Construction of the Real Numbers

A real number is an equivalence class of Cauchy sequences of rationals. Lengthening the decimal expansion of 2\sqrt{2} marches rational terms into the hole — the march itself is the real number 2\sqrt{2}. Completeness is what this construction buys.

(xn)∼(yn)  ⟺  xn−yn→0(x_n) \sim (y_n) \iff x_n - y_n \to 0
2=[(1,1.4,1.41,1.414,… )]\sqrt{2} = [(1, 1.4, 1.41, 1.414, \dots)]

Part of these learning paths

Examples & Exercises

Let xn=⌊2⋅10n⌋10nx_n = \frac{\lfloor \sqrt{2}\cdot 10^n \rfloor}{10^n} be 2\sqrt{2} cut off after nn decimal places. Prove that (xn)(x_n) is a Cauchy sequence of rationals.

∣xm−xn∣<10−N(m,n≥N)|x_m - x_n| < 10^{-N} \quad (m, n \geq N)
Decimal truncations, n=4n = 4 — more digits pull the terms into the hole.
Show proof
  1. Each xnx_n is a fraction with denominator 10n10^n, hence rational. With tn=⌊2⋅10n⌋t_n = \lfloor \sqrt{2}\cdot 10^n \rfloor, for m≥nm \geq n we have tm≥10m−ntnt_m \geq 10^{m-n} t_n, so xm≥xnx_m \geq x_n — adding digits never decreases the value.
  2. Putting y=2⋅10ny = \sqrt{2}\cdot 10^n into ⌊y⌋≤y<⌊y⌋+1\lfloor y \rfloor \leq y < \lfloor y \rfloor + 1 gives xn≤2<xn+10−nx_n \leq \sqrt{2} < x_n + 10^{-n}. So the error satisfies ∣xn−2∣<10−n\lvert x_n - \sqrt{2}\rvert < 10^{-n}.
  3. For m≥nm \geq n we get 0≤xm−xn<10−n0 \leq x_m - x_n < 10^{-n} (combining xm≤2x_m \leq \sqrt{2} with the bound above). Hence ∣xm−xn∣<10−N→0\lvert x_m - x_n\rvert < 10^{-N} \to 0 whenever m,n≥Nm, n \geq N — (xn)(x_n) is a Cauchy sequence of rationals.

Prove that an=1+1na_n = 1 + \frac1n and bn=1b_n = 1 are different rational Cauchy sequences defining the same real number.

an−bn=1n→0⟹[(an)]=[(bn)]=1a_n - b_n = \frac1n \to 0 \quad\Longrightarrow\quad [(a_n)] = [(b_n)] = 1
The Babylonian sequence — a different sequence marching into the same hole 2\sqrt{2}.
Show proof
  1. Both (an)(a_n) and (bn)(b_n) are rational sequences. Their difference is an−bn=1n→0a_n - b_n = \frac1n \to 0, so (an)∼(bn)(a_n) \sim (b_n) by the definition of the equivalence relation.
  2. Equivalent Cauchy sequences define the same class, so [(an)]=[(bn)][(a_n)] = [(b_n)] — different on the surface, one real number underneath.
  3. The class of the constant sequence (1,1,1,… )(1, 1, 1, \dots) is the rational 11 sitting inside the reals. Hence [(an)]=1[(a_n)] = 1 — the thing (1+1n)(1+\frac1n) "converges to" is the real number 11.

Prove that the termwise sum (xn+yn)(x_n + y_n) of two rational Cauchy sequences is again Cauchy.

∣(xm+ym)−(xn+yn)∣≤∣xm−xn∣+∣ym−yn∣|(x_m+y_m)-(x_n+y_n)| \leq |x_m-x_n| + |y_m-y_n|
Decimal truncations, n=5n = 5 — sums of Cauchy sequences stay Cauchy, so addition is well-defined.
Show proof
  1. Let ε>0\varepsilon > 0. Since (xn)(x_n) is Cauchy, there is N1N_1 with ∣xm−xn∣<ε2\lvert x_m - x_n\rvert < \frac{\varepsilon}{2} for m,n≥N1m, n \geq N_1, and likewise N2N_2 for (yn)(y_n).
  2. By the triangle inequality, ∣(xm+ym)−(xn+yn)∣≤∣xm−xn∣+∣ym−yn∣\lvert (x_m+y_m) - (x_n+y_n)\rvert \leq \lvert x_m - x_n\rvert + \lvert y_m - y_n\rvert.
  3. With N=max⁡(N1,N2)N = \max(N_1, N_2), the right-hand side is below ε2+ε2=ε\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon for m,n≥Nm, n \geq N. So (xn+yn)(x_n + y_n) is Cauchy. Changing representatives to (xn′)∼(xn)(x'_n) \sim (x_n) still gives (xn+yn)−(xn′+yn)=(xn−xn′)→0(x_n+y_n) - (x'_n+y_n) = (x_n - x'_n) \to 0, hence the same class — addition is well-defined, independent of the chosen representatives.

2. Sequences