Proving Irrationality of Square Roots - Visual Mathematical Proofs

Welcome to Mathematical Proofs!

Let's explore the fascinating world of irrational numbers
using the powerful technique of proof by contradiction

1/2

3/4

√2

5/3

Theorem 1.2 (Supporting Lemma)

If p is prime and p divides a², then p divides a

Let a =

p₁

×

p₂

× ... ×

pₙ

↓

Then a² =

p₁²

×

p₂²

× ... ×

pₙ²

By uniqueness of prime factorization:
If p divides a², then p must be one of p₁, p₂, ..., pₙ
Therefore, p divides a

Proof: √2 is Irrational

1

Assumption (for contradiction)

Assume √2 is rational

√2 = a/b (where a, b are coprime)

2

Square Both Sides

2 = a²/b²

2b² = a²

3

Apply Theorem 1.2

Since 2 divides a², then 2 divides a
Therefore: a = 2c for some integer c

4

Substitution

2b² = (2c)² = 4c²

b² = 2c²

So 2 divides b² → 2 divides b

!

CONTRADICTION!

Both a and b are divisible by 2
This violates coprime assumption

∴

CONCLUSION

Therefore: √2 must be
IRRATIONAL

Step-by-Step Algebraic Manipulation

√2 = a/b (assumption)

2 = a²/b² (squaring both sides)

2b² = a² (cross multiply)

2b² = (2c)² = 4c² (substitute a = 2c)

b² = 2c² (divide by 2)

Both a and b are even → Contradiction!

Proof: √3 is Irrational

1

Square and Rearrange

3 = a²/b² → 3b² = a²

2

Apply Theorem

3 divides a² → 3 divides a → a = 3c

3

Substitute Back

3b² = (3c)² = 9c² → b² = 3c²

So 3 divides b² → 3 divides b

4

Contradiction Found!

Both a and b divisible by 3
Violates coprime assumption

Therefore: √3 is IRRATIONAL

General Result

For any prime number p:

√p is IRRATIONAL

Examples of Irrational Square Roots:

√2

√3

√5

√7

√11

√13

The same proof technique works for all prime square roots!