SOLUTIONS FOR THE THIRD PROBLEM SET

Transcription

1 SOLUTIONS FOR THE THIRD PROBLEM SET. On the handout about continued fractions, one finds a definition of the function f n (x) for n 0 associated to a sequence a 0,a,... We have discussed the functions f n (x) in class and proved several properties that they satisfy. One should use some of these properties in the following question. Please do not actually compute the numbers listed. Arrange the following four rational numbers in increasing order. s {;,3,,,6}, t {;,3,,,7}, u {;,3,,,6,7}, v {;,3,,,6,7,8} SOLUTION. Let f (x) {;,3,,,x}. Then, as a special case of a general theorem proved in class, we know that f (x) is a strictly decreasing function of x for 0 < x <. We have s f (6), t f (7), u f (6+ 7 ) Since 6 < 6+ < 7, it follows that s > u > t. It remains to decide where to place v in 7 the ordering. We will mention two methods. Notice that v f ({6;7,8}) and {6;7,8} < < obtain that s > v > u. Hence, we get the ordering Also, notice that Using the fact that f (x) is strictly decreasing again, we t < u < v < s Alternatively, we can use the function f 6 (x) {;,3,,,6,7,x} which is also a strictly decreasing function for 0 < x <. As discussed in class, the range of this function is the interval {;,3,,,6,7} < y < {;,3,,,6}. That is, the range of f 6 (x) is the interval u < y < s. Since v f 6 (8) is in the range of f 6 (x), we again obtain the inequalities u < v < s. Thus, the four numbers arranged in increasing order are: t, u, v, s.

2 . Suppose that θ is an irrational number and that θ {a 0 ;a,...,a n,...} is its continued fraction expansion. (a) Assume that a n for all n 000. Prove that θ Q[ 0]. (b) Assume now that θ Q[ 0]. Is it true that a n for all n n 0, where n 0 is some positive integer? (a) SOLUTION. First of, it will be useful to find the continued fraction expansion of α We write α Then a 0 [α 0 ]. Thus, α 0 a We obtain α α 0 a ( 0 7)( 0+7) Hence α 0+7 and a [α ] [ 0+7]. Note that α a 0 7. We obtain α α a just as before. Hence α α. Therefore a a and the continued fraction algorithm will give α n α and a n a for all n. That is, the continued fraction expansion for 7+ 0 is {;,,,...}. We have 7+ 0 α {;,,,...}. Now it is stated that θ {a 0 ;a,...,a 999,,,,,...}. As discussed in class, we have θ {a 0 ;a,...,a 999,θ 000 } and so the continued fraction expansion of θ 000 is {;,,,...}. It follows that θ 000 α Thus, we have θ 000 Q[ 0]. As explained in class, it then follows that we also have θ Q[ 0]. This is what we wanted to show.

3 (b) SOLUTION. The statement is actually false. Notice that 0. Hence, we have Q[ 0]. We know the continued fraction expansion for. It was derived one day in class. It is {;,,,,...} Thus, θ has a continued fraction expansion where none of the a n s is equal to. 3. Suppose that d n +, where n is a positive integer. Find the continued fraction expansion of d. SOLUTION. Let θ d. First of all, we have n < d < n+. Thus, Then, θ 0 a 0 d n. a 0 [θ 0 ] [θ] [ d] n. θ θ 0 a 0 d n d+n ( d n)( d+n) d+n d n d+n. Then, a [θ ] [ d+n] [ d]+n n. Hence, θ a d n. Therefore, θ θ a d n d+n just as in the calculation of θ. Notice that θ θ. We have a repetition. As explained in class, we then have θ k+ θ k for all k. Therefore, we have θ k θ d + n for all k. It follows that a k [θ k ] n for all k. That is, the continued fraction expansion of d is given by d {n;n,n,n,...}.. Find the continued fraction expansion for θ 9. SOLUTION. We have θ 0 θ 9 and a 0 [θ 0 ]. Furthermore, we obtain θ θ 0 a 0 and hence a [θ ]. Then 9 9+ ( 9 )( 9+) 9+ θ θ a 9 3 ( 9+3) ( 9 3)( 9+3) 9+3 3

4 and hence a [θ ]. Then θ 3 θ a 9 ( 9+) ( 9 )( 9+) 9+ and hence a 3 [θ 3 ]. Then θ θ 3 a ( 9+3) ( 9 3)( 9+3) 9+3 and hence a [θ ]. Then θ θ a 9 ( 9+) ( 9 )( 9+) 9+ and hence a [θ ] 0. Then θ 6 θ a 9 9+ ( 9 )( 9+) 9+ and hence θ 6 θ. This is the first repetition. As discussed in class, we have θ n+ θ n for all n. The continued fraction expansion for θ 9 is given by 9 {;,,,,0,,,,,0,,,,,0,...}.. Let θ {;,,3,,,,3,,...}. Find an integer d such that θ Q[ d]. SOLUTION. Notice that θ {;ψ}, where ψ θ {;,3,,,,3,,...}. We have ψ {;,3,,ψ} p 3ψ +p q 3 ψ +q,

5 p 3 where {;,3,} and p {;,3}. Using the formulas in statement from the q 3 q basic facts handout, we have and p 0, p + 3, p 3 3+, p 3 +3 Thus, we have the equation This gives the equation q 0, q, q 3 +, q 3 +. ψ ψ + ψ + ψ +ψ ψ +, or equivalently ψ 0ψ 0. Using the quadratic formula, this gives. ψ 0± ( 0) ( ) 0 Since [ψ], we actually have ± psi ±. It follows that ψ Q[ ]. Hence, θ Q[ ]. We can take d. 6. The sets are A { ±(3+ )ε n 0 n Z } and B { ±(3 )ε n 0 n Z } WewillshowthatA B φ, theemptyset. SupposetothecontrarythatA B isnonempty. Then we have an equation ±(3+ )ε n 0 ±(3 )ε m 0 for some n,m Z and some choice of the signs ±. This gives the equation 3+ 3 ±εm n 0.

6 The right hand side of the above equation is in the set that we denoted by U or U in class. This set is contained in Z[ ]. Thus the right hand side is in Z[ ]. One can also see this by noticing that ε 0 3+, ε 0 3, and m n Z, and then using the fact that Z[ ] is closed under multiplication. However, for the left hand side, we have 3+ 3 (3+ )(3+ ) (3 )(3+ ) which is not in Z[ ]. We have a contradiction. Hence A B φ. 7. We have determined the continued fraction expansion of θ + in class. It is given by θ {;,,,,,...}. We can use the formulas in statement from the basic facts handout to determine the p n s and q n s. We have p 0, p +, p + 3, and p n p n +p n for all n. It is clear that p n F n+ for all n 0. We also have q 0, q, q +, and q n q n +q n for all n. It is clear that p n F n for all n 0. We can now use a result proved in class. We have θ F n+ F n θ p n q n < p n F n for all n. This inequality is also valid for n 0. 6