The irrationality of √2

Theorem

The irrationality of √2

T-004Home MU-101Threads logic · structure

Statement

Let \( p, q \in \mathbb{Z} \) with \( q \neq 0 \). Then \( \left(\dfrac{p}{q}\right)^2 \neq 2 \). Equivalently: there is no rational number \( x \in \mathbb{Q} \) with \( x^2 = 2 \). In particular the real number \( \sqrt{2} \) — the unique positive real with \( \sqrt{2}^2 = 2 \), whose existence is guaranteed by the completeness of \( \mathbb{R} \) — is irrational: \( \sqrt{2} \in \mathbb{R} \setminus \mathbb{Q} \).

Why it matters

This is the first genuine impossibility theorem most mathematicians meet: it asserts that no fraction, out of infinitely many candidates, ever squares to 2 — a claim that cannot be settled by checking cases, only by proof. Historically it shattered the Pythagorean doctrine that all magnitudes are commensurable, and forced the eventual construction of the real numbers. It is the canonical showcase of proof by contradiction and of infinite descent (equivalently, the well-ordering of \( \mathbb{N} \)).

Structurally, it is the seed of a whole family of results: \( \sqrt{n} \) is irrational for every non-square \( n \), \( x^2 - 2 \) is irreducible over \( \mathbb{Q} \), and \( \mathbb{Q}(\sqrt{2}) \) is a degree-2 field extension. Every argument in algebraic number theory that separates "rational" from "algebraic of higher degree" is, at bottom, a descendant of this one.

Hypotheses

Hypothesis. \( p \) and \( q \) are integers.Drop integrality and the claim is false: with real "numerator" \( p = \sqrt{2} \), \( q = 1 \), the ratio \( p/q \) squares to 2. The whole proof lives in the arithmetic of \( \mathbb{Z} \) (parity, divisibility); it says nothing about ratios of arbitrary reals. Hypothesis. \( q \neq 0 \).With \( q = 0 \) the expression \( p/q \) is undefined; there is no element of \( \mathbb{Q} \) to talk about. (This is a non-degeneracy condition, not a loophole: \( \mathbb{Q} \) is by definition the set of ratios with nonzero denominator.) Hypothesis. The radicand is \( 2 \) — a positive integer that is not a perfect square.Replace 2 by 4 and the statement fails: \( (2/1)^2 = 4 \). Trace the proof with 4 to see exactly where it dies: \( p^2 = 4q^2 \) gives \( p = 2k \), then \( 4k^2 = 4q^2 \), i.e. \( k^2 = q^2 \) — no forced descent, no contradiction. Hypothesis (implicit). The ambient ring is \( \mathbb{Z} \), where \( 2 \) is prime (Euclid's lemma holds for 2).The result is not a consequence of the field axioms alone: in \( \mathbb{F}_7 \) we have \( 3^2 = 9 = 2 \), so "\( \sqrt{2} \)" exists there; and in the ring \( \mathbb{Z}[\sqrt{2}] \) the element 2 factors as \( \sqrt{2}\cdot\sqrt{2} \). The proof genuinely uses the specific arithmetic of \( \mathbb{Z} \): that \( 2 \mid ab \) forces \( 2 \mid a \) or \( 2 \mid b \).

Proof

We first isolate the one lemma the proof needs, then run the classical descent. Nothing is used beyond the ring axioms of \( \mathbb{Z} \), the parity lemma proved in Step 2, and the well-ordering principle.

Suppose, for contradiction, that there exist \( p, q \in \mathbb{Z} \), \( q \neq 0 \), with \[ \left(\frac{p}{q}\right)^2 = 2 . \]

Proof by contradiction: we assume the negation of the theorem and derive an absurdity. Replacing \( p, q \) by \( |p|, |q| \) (which leaves the square unchanged), we may and do assume \( p \geq 0 \) and \( q \geq 1 \). A

Lemma (parity of squares). For \( m \in \mathbb{Z} \): if \( m^2 \) is even, then \( m \) is even.
Proof of lemma. Contrapositive: if \( m \) is odd, write \( m = 2t + 1 \) with \( t \in \mathbb{Z} \); then \[ m^2 = 4t^2 + 4t + 1 = 2\left(2t^2 + 2t\right) + 1, \] which is odd. Hence \( m^2 \) even \( \Rightarrow \) \( m \) even. \( \square \)

Every integer is even or odd (division algorithm with divisor 2 — existence and uniqueness of remainder \( r \in \{0, 1\} \)); the displayed computation is ring arithmetic in \( \mathbb{Z} \). This lemma is the special case of Euclid's lemma for the prime 2, proved here from scratch so the argument is self-contained. B

Among all pairs \( (p, q) \) with \( q \geq 1 \) and \( p^2 = 2q^2 \)-type witnesses (i.e. \( (p/q)^2 = 2 \)), choose one with \( q \) minimal. The set \[ S = \left\{\, q \in \mathbb{N} : q \geq 1,\ \exists\, p \in \mathbb{Z},\ \left(\tfrac{p}{q}\right)^2 = 2 \,\right\} \] is nonempty by Step 1, so by the well-ordering principle it has a least element; fix such a \( q \) and a corresponding \( p \geq 0 \).

Well-ordering principle: every nonempty subset of \( \mathbb{N} \) has a least element. This is the honest justification of the usual phrase "in lowest terms" — writing a fraction in lowest terms is exactly an appeal to minimality, and skipping this step is the most common gap in student write-ups. (Equivalently one may divide \( p \) and \( q \) by \( \gcd(p,q) \) and demand \( \gcd(p,q) = 1 \); the two devices are interchangeable, and we will use the minimality form.) C

\[ p^2 = 2q^2 . \]

Multiply both sides of \( p^2/q^2 = 2 \) by \( q^2 \neq 0 \); legal in the field \( \mathbb{Q} \), and both sides are integers. A

\( p^2 \) is even, hence \( p \) is even: write \( p = 2k \) with \( k \in \mathbb{Z} \), \( k \geq 0 \).

\( p^2 = 2q^2 \) exhibits \( p^2 \) as \( 2 \times (\text{integer}) \), the definition of even; the Lemma of Step 2 then forces \( p \) even. This is the key pivot of the whole proof. B

\[ (2k)^2 = 2q^2 \ \Longrightarrow\ 4k^2 = 2q^2 \ \Longrightarrow\ q^2 = 2k^2 . \]

Substitute \( p = 2k \) into Step 4 and cancel the factor 2; cancellation is valid because \( \mathbb{Z} \) is an integral domain and \( 2 \neq 0 \). A

\( q^2 \) is even, hence \( q \) is even: write \( q = 2\ell \) with \( \ell \in \mathbb{Z} \), and note \( \ell \geq 1 \) since \( q \geq 1 \) — indeed \( 1 \leq \ell \lt q \).

Step 6 shows \( q^2 = 2k^2 \) is even; the Lemma of Step 2 applies again. The strict inequality \( \ell = q/2 \lt q \) holds because \( q \geq 1 \) forces \( q \geq 2 \) once \( q \) is even. B

From \( q^2 = 2k^2 \) and \( q = 2\ell \): \[ 4\ell^2 = 2k^2 \ \Longrightarrow\ k^2 = 2\ell^2 \ \Longrightarrow\ \left(\frac{k}{\ell}\right)^2 = 2 , \] so \( \ell \in S \) with \( \ell \lt q \) — contradicting the minimality of \( q \) chosen in Step 3.

This is the descent: from any witness we manufactured a strictly smaller one, which the well-ordering principle forbids. (In the \( \gcd \) formulation: Steps 5 and 7 give \( 2 \mid p \) and \( 2 \mid q \), so \( 2 \mid \gcd(p,q) = 1 \), absurd.) Hence the assumption of Step 1 is untenable, and no rational squares to 2. \( \blacksquare \) C

Result

\[ \forall\, p, q \in \mathbb{Z},\ q \neq 0:\quad \left(\frac{p}{q}\right)^2 \neq 2 \qquad\text{i.e.}\qquad \sqrt{2} \notin \mathbb{Q}. \]

Reading. No fraction — however large its numerator and denominator, however cleverly chosen — squares exactly to 2. Rationals can square to values as close to 2 as you like (e.g. \( 99/70 \)), but never hit it. "Irrational" does not mean \( \sqrt{2} \) fails to exist; it means \( \sqrt{2} \) exists in \( \mathbb{R} \) but lies outside \( \mathbb{Q} \).

Scope. A statement about the field \( \mathbb{Q} \) built on the arithmetic of \( \mathbb{Z} \). It does not by itself prove that a real square root of 2 exists — that requires the completeness of \( \mathbb{R} \) (supremum axiom or intermediate value theorem). The same proof template covers \( \sqrt{n} \) for any \( n \) whose prime factorization contains some prime to an odd power, but it says nothing about fields where 2 is a square, such as \( \mathbb{F}_7 \) or \( \mathbb{R} \) itself.

Corollaries & converses

Irrational numbers exist: \( \mathbb{Q} \subsetneq \mathbb{R} \). The diagonal of a unit square is incommensurable with its side.
For every \( r \in \mathbb{Q} \), both \( r + \sqrt{2} \) and (for \( r \neq 0 \)) \( r\sqrt{2} \) are irrational — otherwise field operations in \( \mathbb{Q} \) would solve for \( \sqrt{2} \) rationally.
The polynomial \( x^2 - 2 \) has no root in \( \mathbb{Q} \), hence (being quadratic) is irreducible over \( \mathbb{Q} \); consequently \( \left[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}\right] = 2 \).
Between any two rationals lies an irrational (translate and scale \( \sqrt{2} \)): the irrationals are dense in \( \mathbb{R} \).
Converse-type statements. "\( x \) irrational \( \Rightarrow x^2 \) irrational" is false: \( x = \sqrt{2} \) itself has \( x^2 = 2 \in \mathbb{Q} \). The correct two-way statement is: for \( n \in \mathbb{N} \), \( \sqrt{n} \in \mathbb{Q} \iff n \) is a perfect square (the forward direction is Problem 4 below; the reverse is trivial: \( n = m^2 \Rightarrow \sqrt{n} = m \)).
Sum/product warnings: irrational + irrational can be rational \( \left(\sqrt{2} + (1 - \sqrt{2}) = 1\right) \), and irrational \( \times \) irrational can be rational \( \left(\sqrt{2}\cdot\sqrt{2} = 2\right) \). Irrationality is not closed under arithmetic.

Fails without

Radicand a perfect square. For \( n = 4 \): \( (2/1)^2 = 4 \). The descent machinery stalls: \( p^2 = 4q^2 \Rightarrow p = 2k \Rightarrow k^2 = q^2 \), which is perfectly consistent \( (k = q) \). Perfect squares escape because the prime 2 appears to an even power.
Integrality of \( p, q \). Allow \( p \in \mathbb{R} \): then \( p = \sqrt{2}, q = 1 \) satisfies \( (p/q)^2 = 2 \). The theorem is a statement about \( \mathbb{Z} \)-arithmetic, not about ratios in general.
The ambient field being \( \mathbb{Q} \). In \( \mathbb{F}_7 \), \( 3^2 = 2 \); in \( \mathbb{F}_{17} \), \( 6^2 = 36 = 2 \). "2 has no square root" is false in infinitely many fields, so no proof from field axioms alone can exist — any correct proof must invoke the specific arithmetic (order, parity, primality) of \( \mathbb{Z} \).
Euclid's lemma for the prime 2 (Step 2's lemma). The inference \( 2 \mid m^2 \Rightarrow 2 \mid m \) uses that 2 is prime. Run the same skeleton with the composite modulus 4: from \( 4 \mid m^2 \) one may not conclude \( 4 \mid m \) (take \( m = 2 \)); arguments that borrow the skeleton without re-proving the divisibility lemma for their own modulus are unsound.

Common errors

The unjustified "lowest terms". Writing "let \( \sqrt{2} = p/q \) in lowest terms" without knowing why lowest terms exist. The existence of a reduced representative is a theorem, proved from well-ordering (or from existence of \( \gcd \)); at degree level it must be cited, not assumed silently.
Asserting \( p^2 \) even \( \Rightarrow p \) even with no proof. The parity lemma is the load-bearing step; it needs the contrapositive computation \( (2t+1)^2 = 2(2t^2+2t)+1 \) or an appeal to Euclid's lemma. "Obvious" is not a justification.
Wrong direction of the parity lemma. Using "\( p \) even \( \Rightarrow p^2 \) even" (true but useless here) in place of the converse-form statement actually needed.
Concluding non-existence. Ending with "therefore \( \sqrt{2} \) does not exist." The proof shows \( \sqrt{2} \notin \mathbb{Q} \); existence in \( \mathbb{R} \) is a separate (completeness) theorem.
Contradiction hygiene. Declaring "contradiction!" after Steps 5 and 7 without stating what is contradicted (minimality of \( q \), or \( \gcd(p,q) = 1 \)). Both \( p \) and \( q \) being even contradicts nothing unless reducedness was fixed beforehand.
Blind generalization. Applying the identical script to \( \sqrt{4} \) and "proving" it irrational. Sanity-check any adaptation on a perfect square: the proof must visibly break there.
Circular overkill. Invoking uniqueness of prime factorization (FTA) in a course that will later prove FTA using this very style of descent. The proof above needs only parity, which is safely upstream of FTA.

Discussion

The discovery is attributed to the Pythagorean school, traditionally to Hippasus of Metapontum (5th century BCE), and legend — almost certainly embellished — has him drowned for revealing it. The original form was geometric: the diagonal and side of a square admit no common measuring unit. Euclid's Elements transmits the arithmetic proof (essentially the one above) and Book X develops the general theory of incommensurables; Plato's Theaetetus reports Theodorus proving the irrationality of \( \sqrt{3}, \sqrt{5}, \ldots, \sqrt{17} \) case by case, which suggests the general criterion was not yet available — a vivid snapshot of mathematics mid-crystallization.

Logically the theorem is a first encounter with two permanent tools. First, proof by contradiction with descent: assume a solution, manufacture a strictly smaller one, and let the well-ordering of \( \mathbb{N} \) deliver the absurdity. Fermat later industrialized this as "infinite descent" to prove, e.g., that \( x^4 + y^4 = z^4 \) has no positive solutions. Second, the local-to-impossible pattern: a single well-chosen invariant (here, parity — more precisely the exponent of 2) obstructs all infinitely many candidates at once. Modern number theory is largely the systematic study of such invariants.

Algebraically, the theorem says \( x^2 - 2 \) is irreducible over \( \mathbb{Q} \) (a quadratic with no rational root), so \( \mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\} \) is a field of degree 2 over \( \mathbb{Q} \) — the simplest nontrivial field extension, and the entry point to Galois theory. Analytically, it forces the question the Greeks could not answer: if \( \sqrt{2} \) is not a ratio, what is it? Dedekind's cuts (1872) answer by defining \( \sqrt{2} \) as the cut \( \left(\{x \in \mathbb{Q} : x \leq 0 \text{ or } x^2 \lt 2\},\ \{x \in \mathbb{Q} : x \gt 0 \text{ and } x^2 \gt 2\}\right) \); the theorem is exactly the statement that this cut is not generated by any rational point.

The slickest modern proof is valuation-theoretic and shows why the result is really about parity of exponents. Let \( v_2 : \mathbb{Q}^\times \to \mathbb{Z} \) be the 2-adic valuation, \( v_2(2^a m/n) = a \) for odd \( m, n \); it satisfies \( v_2(xy) = v_2(x) + v_2(y) \). If \( x \in \mathbb{Q}^\times \) had \( x^2 = 2 \), then \( 2v_2(x) = v_2(2) = 1 \), impossible since the left side is even and the right side odd. One line — and it generalizes instantly: \( \sqrt{n} \in \mathbb{Q} \iff v_p(n) \) is even for every prime \( p \), i.e. \( n \) is a perfect square. The same statement shows \( 2 \) is not a square in the field \( \mathbb{Q}_2 \) of 2-adic numbers, since \( v_2 \) extends there; irrationality of \( \sqrt{2} \) is thus a "local" fact at the prime 2. Common misconceptions. (i) Irrationality is not non-existence: \( \sqrt{2} \) is a perfectly definite real number, approximately \( 1.41421356 \). (ii) A non-repeating decimal expansion is a consequence of irrationality (via the theorem that eventually periodic expansions are exactly the rationals), not its definition and not something the proof above establishes directly. (iii) The proof is not "about" the symbol \( \sqrt{2} \); it is a theorem of pure integer arithmetic — \( p^2 = 2q^2 \) has no solution with \( q \neq 0 \) — and would stand even before any real numbers are constructed.

Worked examples

Example 1. Show that \( 3 - 5\sqrt{2} \) is irrational.

Suppose for contradiction \( r := 3 - 5\sqrt{2} \in \mathbb{Q} \).

Proof by contradiction; \( r \) is a well-defined real number since \( \sqrt{2} \in \mathbb{R} \). A

\[ \sqrt{2} = \frac{3 - r}{5} . \]

Solve the defining equation for \( \sqrt{2} \): subtract \( r \) from 3 and divide by \( 5 \neq 0 \). Both operations are field operations in \( \mathbb{R} \). A

\( 3, r, 5 \in \mathbb{Q} \) and \( \mathbb{Q} \) is a field, so \( \dfrac{3 - r}{5} \in \mathbb{Q} \); hence \( \sqrt{2} \in \mathbb{Q} \).

Closure of \( \mathbb{Q} \) under subtraction and division by a nonzero element — the field axioms for \( \mathbb{Q} \). B

This contradicts T-004. Hence \( r \notin \mathbb{Q} \).

Direct application of the theorem: \( \sqrt{2} \notin \mathbb{Q} \). A

\[ 3 - 5\sqrt{2} \ \notin\ \mathbb{Q}. \]

Reading. A single irrational contaminates any nontrivial rational-affine combination: for \( a, b \in \mathbb{Q} \), \( b \neq 0 \), the number \( a + b\sqrt{2} \) is always irrational, by the same three moves.

Example 2. Show that there exist irrational numbers \( a, b \) such that \( a^b \) is rational.

Consider \( c = \sqrt{2}^{\sqrt{2}} \). Exactly one of two cases holds: \( c \in \mathbb{Q} \) or \( c \notin \mathbb{Q} \).

\( c \) is a well-defined positive real (\( x^y := e^{y \ln x} \) for \( x \gt 0 \)); the case split is the law of excluded middle. B

Case 1: \( c \in \mathbb{Q} \). Take \( a = b = \sqrt{2} \). Then \( a, b \) are irrational and \( a^b = c \in \mathbb{Q} \).

Irrationality of \( a \) and \( b \) is exactly T-004. A

Case 2: \( c \notin \mathbb{Q} \). Take \( a = c \), \( b = \sqrt{2} \). Then \[ a^b = \left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2}\cdot\sqrt{2}} = \sqrt{2}^{\,2} = 2 \in \mathbb{Q}. \]

The exponent law \( (x^y)^z = x^{yz} \) for \( x \gt 0 \) (valid for the real exponential); \( a \) is irrational by the case hypothesis, \( b \) by T-004. B

In either case a valid pair \( (a, b) \) exists — even though the argument does not tell us which case obtains.

A classic non-constructive existence proof (rejected by intuitionists precisely for its use of excluded middle without a witness). In fact the Gelfond–Schneider theorem settles it: \( \sqrt{2}^{\sqrt{2}} \) is transcendental, so Case 2 is the true one — but T-004 alone already suffices for existence. C

\[ \exists\, a, b \in \mathbb{R} \setminus \mathbb{Q} \ \text{ with }\ a^b \in \mathbb{Q}. \]

Reading. Irrationality is spectacularly non-closed: even exponentiation of irrationals can land back in \( \mathbb{Q} \). Both branches of the proof lean on T-004 to certify their irrational inputs.

Problems

Prove that \( \sqrt{3} \) is irrational. (Identify precisely which lemma replaces the parity lemma.)
Solution

Lemma. If \( 3 \mid m^2 \) then \( 3 \mid m \). Proof: every integer is \( m \equiv 0, 1, 2 \pmod 3 \); then \( m^2 \equiv 0, 1, 4 \equiv 1 \pmod 3 \) respectively. So \( m^2 \equiv 0 \pmod 3 \) occurs only when \( m \equiv 0 \pmod 3 \). \( \square \)

Suppose \( \sqrt{3} = p/q \) with \( p, q \in \mathbb{Z} \), \( q \geq 1 \) minimal. Then \( p^2 = 3q^2 \), so \( 3 \mid p^2 \), and the lemma gives \( p = 3k \). Substituting: \( 9k^2 = 3q^2 \), so \( q^2 = 3k^2 \), so \( 3 \mid q^2 \), so \( q = 3\ell \) with \( 1 \leq \ell \lt q \). Then \( 9\ell^2 = 3k^2 \) gives \( k^2 = 3\ell^2 \), i.e. \( (k/\ell)^2 = 3 \) with \( \ell \lt q \), contradicting minimality of \( q \). Hence \( \sqrt{3} \notin \mathbb{Q} \). The parity lemma has been replaced by Euclid's lemma for the prime 3, proved by exhausting residues mod 3. \( \blacksquare \)
Prove that \( \sqrt[3]{2} \) is irrational.
Solution

Lemma (parity of cubes). If \( m^3 \) is even then \( m \) is even. Proof: contrapositive — if \( m = 2t + 1 \) then \( m^3 = 8t^3 + 12t^2 + 6t + 1 = 2(4t^3 + 6t^2 + 3t) + 1 \), odd. \( \square \)

Suppose \( \sqrt[3]{2} = p/q \), \( q \geq 1 \) minimal, so \( p^3 = 2q^3 \). Then \( p^3 \) is even, so \( p = 2k \) by the lemma; substituting, \( 8k^3 = 2q^3 \), so \( q^3 = 4k^3 = 2(2k^3) \) is even, so \( q = 2\ell \) with \( \ell \lt q \). Then \( 8\ell^3 = 4k^3 \) gives \( k^3 = 2\ell^3 \), a strictly smaller witness — contradiction with minimality. Hence \( \sqrt[3]{2} \notin \mathbb{Q} \). (Note the descent survives the change of exponent because the exponent 3 of the cube is still incompatible with the single factor 2 on the right: in valuation language, \( 3 v_2(p) = 1 + 3 v_2(q) \) has no integer solution since the left side is \( \equiv 0 \) and the right \( \equiv 1 \pmod 3 \).) \( \blacksquare \)
Prove that \( \sqrt{2} + \sqrt{3} \) is irrational. (You may use Problem 1.)
Solution

First, \( \sqrt{6} \) is irrational: if \( p^2 = 6q^2 \) with \( q \geq 1 \) and \( \gcd(p, q) = 1 \), then \( p^2 \) is even, so \( p = 2k \) (parity lemma, T-004 Step 2); then \( 4k^2 = 6q^2 \), so \( 2k^2 = 3q^2 \), so \( q^2 \) is even (since \( 2 \mid 3q^2 \) and \( 2 \nmid 3 \), Euclid's lemma for 2 gives \( 2 \mid q^2 \)), so \( q \) is even — contradicting \( \gcd(p, q) = 1 \).

Now suppose \( s = \sqrt{2} + \sqrt{3} \in \mathbb{Q} \). Then \[ s^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}, \qquad\text{so}\qquad \sqrt{6} = \frac{s^2 - 5}{2}. \] The right side is rational (field operations on rationals), contradicting the first paragraph. Hence \( \sqrt{2} + \sqrt{3} \notin \mathbb{Q} \). \( \blacksquare \)
Prove: for \( n \in \mathbb{N} \), if \( \sqrt{n} \in \mathbb{Q} \) then \( \sqrt{n} \in \mathbb{N} \). Deduce that \( \sqrt{n} \) is irrational for every \( n \) that is not a perfect square.
Solution

Suppose \( \sqrt{n} = p/q \) with \( p, q \in \mathbb{N} \), \( q \geq 1 \), \( \gcd(p, q) = 1 \). Then \( p^2 = n q^2 \), so \( q \mid p^2 \).

Claim: \( \gcd(p, q) = 1 \Rightarrow \gcd(p^2, q) = 1 \). Proof: if some prime \( r \) divides both \( q \) and \( p^2 \), then by Euclid's lemma \( r \mid p \), so \( r \mid \gcd(p, q) = 1 \), absurd. (Euclid's lemma: if a prime divides a product it divides a factor — provable from Bézout's identity, itself a consequence of well-ordering.) \( \square \)

Now \( q \mid p^2 \) means \( q \) is a common divisor of \( p^2 \) and \( q \), so \( q \mid \gcd(p^2, q) = 1 \), forcing \( q = 1 \). Hence \( \sqrt{n} = p \in \mathbb{N} \), and \( n = p^2 \) is a perfect square.

Contrapositive: if \( n \) is not a perfect square, \( \sqrt{n} \notin \mathbb{Q} \). This recovers T-004 (\( n = 2 \)), Problem 1 (\( n = 3 \)), and all of Theodorus's cases at once. \( \blacksquare \)
(Harder.) Let \( \mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a, b \in \mathbb{Q}\} \). Prove that \( \sqrt{3} \notin \mathbb{Q}(\sqrt{2}) \). (Hence \( \mathbb{Q}(\sqrt{2}) \) and \( \mathbb{Q}(\sqrt{3}) \) are distinct quadratic extensions of \( \mathbb{Q} \).)
Solution

Suppose \( \sqrt{3} = a + b\sqrt{2} \) with \( a, b \in \mathbb{Q} \). Squaring, \[ 3 = a^2 + 2b^2 + 2ab\sqrt{2}. \] Rearranging: \( 2ab\sqrt{2} = 3 - a^2 - 2b^2 \in \mathbb{Q} \). We split on whether the coefficient \( 2ab \) vanishes.

Case \( ab \neq 0 \). Then \( \sqrt{2} = \dfrac{3 - a^2 - 2b^2}{2ab} \in \mathbb{Q} \), contradicting T-004.

Case \( b = 0 \). Then \( \sqrt{3} = a \in \mathbb{Q} \), contradicting Problem 1.

Case \( a = 0 \). Then \( 3 = 2b^2 \). Write \( b = p/q \) in lowest terms: \( 3q^2 = 2p^2 \). The right side is even, so \( 3q^2 \) is even; since \( 2 \nmid 3 \), Euclid's lemma for 2 gives \( 2 \mid q^2 \), hence \( 2 \mid q \) (parity lemma). Write \( q = 2\ell \): \( 12\ell^2 = 2p^2 \), so \( p^2 = 6\ell^2 \) is even, so \( 2 \mid p \) — contradicting \( \gcd(p, q) = 1 \). (Equivalently: \( v_2(3q^2) \) is even while \( v_2(2p^2) \) is odd.)

All three cases are impossible, so \( \sqrt{3} \notin \mathbb{Q}(\sqrt{2}) \). Consequently \( x^2 - 3 \) remains irreducible over \( \mathbb{Q}(\sqrt{2}) \), and \( \left[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}\right] = 4 \) — the first step into the lattice of multiquadratic extensions and their Galois theory. \( \blacksquare \)