subtracting the smaller positive integer from the larger preserves their common divisors. Repeating the operation leads to a much smaller equivalent gcd problem.

An integer divides both a and b exactly when it divides both a and b−ab-a. Consequently the sets of common divisors, and hence the gcd, stay the same.

For b=aq+r with 0≤r<a0\le r<a, replace (a,b) by (r,a). Continue until one entry is zero; the other is the gcd. For nonnegative n, gcd(n,0)=n.

Knowledge cards

01

Reduce the larger pile

Subtracting the smaller positive integer from the larger preserves their common divisors. Repeating the operation leads to a much smaller equivalent gcd problem.

02

Why the gcd is preserved

An integer divides both a and b exactly when it divides both a and b−ab-a. Consequently the sets of common divisors, and hence the gcd, stay the same.

03

Replace many subtractions by division

For b=aq+r with 0≤r<a0\le r<a, replace (a,b) by (r,a). Continue until one entry is zero; the other is the gcd. For nonnegative n, gcd(n,0)=n.

Explore the knowledge in this video

Open video knowledge graph →

  • Euclidean algorithm ExplanationAt 3:00
    Why this connection?

    Subtracting the smaller positive integer from the larger preserves their common divisors. Repeating the operation leads to a much smaller equivalent gcd problem. An integer divides both a and b exactly when it divides both a and b−ab-a. Consequently the sets of common divisors, and hence the gcd, stay the same. For b=aq+r with 0≤r<a0\le r<a, replace (a,b) by (r,a). Continue until one entry is zero; the other is the gcd. For nonnegative n, gcd(n,0)=n.

Questions this video answers

Know when to use it

↗
Understand why

↗
Understand why

↗
Find a method

↗
Meet the concept

↗
Understand why

↗
Find a method

↗