√2 is Irrational

Plotting the rationals p/qp/q with denominator at most NN on [1,2][1, 2] leaves a hole at 2≈1.4142\sqrt{2} \approx 1.4142 — no NN ever fills it. Proposition 4.4.4: no rational squares to 22.

(pq)2≠2∀ pq∈Q\left(\frac{p}{q}\right)^2 \ne 2 \quad \forall\, \frac{p}{q} \in \mathbb{Q}
∣1712−2∣≈0.00245>0\left|\frac{17}{12} - \sqrt{2}\right| \approx 0.00245 > 0

Examples & Exercises

Prove completely that no rational number p/qp/q squares to 22 — the equation x2=2x^2 = 2 has no rational solution (Tao, Proposition 4.4.4).

(pq)2=2 has no solution pq∈Q\left(\frac{p}{q}\right)^2 = 2 \text{ has no solution } \frac{p}{q} \in \mathbb{Q}
N=5N = 5: among rationals with denominator at most 55, the closest to 2\sqrt{2} is 7/5≈1.47/5 \approx 1.4 at distance about 0.01420.0142 — but the proof below shows this distance can never be 00.
Show proof
  1. Suppose a positive rational p/qp/q with (pq)2=2\left(\frac{p}{q}\right)^2 = 2 exists. By problem 2 below, we may take p/qp/q in lowest terms, i.e. gcd⁡(p,q)=1\gcd(p, q) = 1.
  2. Clearing denominators gives p2=2q2p^2 = 2q^2, so p2p^2 is even. If pp were odd then p2p^2 would be odd (odd × odd = odd), so pp must be even: p=2kp = 2k for a positive integer kk.
  3. Substituting gives 4k2=2q24k^2 = 2q^2, i.e. q2=2k2q^2 = 2k^2, so q2q^2 is even too — and the same argument makes qq even. But pp and qq both even means gcd⁡(p,q)≥2\gcd(p, q) \ge 2, contradicting the lowest-terms assumption.
  4. Hence no rational squares to 22. However close a square gets to 22 — like 75\frac{7}{5} or 1712\frac{17}{12} (problem 3) — no fraction lands exactly on 22: the rational line has a hole where 2\sqrt{2} should be.

Prove completely, by the well-ordering principle, that every positive rational has a lowest-terms representation.

∀ r>0 rational, ∃ p,q∈N: r=pq, gcd⁡(p,q)=1\forall\, r > 0\ \text{rational},\ \exists\, p, q \in \mathbb{N}:\ r = \frac{p}{q},\ \gcd(p, q) = 1
N=5N = 5: every dot is drawn in lowest terms (e.g. 3/23/2, never 6/46/4) — of all fractions naming the same value, the one with the smallest denominator is chosen as the representative.
Show proof
  1. Fix a positive rational rr. Let S={n∈N:r=m/n for some m∈N}S = \{n \in \mathbb{N} : r = m/n \text{ for some } m \in \mathbb{N}\}; since r=a/br = a/b for positive integers a,ba, b, we have b∈Sb \in S — so SS is a nonempty subset of N\mathbb{N}. (Here N={1,2,3,… }\mathbb{N} = \{1, 2, 3, \dots\}.)
  2. By the well-ordering principle, SS has a least element q0q_0. Pick p0∈Np_0 \in \mathbb{N} with r=p0/q0r = p_0/q_0. We now show gcd⁡(p0,q0)=1\gcd(p_0, q_0) = 1.
  3. If d=gcd⁡(p0,q0)>1d = \gcd(p_0, q_0) > 1, then r=p0/dq0/dr = \frac{p_0/d}{q_0/d} with q0/d<q0q_0/d < q_0, so q0/d∈Sq_0/d \in S — contradicting the minimality of q0q_0. Hence d=1d = 1, and r=p0/q0r = p_0/q_0 is in lowest terms.
  4. This justifies the "we may take lowest terms" step in problem 1: if some rational squared to 22, we could look at one with the smallest denominator — and it is exactly that minimal-denominator fraction that falls into the even/even contradiction.

Verify by direct computation that (75)2=4925<2<289144=(1712)2\left(\frac{7}{5}\right)^2 = \frac{49}{25} < 2 < \frac{289}{144} = \left(\frac{17}{12}\right)^2, and show the two fractions differ by 160<0.06\frac{1}{60} < 0.06 — numerical evidence that rationals get arbitrarily close to the 2\sqrt{2} slot.

(75)2=4925<2<289144=(1712)2\left(\frac{7}{5}\right)^2 = \frac{49}{25} < 2 < \frac{289}{144} = \left(\frac{17}{12}\right)^2
1712−75=160≈0.0167<0.06\frac{17}{12} - \frac{7}{5} = \frac{1}{60} \approx 0.0167 < 0.06
N=12N = 12: the closest rational to 2\sqrt{2} is 17/12≈1.4166717/12 \approx 1.41667 (distance about 0.002450.00245), and 7/5≈1.47/5 \approx 1.4 is drawn too — the two fractions squeeze 22 from both sides.
Show proof
  1. Compute directly: (75)2=4925=1.96<2\left(\frac{7}{5}\right)^2 = \frac{49}{25} = 1.96 < 2 and (1712)2=289144=2.0069…>2\left(\frac{17}{12}\right)^2 = \frac{289}{144} = 2.0069\ldots > 2 (verified exactly by integer comparison: 49<50=2×2549 < 50 = 2 \times 25 gives 4925<2\frac{49}{25} < 2, and 289>288=2×144289 > 288 = 2 \times 144 gives 289144>2\frac{289}{144} > 2).
  2. The fractions differ by 1712−75=85−8460=160≈0.0167<0.06\frac{17}{12} - \frac{7}{5} = \frac{85 - 84}{60} = \frac{1}{60} \approx 0.0167 < 0.06. Their squares differ by 289144−4925=7225−70563600=1693600<0.05\frac{289}{144} - \frac{49}{25} = \frac{7225 - 7056}{3600} = \frac{169}{3600} < 0.05 — both squares sit inside an interval of width under 0.050.05 straddling 22.
  3. By problem 1, no rational in this interval squares to exactly 22. Yet rationals squeezing 22 from both sides, like 75\frac{7}{5} and 1712\frac{17}{12}, are easy to find (in the model the shortest distance shrinks 0.0142→0.00245→⋯0.0142 \to 0.00245 \to \cdots as NN grows) — the rationals get as close as one likes to the 2\sqrt{2} slot, but never touch it. The number filling this hole is the 2\sqrt{2} whose existence the least-upper-bound axiom guarantees (the n-th root scene).

1. Number Systems