## Table of Contents

## Exercise 3.2

*Medium.* Assuming the eventual periodicity of [the continued fractions \([1; \overline 2]\) and \([5; 3, \overline{1, 2}]\)], show that they must represent [\(\sqrt2\) and \(7 - \sqrt3\)] respectively. (*Hint:* Find a quadratic equation that these quantities must satisfy, and use the quadratic formula.)

### Solution

The idea is to “unravel” the continued fraction. In the first case, we have \[\begin{aligned} x &= 1 + (2 + (2 + \dots )^{-1})^{-1} \newline &= 1 + (1 + [1 + (2 + \dots)^{-1}])^{-1} \newline &= 1 + (1 + x)^{-1}, \end{aligned}\] which gives the quadratic equation \(x^2 - 2 = 0.\) In the second case, left to the reader, we obtain the quadratic equation \(x^2 - 14x + 46 = 0.\) Resolving the plus/minus appropriately (noting, for instance, that \([1; \overline2] > 0,\)) we obtain the desired.

## Exercise 3.3

*Easy.* The first step in the Eudoxan theory was to supply a criterion as to when a length ratio \(a : b\) would be greater than another such ratio \(c : d.\) This criterion is that some positive integers \(M, N\) exist such that \(Ma > Nb\) and \(Nd > Mc.\)

Can you see why this works?

### Solution

Rearranging gives \(\dfrac{a}{b} > \dfrac{N}{M} > \dfrac{c}{d}.\)

## Exercise 3.4

*Hard.* Generalize rigorously the preceding length ratios to real numbers, defining the sum and product.

# FIX THIS TEXT, S.

### Solution

RP is essentially asking for a construction of the reals. The nicest way to do this, in my opinion, involves defining a real number as an equivalence class of Cauchy sequences of rationals. This appears as an exercise in Chapter 2 of Baby Rudin and probably somewhere in any decent real analysis book you consult.

Having shown that this construction is well-defined, the sum and product of two real numbers (=sequences) can simply be taken termwise without issue, since limits preserve sums and products.