By Peter J. Cameron

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**Extra resources for A Course on Number Theory [Lecture notes]**

**Sample text**

Now let d = [3; 5, 2, 1]. Then d = [3; 5, c] [3, 5, c] = [5, c] 16c + 3 = 5c + 1 √ 19 + 16 3 √ = 6+5 3 √ √ (19 + 16 3)(6 − 5 3) √ √ = (6 + 5 3)(6 − 5 3) √ 126 − 3 . = 39 Note that d, like c, is a “quadratic irrational”, an algebraic integer satisfying a quadratic equation. ) In this chapter we are going to show that the result suggested by these examples is true in general. A real number has a periodic continued fraction if and only if it is a quadratic irrational. We will also find which numbers have purely periodic continued fractions.

An ] for n ≥ 0. Then the sequence c0 , c1 , c2 of rational numbers converges to a limit. 2. THE DEFINITION 31 Remark This explains why we called the numbers c0 , c1 , . . “convergents”. Proof Since c0 , c1 , . . , cn are the convergents to the finite continued fraction [a0 ; a1 , . . , an ], all the results of Chapter 2 apply here. We have cn = pn /qn , where pn = [a0 , a1 , . . , an ] and qn = [a1 , . . , an ]. Now c0 < c2 < c4 < · · · < c5 < c3 < c1 and ck − ck−1 = (−1)k−1 qk−1 qk for k ≥ 1.

Am ] = C, [a1 , . . , am−1 ] = D (these are all positive integers). Then Az + B Cz + D √ Au + B + Av d √ = Cu + D +Cv d √ √ (Au + B + Av d)(Cu + D −Cv d) = , (Cu + D)2 − (Cv)2 d y = √ which is a quadratic irrational since it has the form x + y d for some rational numbers x and y. Now our goal is to prove the converse of the last two results: if y is a (reduced) quadratic irrational, then its continued fraction is (purely) periodic. Let us begin with an example. 3. THE MAIN THEOREM 47 √ √ Example Find the continued fraction of 2 + 7.