By Peter J. Cameron

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

**Example text**

The best approximations occur when we stop just before a large number in the continued fraction. 000000266764189 . .. 3 Every irrational number greater than 1 is the limit of a unique infinite continued fraction. Proof Suppose that y is irrational and y = [b0 ; b1 , b2 , . ]. Then y = b0 + 1/y1 , where y1 > 1; so b0 = y and y1 = 1/(y − b0 ) are determined by y. Similarly b1 = y1 and y2 = 1/(y1 − b1 ) are determined, and so on. Remark We have considered numbers greater than 1 so far. The results can be easily extended to arbitrary irrational numbers.

1] are ratios of consecutive Fibonacci numbers. So, if Fn is the nth Fibonacci number, we have √ Fn+1 1 + 5 lim = . 3. 3 35 Approximation by convergents We have seen that, if y = [a0 ; a1 , a2 , . ], and cn = [a0 ; a1 , . . , an ] is the nth convergent to y, then the numbers cn are rational numbers which tend to the limit y. In this section, we will see that they give the best possible approximations to y. What should a good rational approximation p/q to y be? First, of course, it should be close to y.

But c > 2, so we must take the plus sign; c = 1 + 3. 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.

### A Course on Number Theory [Lecture notes] by Peter J. Cameron

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