*In which we explore infinite continued fractions.*

- Definition of
*convergents*. - Definition of sequences and .
- Lemma 58:
*For , we have . That is, is equal to the convergent.*We proved this by induction, using the fact that . - Lemma 59:
*For , we have .*This was an easy induction. - Lemma 60:
*For , if , , , …, are integers, then the terms and are coprime integers.*This follows immediately from the previous lemma. - We reminded ourselves that if we are to find a continued fraction for the irrational number , then we must have , and , and and for .
- Proposition 61:
*The convergents converge. That is, if we define and and as above, then the sequence , , , … converges to .*We proved this by using Lemma 58 to show that , and then computing . - Lemma 62:
*We have for all , and so we also have for .*We used the expression we obtained for in the previous result, together with a couple of easy estimates using the recurrence for .

#### Further reading

The usual selection. Hardy and Wright (*An Introduction to the Theory of Numbers*) and Baker (*A concise introduction to the theory of numbers*) both discuss Diophantine approximation via continued fractions, and have some nice material going beyond the topics that we’ll see in lectures.

#### Preparation for Lecture 20

We’ve seen that continued fractions give a sequence of rational numbers that converge to a given real number (namely the convergents). Next time we’ll consider these as rational approximations of that real number. I suggest that you compute some convergents (for some examples that we haven’t seen in lectures), to see whether they give good approximations. How do they compare with the rational approximations arising from truncating decimal expansions?

Next time we’ll see that in some sense the best rational approximations come from continued fractions. You could usefully think about how we might formulate that result in a precise way (and, of course, how we might prove it).

We are then going to turn our attention to the Diophantine equation (where is a fixed integer that isn’t a square, and where we are looking for integer solutions and ). Can you find some solutions when or , for example? It would be good to get a feeling for whether there are no solutions, some but only finitely many solutions, or infinitely many solutions, for different values of .

(And what happens to the equation when *is* a square? Why have I ruled out that case above?)

November 30, 2011 at 12:11 pm

[…] Lemma 76: Let be a convergent to . Then , and moreover we have . (The angle bracket notation is the same as that introduced in Lecture 7). We proved this using our upper bound on , which in turn we proved in Lecture 19. […]

March 13, 2012 at 7:21 pm

[…] see how continued fractions can fit into a number theory course, see these three lecture […]