Revisiting Irrational Numbers

Build a perfectly square room exactly 1 metre on each side and stretch a tape across the diagonal — no matter how finely your tape is marked, you can never read the diagonal exactly. Its length is √2 metres, a number that refuses to be written as any ratio of whole numbers. Numbers like this are not mistakes or approximations; they are a whole different species called irrational numbers, and Class 10 asks you to prove that they exist.

Definition: A rational number is any number that can be written as p/q, where p and q are integers and q ≠ 0. Its decimal expansion is either terminating (like 0.75) or non-terminating but repeating (like 0.333…). Examples: 3, −7, 1/2, 11/13, 22/7.

Definition: An irrational number is a real number that cannot be written as p/q with integer p and q (q ≠ 0). Its decimal expansion is non-terminating AND non-repeating. Classic examples: √2, √3, √5, √7, π, e.

Definition: Proof by contradiction (reductio ad absurdum) is a method where we assume the statement we want to prove is false, follow the logic honestly, and reach an impossibility. Since the impossibility grew out of our assumption, the assumption itself must be wrong — so the original statement is true.

The lemma that powers every irrationality proof

Every irrationality proof in NCERT Class 10 rests on one small but mighty result that comes straight from the Fundamental Theorem of Arithmetic (every integer greater than 1 factorises into primes uniquely):

Theorem 1.3 (NCERT): If a prime number p divides a², then p divides a (where a is a positive integer).

Why is this true? If p divides a², then p must appear in the prime factorisation of a². But the primes of a² are exactly the primes of a, each appearing twice (paired up). So a prime sitting inside a² had to be sitting inside a in the first place. There is nowhere else for it to have come from.

You will use this lemma twice in a single proof — once to catch a factor in the numerator, and once more to catch the same factor in the denominator. Keep it at your fingertips: p prime and p | a² ⇒ p | a.

The strategy in plain words

We want to show √2 is irrational. Proving this directly is painful, because "irrational" means "cannot be written as p/q" — we would have to rule out every possible pair p, q. So we flip the script. We assume √2 is rational, write it as p/q in lowest terms, and then squeeze the equation until it breaks the very "lowest terms" condition we started with. That breakage is the contradiction. The only way out is to admit the first assumption was wrong.

Every such proof follows the same four moves:

1. Assume the number is rational: it equals p/q with integers p, q, q ≠ 0, in lowest terms (gcd(p, q) = 1).
2. Square or rearrange to link p, q and the prime you care about.
3. Use the lemma to show the prime divides both p and q.
4. Spot the contradiction — p and q were supposed to share no factor, yet they share one. Hence the assumption was false.

That last line is the punchline. The "lowest terms" opening is the cliff over which the contradiction throws the assumption. If you forget to state "in lowest terms" at the very start, the proof has nothing to contradict.

The proof that √2 is irrational, step by step

We use proof by contradiction.

Step 1 — Assume the opposite. Suppose √2 is rational. Then √2 = p/q, where p and q are integers, q ≠ 0, and the fraction is in its lowest terms, i.e. gcd(p, q) = 1.

Step 2 — Rearrange and square.
√2 = p/q ⇒ q√2 = p ⇒ squaring both sides, 2q² = p². …(i)

Step 3 — Deduce a factor of p. Equation (i) shows 2 divides p². By the lemma (2 is prime), 2 divides p. So write p = 2m for some integer m.

Step 4 — Substitute back. Put p = 2m into (i):
2q² = (2m)² = 4m² ⇒ q² = 2m². …(ii)

Step 5 — Deduce a factor of q. Equation (ii) shows 2 divides q², so by the same lemma, 2 divides q.

Step 6 — Reach the contradiction. From Steps 3 and 5, both p and q are divisible by 2, so 2 is a common factor of p and q. This contradicts our assumption that gcd(p, q) = 1 (lowest terms).

Step 7 — Conclusion. The assumption that √2 is rational must be false. Therefore √2 is irrational. Proved. ∎

Notice what we did not do. We did not compute √2 to many decimals and note it never repeats — that is only an observation, not a proof. We used pure logic: assume, derive, contradict. The very same template proves √3, √5, √7, and √p for any prime p.

The picture below places √2 exactly on the number line as the diagonal of a unit square — a length you can draw precisely but never write as a fraction.

Extending the idea: sums and products with irrationals

Once you know √2, √3 or √5 is irrational, you rarely need a fresh proof for expressions built from them. Two facts, both provable by the same contradiction trick, do the heavy lifting:

Fact A — rational + irrational = irrational. If their sum were rational, then irrational = rational − rational = rational, a contradiction.

Fact B — (non-zero rational) × irrational = irrational. If the product were rational, dividing by the non-zero rational would make the irrational rational — impossible.

These instantly tell you that 5 + √3, 2√7, and (3 − √5)/2 are all irrational. Do note the caution in Fact B: the rational multiplier must be non-zero, because 0 × √2 = 0 is rational. And the "closure" fails in general: √2 + (−√2) = 0 is rational, and √2 × √2 = 2 is rational, so irrationals are not closed under addition or multiplication. That is a favourite trap in assertion-reason questions.

Why irrational numbers matter

Irrational numbers are the rule, not the exception. Between any two rationals on the number line lie infinitely many irrationals. Without them you could not describe the diagonal of a unit square (√2), the circumference-to-diameter ratio of a circle (π), or the natural growth constant in compound interest (e). When the Pythagoreans first proved √2 irrational, it shattered their belief that "all numbers are ratios of whole numbers" — legend says the discoverer, Hippasus, was drowned for revealing it.

For your board exam, expect a direct 3-mark question ("Prove √2 is irrational" or "Prove √5 is irrational") and objective questions that lean on Facts A and B to settle expressions like 7 − 2√3 in a single line. Master the structure once and you unlock several marks across the Real Numbers chapter.

Common misconception: "My calculator shows √2 = 1.41421356, a finite decimal, so it is rational." The calculator merely truncates at its memory limit. The true expansion of √2 never terminates and never repeats — what you see is an approximation, not the number. Also beware: 22/7 is a rational approximation of π, but 22/7 ≠ π. And not all square roots are irrational — √4 = 2, √9 = 3 are perfectly rational. Only the square root of a non-perfect-square positive integer is irrational.

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Work through these examples. They begin with the core board proofs (√2, √3, √5) and build up to the multi-step twists NCERT loves — sums, products, assertion-reason, reading the figure, and the classic sign/zero traps. Keep the lemma handy: p prime and p | a² ⇒ p | a, and the two shortcuts: rational + irrational = irrational, non-zero rational × irrational = irrational.

Example 1 — Prove √2 is irrational (basic, board 3-marker)

Q: Prove that √2 is irrational.
Solve:
Assume √2 = p/q in lowest terms (gcd(p, q) = 1, q ≠ 0).
Then q√2 = p ⇒ 2q² = p², so 2 | p² ⇒ 2 | p. Write p = 2m.
Substitute: 2q² = 4m² ⇒ q² = 2m², so 2 | q² ⇒ 2 | q.
Now 2 divides both p and q, contradicting gcd(p, q) = 1.
Answer: The assumption fails, so √2 is irrational. ∎ (Sanity check: the contradiction lands precisely on the "lowest terms" line — exactly where it should.)

Example 2 — Prove √3 is irrational

Q: Prove that √3 is irrational.
Solve:
Assume √3 = p/q in lowest terms.
Squaring: 3 = p²/q² ⇒ 3q² = p², so 3 | p² ⇒ 3 | p (3 is prime). Write p = 3m.
Substitute: 3q² = 9m² ⇒ q² = 3m², so 3 | q² ⇒ 3 | q.
Then 3 divides both p and q — contradiction with gcd(p, q) = 1.
Answer: √3 is irrational. ∎ Note how only the prime changed from 2 to 3; the skeleton is identical.

Example 3 — Prove 5 − √3 is irrational (given √3 is irrational)

Q: Given that √3 is irrational, prove that 5 − √3 is irrational.
Solve:
Assume 5 − √3 is rational, say 5 − √3 = a/b (integers a, b; b ≠ 0).
Isolate the surd: √3 = 5 − a/b = (5b − a)/b.
Since a, b are integers, (5b − a)/b is a ratio of integers, i.e. a rational number.
So √3 would be rational — but it is given irrational. Contradiction.
Answer: 5 − √3 is irrational. ∎ (This is just Fact A: rational − irrational = irrational.)

Example 4 — Tricky: prove 3 + 2√5 is irrational

Q: Prove that 3 + 2√5 is irrational, given that √5 is irrational.
Solve:
Assume 3 + 2√5 is rational, say 3 + 2√5 = a/b (integers a, b; b ≠ 0).
Then 2√5 = a/b − 3 = (a − 3b)/b, so √5 = (a − 3b)/(2b).
The right side is a ratio of integers (denominator 2b ≠ 0), hence rational.
So √5 would be rational — contradicting the given fact. Contradiction.
Answer: 3 + 2√5 is irrational. ∎ Trap: you must divide by 2 as well as b; the multiplier 2 does not disturb rationality since it is a non-zero rational (Fact B in reverse).

Example 5 — Tricky: prove √5 is irrational, then use it for 1/√5

Q: Prove √5 is irrational, and hence show that 1/√5 is irrational.
Solve:
Part 1: Assume √5 = p/q in lowest terms. Then 5q² = p² ⇒ 5 | p² ⇒ 5 | p. Write p = 5m.
So 5q² = 25m² ⇒ q² = 5m² ⇒ 5 | q² ⇒ 5 | q. Then 5 | p and 5 | q — contradiction. Hence √5 is irrational.
Part 2: Rationalise: 1/√5 = (1/√5) × (√5/√5) = √5/5. Now 1/5 is a non-zero rational and √5 is irrational, so by Fact B, (1/5) × √5 = √5/5 is irrational.
Answer: 1/√5 = √5/5 is irrational. ∎ (Check: if 1/√5 were rational, its reciprocal √5 would be rational too — impossible.)

Example 6 — Tricky (assertion–reason): are irrationals closed under addition?

Q: Assertion (A): The sum of two irrational numbers is always irrational. Reason (R): √2 is irrational. Choose: (a) both true, R explains A; (b) both true, R does not explain A; (c) A false, R true; (d) A true, R false.
Solve:
Test A with a counter-example: √2 and (−√2) are both irrational, but √2 + (−√2) = 0, which is rational. So A is false. R (√2 is irrational) is a true statement.
Answer: (c) — A is false, R is true. The trap: "irrational + irrational" is NOT always irrational; only "rational + irrational" is guaranteed irrational.

Example 7 — Tricky (figure reading): is the labelled diagonal rational?

Q: In the number-line figure above, a unit square (side 1) sits on the line and its diagonal is swung down with a compass to mark a point P. Is the length OP rational or irrational? Justify using the figure.
Solve:
By Pythagoras on the unit square, diagonal² = 1² + 1² = 2, so the diagonal (and hence OP) = √2.
Since 2 is not a perfect square, √2 is the square root of a non-perfect-square integer, which is irrational (proved in Example 1). The compass only transfers this exact length onto the line; it does not make it a fraction.
Answer: OP = √2, which is irrational. The figure shows a length you can construct precisely yet never express as p/q. (Sanity check: OP ≈ 1.414 lies between the ticks 1 and 2, as the dashed arc lands.)

Example 8 — Tricky: spot the flaw / classify the number

Q: A student claims "0.101001000100001… is rational because it has a clear pattern, while 22/7 is irrational because it is π." Identify and correct both errors.
Solve:
Error 1: 0.101001000100001… has a pattern, but the pattern is non-repeating (the block of 0s keeps growing), and the decimal is non-terminating. Non-terminating AND non-repeating ⇒ it is irrational, not rational. "Has a pattern" is not the test; "repeating block" is.
Error 2: 22/7 is a ratio of two integers (7 ≠ 0), so 22/7 is rational by definition. It is only an approximation of π; 22/7 ≠ π, and π itself is irrational.
Answer: 0.101001000100001… is irrational; 22/7 is rational. Both of the student's claims are wrong. The rigorous criterion is "non-terminating AND non-repeating", never "looks messy" or "has a pattern".

What irrational numbers are and how to prove numbers like √2 are irrational.

• A number is irrational if it cannot be written as p/q where p, q are integers and q ≠ 0.
• Examples: √2, √3, √5, π, and non-terminating non-repeating decimals.
• To prove √2 is irrational we use proof by contradiction: assume √2 = p/q in lowest terms, then show 2 divides both p and q, contradicting the assumption.
• Theorem: if a prime p divides a², then p divides a.
• The sum or product of a rational and an irrational number is always irrational (e.g. 2 + √3, 5√2).
• √2 is irrational — a classic exam proof to remember.

Cover the answer, recall, then check. 7 cards on irrational numbers.

Q1. What is an irrational number?
A1. A number that cannot be written in the form p/q with integers p, q (q ≠ 0), e.g. √2, π.

Q2. What method is used to prove √2 is irrational?
A2. Proof by contradiction — assume it is rational and derive a contradiction.

Q3. State the theorem used in these proofs about primes.
A3. If a prime p divides a², then p also divides a.

Q4. Is 2 + √3 rational or irrational? Why?
A4. Irrational — the sum of a rational and an irrational number is always irrational.

Q5. Is the product 5 × √2 rational or irrational?
A5. Irrational — a non-zero rational times an irrational is irrational.

Q6. Give an example of a number whose decimal is non-terminating and non-repeating.
A6. √2 = 1.41421356... (its digits never terminate or repeat).

Q7. In proving √2 irrational, what contradiction appears?
A7. Both p and q turn out to be divisible by 2, contradicting that p/q was in lowest terms.