Rational and irrational numbers
A rational number can be written as p/q, where p and q are integers and q ≠ 0, like 3/4, −5, 0.25. Its decimal either ends or repeats (1/3 = 0.333…).
An irrational number cannot be written as p/q. Its decimal goes on forever without a repeating pattern: √2, √3, √5, π. Rational and irrational numbers together are the real numbers.
The key fact (a theorem we use)
Let p be a prime. If p divides a², then p divides a (a is a positive integer). Why: by the Fundamental Theorem of Arithmetic, the primes of a² are the primes of a, each used twice. So if p appears in a², it was already in a.
Proof that √2 is irrational
This is a proof by contradiction: we assume the opposite and show it leads to something impossible.
- Assume √2 is rational: √2 = a/b, where a, b are co-prime integers, b ≠ 0.
- Square: 2 = a²/b², so a² = 2b². Hence 2 divides a², so 2 divides a.
- Write a = 2c. Then 4c² = 2b², so b² = 2c². Hence 2 divides b², so 2 divides b.
- Now 2 divides both a and b. But a, b are co-prime. Contradiction.
- So our assumption was wrong: √2 is irrational.
Proof that √3 and √5 are irrational
Same steps, with 3 (or 5) instead of 2. Assume √3 = a/b (co-prime). Then a² = 3b², so 3 divides a. Put a = 3c: 9c² = 3b², so b² = 3c², so 3 divides b. Both share 3: contradiction. For √5, use 5 everywhere.
Note: this works because 2, 3, 5 are prime. For √4 = 2 the proof fails, which is correct since 2 is rational.
Proving numbers like 3 + 2√5 are irrational
Assume 3 + 2√5 = r, a rational number. Then √5 = (r − 3)/2. The right side is rational (rationals are closed under −, ÷). So √5 would be rational, which is false. Hence 3 + 2√5 is irrational.
Rules: rational + irrational = irrational; non-zero rational × irrational = irrational. But irrational + irrational can be rational: √2 + (−√2) = 0.
Try it: a practical
Draw a 1 cm × 1 cm square and measure its diagonal as carefully as you can (about 1.41 cm). Now in the 3D free play, hunt for p and q so that p² = 2q². Write down your closest tries (like 7/5, 17/12). They get close, never exact. That is the proof you can see.
Board exam corner
A 3-mark question almost every year: prove √2, √3 or √5 is irrational, or prove a form like 5 − 2√3 or 1/√2 is irrational. Write every line: the assumption with co-prime a, b, the squaring, the "prime divides a² ⇒ divides a" step, and the clear contradiction statement.
Key formulas and definitions
- Rational: p/q, p, q integers, q ≠ 0
- Prime p | a² ⇒ p | a
- √2 = a/b ⇒ a² = 2b² ⇒ a = 2c ⇒ b² = 2c² ⇒ contradiction
- rational + irrational = irrational
- non-zero rational × irrational = irrational
Worked examples
1. Is 0.101001000100001… rational or irrational?
The pattern of zeros keeps growing, so the decimal never repeats a fixed block. It does not end either. So it is irrational.
2. Prove that √5 is irrational.
Assume √5 = a/b, a and b co-prime. Then a² = 5b², so 5 | a² ⇒ 5 | a. Let a = 5c: 25c² = 5b² ⇒ b² = 5c² ⇒ 5 | b. So 5 divides both a and b, contradicting co-prime. Hence √5 is irrational.
3. Prove that 3 + 2√5 is irrational.
Assume 3 + 2√5 = r (rational). Then √5 = (r − 3)/2, which is rational. But √5 is irrational. Contradiction, so 3 + 2√5 is irrational.
4. Prove that 1/√2 is irrational.
Assume 1/√2 = r (rational, r ≠ 0). Then √2 = 1/r, which is rational. Contradiction. So 1/√2 is irrational.
5. Prove that 7√3 is irrational.
Assume 7√3 = r. Then √3 = r/7 is rational, which is false. So 7√3 is irrational.
6. Prove that √2 + √3 is irrational.
Assume √2 + √3 = r. Then √3 = r − √2. Square: 3 = r² − 2r√2 + 2, so √2 = (r² − 1)/(2r), which is rational. Contradiction. Hence √2 + √3 is irrational.
Common mistakes
- Forgetting to say a and b are co-prime. Without it there is no contradiction.
- Jumping from "2 divides a²" to the end without saying "so 2 divides a".
- Thinking every square root is irrational: √9 = 3 and √16 = 4 are rational.
- Saying irrational + irrational is always irrational: √2 + (−√2) = 0 is rational.