Irrational numbers — √2, π, e — proof that √2 is irrational: Worked examples

This page hunts down every kind of question the √2 irrationality topic can throw at you. We first lay out a scenario matrix — a checklist of every "case class" — then work an example for each cell. Nothing is left uncovered: perfect squares, non-square integers, the geometric version, the "close approximation" trap, and the exam twist that generalises to $3$ , $p$ , and beyond.

Everything here leans on three prerequisite ideas: Proof by Contradiction, Greatest Common Divisor (GCD), and Prime Factorization. If any word below feels new, its meaning is rebuilt in place.

The scenario matrix

Think of "prove/disprove irrationality" problems as living in a grid. Each row is a case class — a genuinely different situation the proof machinery has to survive. If we work at least one example per row, a reader can never meet a situation we didn't show.

#	Case class	What's special about it	Does the contradiction fire?	Example
A	Number is a non-square integer ( $2$ )	Prime factor appears an odd number of times	Yes → irrational	Ex 1
B	Number is a perfect square ( $4$ )	The parity chain closes consistently	No → rational	Ex 2
C	Odd prime under the root ( $3$ )	Parity → "divisible by 3"	Yes → irrational	Ex 3
D	Non-prime, non-square ( $6$ )	Uses prime factorization, not just "even"	Yes → irrational	Ex 4
E	Geometric / word problem (square tile diagonal)	Contradiction becomes visual scaling	Yes → irrational	Ex 5
F	Degenerate / boundary inputs ( $0$ , $1$ )	Smallest cases — must not misfire	No → rational	Ex 6
G	Limiting / approximation trap ( $\frac{1414}{1000}$ )	"Close to 2" is not "equals 2"	N/A — approximation ≠ equality	Ex 7
H	Exam twist: sum/combination ( $2 + 3$ , $1 + 2$ )	Rational ± irrational	Stays irrational	Ex 8
I	General theorem ( $n$ , $n$ not a perfect square)	One proof to rule them all	Yes → irrational	Ex 9

Row A — Non-square integer: $2$

Worked example Ex 1 — Re-run the classic, but

watch the primes Prove $2$ is irrational, tracking exactly where the number 2 forces the crash.

Forecast: Guess now — after we write $2 = \frac{p}{q}$ in lowest terms, how many times must the factor 2 divide $p$ ? And why does that number never balance?

Assume $2 = \frac{p}{q}$ with $g cd (p, q) = 1$ . Why this step? Proof by Contradiction needs a concrete thing to break; "lowest terms" (via Greatest Common Divisor (GCD)) means $p, q$ share no factor — our tripwire.
Square and clear: $2 q^{2} = p^{2}$ . Why this step? Turns a statement about $2$ into one about integers, where prime counting works.
Count the 2's. Let $a$ = number of times 2 divides $p$ , and $b$ = number of times 2 divides $q$ . Squaring doubles each: $p^{2}$ has $2 a$ twos, $q^{2}$ has $2 b$ twos. Why this step? This is the fingerprint idea — squaring can only give even counts.
Balance the equation $2 q^{2} = p^{2}$ : left side has $1 + 2 b$ twos, right side has $2 a$ twos. So $1 + 2 b = 2 a$ . Why this step? Both sides are the same integer, so their prime-2 counts must match.
Crash: $1 + 2 b$ is odd, $2 a$ is even. An odd number can never equal an even number. Why this step? This is the contradiction — sharper than "both even": it's "an odd equals an even."

Verify: Try any candidate, e.g. $p = 7, q = 5$ : $2 q^{2} = 50$ , $p^{2} = 49$ — off by one, never equal. Try $p = 99, q = 70$ : $2 q^{2} = 9800$ , $p^{2} = 9801$ . The eternal "off by 1" is the parity mismatch made visible. $□$

Row B — Perfect square: $4$ (the proof must not misfire)

Worked example Ex 2 — Why

$4$ escapes Show that running the same machinery on $4$ produces no contradiction.

Forecast: Guess where the odd-vs-even clash from Ex 1 gets replaced by an even-vs-even balance.

Assume $4 = \frac{p}{q}$ , $g cd (p, q) = 1$ , square: $4 q^{2} = p^{2}$ . Why? Same start as Ex 1 — fairness demands identical treatment.
Count 2's: left side has $2 + 2 b$ twos, right side has $2 a$ twos, so $2 + 2 b = 2 a$ . Why? Same fingerprint step. But note $2 + 2 b$ is now even.
No crash: $2 + 2 b = 2 a \Rightarrow a = b + 1$ . Perfectly solvable! e.g. $b = 0, a = 1$ . Why? The extra factor here is $4 = 2^{2}$ , contributing an even count of 2's, so both sides stay even.
Read off the answer: $p = 2 q$ , so $\frac{p}{q} = 2$ . Rational, as expected. Why? A consistent equation means √4 genuinely is the fraction $\frac{2}{1}$ .

Verify: $4 = 2 = \frac{2}{1}$ , and $2^{2} = 4$ . ✓ Cell B confirmed: perfect squares stay rational. (Case B)

Row C — Odd prime: $3$

Worked example Ex 3 — Parity becomes "divisibility by 3"

Prove $3$ is irrational.

Forecast: In Ex 1 the magic word was even. What word replaces it here?

Assume $3 = \frac{p}{q}$ , $g cd (p, q) = 1$ ; square: $3 q^{2} = p^{2}$ . Why? Same opening move.
$p^{2}$ is a multiple of 3, so $p$ is a multiple of 3. Why this step? Because 3 is prime: if $3 ∣ p^{2} = p \cdot p$ , then $3$ divides one of the factors, i.e. $3 ∣ p$ . (This prime property is exactly what Prime Factorization guarantees — it would fail for a composite like 4.)
Write $p = 3 m$ : then $3 q^{2} = 9 m^{2} \Rightarrow q^{2} = 3 m^{2}$ , so $3 ∣ q$ too. Why? Substitute and simplify — now $q$ inherits the factor 3.
Crash: $3 ∣ p$ and $3 ∣ q$ contradicts $g cd (p, q) = 1$ . Why? They share the factor 3, but we assumed lowest terms.

Verify: Closest fractions: $\frac{7}{4}$ gives $3 q^{2} = 48$ , $p^{2} = 49$ ; $\frac{26}{15}$ gives $3 \cdot 225 = 675$ , $2 6^{2} = 676$ . Always off by 1 again — never exact. $□$ (Case C)

Row D — Composite non-square: $6$

Row E — Geometric / word problem

Worked example Ex 5 — The tile-floor problem

A square floor tile has side exactly $1 m$ . A builder claims the diagonal can be cut as a whole number of "diagonal-units" that also fit a whole number of times along the side. Show this is impossible.

Forecast: By the Pythagorean Theorem the diagonal is $2 m$ . If a common unit fitted both, what would $2$ have to be?

A common unit fitting both means side $= q$ units and diagonal $= p$ units, whole numbers, with $g cd (p, q) = 1$ (else use a bigger unit). Why? "Fits a whole number of times" = integer lengths; smallest unit = lowest terms.
Pythagoras: $p^{2} = q^{2} + q^{2} = 2 q^{2}$ . Why? The diagonal of a square with side $q$ satisfies $p^{2} = q^{2} + q^{2}$ — the geometric source of the "2".
This is exactly Ex 1's equation, so $p, q$ must both be even — contradicting lowest terms. Why? Geometry has handed us the same integer equation; the parity crash follows.
Picture (figure): the red diagonal and mint side can never both be measured whole by one ruler; whatever unit you pick, the diagonal falls between marks forever.

Verify: Side $1$ , diagonal $2 = 1.41421 \dots$ Try unit $= 0.5$ : side $= 2$ units, diagonal $= 2.828 \dots$ units — not whole. No unit ever works. $□$ (Case E)

Row F — Degenerate / boundary inputs: $0$ , $1$

Row G — Limiting / approximation trap

Worked example Ex 7 — "

$\frac{1414}{1000}$ is basically √2" A student insists $2 = \frac{1414}{1000}$ because a calculator shows $1.414$ . Refute it, and explain why no finite decimal ever works.

Forecast: Guess what $(\frac{1414}{1000})^{2}$ actually equals — is it $2$ ?

Square the claim: $(\frac{1414}{1000})^{2} = \frac{1 999 396}{1 000 000} = 1.999396$ . Why? If it equalled $2$ , its square would be exactly $2$ . It isn't.
Any finite decimal $\frac{p}{1 0 ^{k}}$ is a fraction, hence rational; by Ex 1, $2$ isn't. So no such decimal can equal $2$ . Why? A terminating decimal is literally a fraction over a power of ten — a member of the Rational Numbers that $2$ is forbidden from joining.
Approximation ≠ equality. Rationals are dense on the Real Number Line — you can get arbitrarily close — but "close" leaves a nonzero gap for every finite fraction. Why? Density explains the feeling of closeness without granting equality.

Verify: $1.999396 \neq = 2$ , gap $= 0.000604$ . Better try $\frac{99}{70}$ : square $= \frac{9801}{4900} = 2.0002 \dots \neq = 2$ . Never exact. $□$ (Case G)

Row H — Exam twist: rational combined with irrational

Worked example Ex 8 — Prove

$1 + 2$ is irrational Show that adding the rational number $1$ to $2$ can't repair it into a fraction.

Forecast: Guess the trick — can we isolate $2$ and reach a contradiction with what we already proved?

Assume $1 + 2 = r$ is rational. Why? Proof by Contradiction again — suppose the thing we doubt is true.
Isolate: $2 = r - 1$ . Why? Subtract the rational $1$ from both sides.
Rational minus rational is rational. So $r - 1$ is rational, forcing $2$ to be rational. Why? The Rational Numbers are closed under subtraction: $\frac{a}{b} - \frac{c}{d} = \frac{a d - b c}{b d}$ .
Crash: but Ex 1 proved $2$ is irrational. Contradiction. Why? We reached a statement contradicting an established theorem.

Verify: $1 + 2 = 2.41421 \dots$ — non-terminating, non-repeating. If it were $\frac{a}{b}$ , then $2 = \frac{a - b}{b}$ would be too. Same argument kills $2 + 3$ , $52$ , $\frac{2}{4}$ . $□$ (Case H)

Row I — The general theorem

Worked example Ex 9 —

$n$ is irrational whenever $n$ is not a perfect square Prove the master result that generates rows A, C, D at once.

Forecast: Guess the single feature of $n$ 's Prime Factorization that decides everything.

Setup. Suppose $n = \frac{p}{q}$ , $g cd (p, q) = 1$ , so $n q^{2} = p^{2}$ .
Since $n$ is not a perfect square, some prime $π$ divides $n$ an odd number of times, say $2 t + 1$ . Why? A number is a perfect square iff every prime exponent is even — so "not a square" means at least one odd exponent.
Count that prime. Right side $p^{2}$ has an even count of $π$ . Left side $n q^{2}$ has $(2 t + 1) + (even)$ = odd count. Why? Squaring doubles counts (even); the odd exponent in $n$ tips the total to odd.
Crash: even $=$ odd, impossible. Why? Same fingerprint mismatch — now stated once, forever.

Verify: $n = 2, 3, 5, 6, 7, 8, \dots$ (all non-squares) are irrational; $n = 1, 4, 9, 16, \dots$ (squares) are rational. Spot check: $8 = 2^{3}$ has odd exponent 3 → $8 = 2.828 \dots$ irrational; $9 = 3^{2}$ all even → $9 = 3$ rational. $□$ (Case I)

Recall Self-test — one per row

A number is a perfect square iff every prime exponent is what? ::: Even For $3$ , the divisibility that replaces "even" is divisibility by ::: 3 Why does the $2$ proof NOT break $4$ ? ::: The extra factor $4 = 2^{2}$ contributes an even count of 2's, so both sides stay even — no clash Is $1 + 2$ rational or irrational? ::: Irrational (rational − rational would force $2$ rational) Does $(\frac{1414}{1000})^{2}$ equal 2? ::: No, it equals $1.999396$ Single feature of $n$ that makes $n$ irrational? ::: At least one prime appears an odd number of times