1.3 The Real Numbers.

Transcription

1 24 CHAPTER. NUMBERS.3 The Rea Numbers. The rea numbers: R = {numbers on the number-ine} require some rea anaysis for a proper definition. We sidestep the anaysis, reying instead on our ess precise notions of continuity from cacuus. Notice that the rea numbers are ordered (from eft to right) and come in three types: where R = R {0} R + R + = {positive rea numbers} are the rea numbers that measure engths (just as the natura numbers count). Notice aso that rationa numbers are exampes of rea numbers. We didn t define the rationa numbers to be numbers on the number-ine, but since the sope of a ine through (0, 0) and (b, a) is the y-coordinate of its intersection with the vertica ine x =, we may think about our number-ine in that way (as the vertica ine x = ), and then Q R = {sopes of a ines through (0, 0) (except the y-axis)} We want to see that the rea numbers are a fied (see..2) and that most of the rea numbers are not rationa (remember 2). In fact, we wi be abe to find penty of irrationa numbers using: Decimas: An infinite decima is a sequence of the foowing form: q.d d 2 d 3 where q is a whoe number (a natura number or zero), and each d i is a digit (whoe number between 0 and 9). A the decimas we wi use wi be infinite. A terminating decima is a decima q.d d 2 d n which we wi make infinite by padding it with zeroes: q.d d 2 d n 00 A terminating decima aways represents a rationa number: q.d d 2 d n 00 = q + d 0 + d d n 0 n (Remember, we no onger use the brackets when writing rationa numbers!) A non-terminating decima represents a rea number in the same way, except that we need the notion of convergence from cacuus to make sense of the infinite sum: q + d 0 + d d

2 .3. THE REAL NUMBERS. 25 Conversey, every (positive) rea number has a decima expansion. Definition of Decima Expansions: Given a positive rea number r R +, the (infinite) decima expansion of r is defined as foows: q is chosen so that q r < q +. The digits are then chosen by induction: (i) d is chosen so that: q + d 0 r < q + d It is between 0 and 9 because q r < q +. (ii) Each d n+ is chosen so that: q + d d n+ 0 n+ r < q + d d n+ 0 n+ + 0 n+ It is between 0 and 9 because: q + d d n 0 n r < q + d d n 0 n + 0 n This gives a sequence of terminating decimas (= rationa numbers): q, q.d, q.d d 2, q.d d 2 d 3, etc. that converges to r. Thus, r = q.d d 2 d 3. Exampes: (a) The natura number m expands as the terminating decima: The infinite decima: m (m ) aso represents m, but you wi never get it as the decima expansion. A the terminating decimas (and ony the terminating decimas) have this ambiguity. (b) The decima expansion of /3 is because < 3 < no matter how many digits we take (mutipy through by 3 to see this). (c) We can decima expand 2 as far as we want by squaring: 2 = < 2 < 4 = 2 2, so < 2 < 2 so q =. (.4) 2 =.96 < 2 < 2.25 = (.5) 2, so d = 4. (.4) 2 =.988 < 2 < = (.42) 2, so d 2 =. (.44) 2 = < 2 < = (.45) 2, so d 3 = 4.

3 26 CHAPTER. NUMBERS Remark: You wi frequenty see: 2 =.44. Unike the decima above, this use of means ony that.44 are the first three digits of the infinite decima expansion of 2. It does not mean that there is a pattern! Expanding Rationas. Given a positive rationa number m, perform the foowing divisions with remainders to define the digits of a decima: First, set = mq + r (this defines q and a whoe number r < m) Next, define the digits by induction: (i) Set 0r = md + r (this defines d, which is a digit, and r < m) (ii) Set each 0r n = md n+ + r n+ (this defines d n+, which is a digit, as we as r n+ < m) and this defines digits d n (and remainders r n < m) for a n by induction. This ooks sort of ike Eucid s agorithm, except this one never ends. But if r n = 0 for some n, then 0 = d n+ = d n+2 = and the decima terminates. You shoud convince yoursef that this agorithm for expanding rationas is exacty how you were taught to find the decima of a rationa number as the ong divison of by m. But now we are in a position to prove that this expansion gives the correct decima expansion of m! Proposition.3.. The infinite decima in the rationa expansion of m is equa to its decima expansion. Proof: To get started, divide = mq + r by m to get: ( ) m = q + r m and then q m < q + (because 0 r m < ) so this is the correct q. Next, a proof by induction checks the decimas: (i) Divide 0r = md + r by 0m to get r m = d 0 + r 0m, and substitute into ( ) to get: m = q + d 0 + r 0m which proves that d is the correct digit (because 0 r m < )! (ii) Once we know that d,..., d n are the correct first n digits, and in fact that: ( ) m = q.d d 2 d n + r n 0 n m then divide 0r n = md n+ + r n+ by 0 n+ m to get and substitute into ( ) to get: r n 0 n m = dn+ 0 n+ + rn+ 0 n+ m, m = q.d d 2 d n + d n+ 0 n+ + r n+ 0 n+ m = q.d d 2 d n+ + r n+ 0 n+ m This proves that d n+ is aso correct, and competes the proof by induction.

4 .3. THE REAL NUMBERS. 27 Definition: A repeating decima is any decima of the form: q.d d 2 d k d k+ d n d k+ d n d k+ d n for some pair of natura numbers k < n. We write this as: q.d d 2 d k d k+ d n to avoid the ambiguity remarked upon earier. Exampe: Your cacuator s output for /35 wi convince you that: = (k =, n = 7) 35 We wi prove this as we prove the foowing: Proposition.3.2. A the decima expansions of rationa numbers repeat. Proof: Consider again step (ii) in the rationa expansion of /m above: (ii) 0r n = md n+ + r n+ From this step, it foows that if r k = r n for some k < n, then: d k+ = d n+ and r k+ = r n+ because they are the quotients and remainders when the same numbers 0r k = 0r n are divided by m (with remainders). But since r k+ = r n+ it wi then foow that d k+2 = d n+2 and r k+2 = r n+2 and so on (this coud be proved by induction, but I think it is cear). Thus when the remainder repeats for the first time, the decima repeats! How do we know that the remainders eventuay repeat? Because a remainders are between 0 and m. So by the time we have done m divisions with remainders, we must have come across a repeat of the remainders!! Exampe: The expansion of /35 reay does repeat as indicated above. = 35(0) + (q = 0 and r = ) Decima so far: 0 0() = 35(0) + 0 (d = 0 and r = 0) Decima so far: 0.0 0(0) = 35(2) + 30 (d 2 = 2 and r 2 = 30) Decima so far: (30) = 35(8) + 20 (d 3 = 8 and r 3 = 20) Decima so far: (20) = 35(5) + 25 (d 4 = 5 and r 4 = 25) Decima so far: (25) = 35(7) + 5 (d 5 = 7 and r 5 = 5) Decima so far: (5) = 35() + 5 (d 6 = and r 6 = 5) Decima so far: (5) = 35(4)+0 (d 7 = 4 and r 7 = 0 = r. Repeat!) Decima:

5 28 CHAPTER. NUMBERS Remarks: The proposition is again teing us something we ve aready earned. On the other hand, now we ve proved it! Aso notice that the proposition tes us that any non-repeating decima gives a rea number which is not rationa. My persona favorite has a simpe pattern, but not a repeating one: There are many more non-repeating decimas than repeating ones! This may seem a strange statement to make since there are infinitey many of both. One good way to think about this is that if a decima coud be chosen at random, then the chances of it repeating are ess than the chances of winning the biggest ottery you coud imagine! Addition: We coud try to define the addition of rea numbers as an addition of infinite decimas, but this woud be messy, as such an addition wi typicay invove infinitey many carries. Instead, we define addition geometricay, via transations. If r is a rea number (possiby negative or 0), then transation by r is the side of the number-ine that is required to move 0 to r. Thus, for instance, transation by sides the number-ine one unit to the right, and transation by sides it one unit to the eft. Transation definition of addition: If s is a rea number, then s + r is the resting pace of s after transating it by r. This sounds fancy, but I caim that it does the same thing as our earier definitions of addition for integers. Why? Because the next integer after a is the transation of a by unit to the right, and the previous integer before a is the transation of a by unit to the eft, whie transating by 0 does nothing (induction takes care of the rest). It takes a bit more work to see: Proposition.3.3. The transation definition of addition does the same thing to rationa numbers as the earier definition. Proof: First of a, notice that the ine of sope n meets the ine x = at a point whose y-coordinate is one nth of the way to. That is, it takes n of the transations by n to move 0 to. This is seen by considering the triange with vertices (0, 0), (n, 0), (n, ) and the simiar triange cut out by the ine x =. Since it takes n (horizonta) transations by to move from 0 to n, simiar trianges te us that the same is true of the vertica transations. We wi refer to n as a fractiona unit. The sum of two rationa numbers was: a b + c ad + bc = d bd and we can assume that the fractions are in owest terms, so b > 0 and d > 0. We can then put the fractions over a common denominator: a b = ad bd and c d = bc bd

6 .3. THE REAL NUMBERS. 29 using Proposition.2.4. Now, transation by c d is transation by c of the fractiona units d, which is aso transation by bc of the fractiona units bd, and ikewise for a b. Thus, addition of these rationa numbers is transation of 0 by ad+bc of the fractiona units bd, which agrees with the addition definition above for rationas. Laws of Addition: One coud prove these with Eucidean geometry, but I woud rather remind you that cacuus does the job. The necessary ingredients are: (a) The transation definition of addition is continuous and (b) Every rea number is a imit of rationa numbers because with these two ingredients, we can use the aws for addition of rationa numbers to deduce the aws for the addition of rea numbers. For exampe, rea numbers r, s, t are imits of rationa numbers: r = im { an b n }, s = im { cn d n }, t = im { en and because addition is continuous, we can pu out imits(!) ( { } { }) an cn (r + s) + t = im + im + im b n d n {( an = im + c ) n + e } n b n d n f n f n } { en and because addition is associative for rationas we can substitute: ( an + c ) n + e n = a ( n cn + + e ) n b n d n f n b n d n f n for each n, and pug back into the imits to get: (r + s) + t = r + (s + t) From the point of view of transations, it is obvious that: 0 is the additive identity and we can get mieage out of the: Negation Transformation: This is defined the same way as before! : R R takes a transation to the equa transation in the opposite direction. From this it is cear that r is the additive inverse of r, and arguing as in Proposition.2.3, we concude that the negation transformation is a inear transformation: (r + s) = r + ( s). Subtraction is defined as usua, to be addition of the additive inverse. f n }

7 30 CHAPTER. NUMBERS Area definition of mutipication: For positive reas r and s, define: rs is the area of the r s rectange and then r( s) = (rs), ( r)s = (rs), ( r)( s) = rs and r 0 = 0 = 0 r. Again, I am appeaing to your geometric intuition of the meaning of area. It can be carefuy defined using imits and cacuus, if you prefer. Proposition.3.4. The area definition for rea numbers does the same thing to rationa numbers as the earier definition. Proof: Because the definitions incorporate negatives in the same way, it is enough to see that the definitions are the same for positive rationa numbers. Of course, the area of an m rectange is m. The area of an m (n + ) rectange is m more than the area of an m n rectange because an m (n + ) rectange is the union of m n and m rectanges! Thus the area definition agrees with the earier definition for natura numbers. As for positive rationa numbers, it again comes down to the fact that n of the fractiona n units are equa to unit. From this it foows that mn of the m n squares exacty fi a square, so mn = m n in both definitions, and again we see that k m n is k of the fractiona squares mn, so it has the appropriate area k mn. The rest of the aws of arithmetic have a very pretty geometric interpretation: The Distributive Law: r(s + t) = rs + rt because an r (s + t) rectange is a union of r s and r t rectanges. The Commutative Law: rs = sr because an r s rectange has the same area as an s r rectange. The Associative Law: r(st) = (rs)t because both are the voumes of an r s t box. is the mutipicative identity. The area of an r rectange is r. We finay want to prove that every rea number except 0 has a mutipicative inverse. This can be done either using cacuus or using geometry. For the cacuus approach, et r be a positive rea number. Then the function: f(x) = rx is continuous (in fact, differentiabe with derivative f (x) = r). Since f(0) = 0 and im x + f(x) = +, the intermediate vaue theorem tes us that there must be some positive rea number s so that f(s) =. In other words, rs = so s = /r. And then, of course, s is the mutipicative inverse of r.

8 .3. THE REAL NUMBERS. 3 The cacuus proof ony says that the inverse exists, not how to find it. For a geometric construction of the mutipicative inverse, et L be the ine through (0, 0) and (r, ). This has sope /r (uness r = 0, in which case it is vertica!). In particuar, the intersection of L with the vertica ine x = is the point (, /r). That is, by drawing L and intersecting with x =, we have constructed the mutipicative inverse. This is more satisfying that just proving that it exists! So R is a fied. As we said earier, addition and mutipication of decimas is messy. There are, however, a coupe of usefu exceptions to this. Mutipying by powers of 0: If r = q.d d 2 d 3, then: 0r = (0q + d ).d 2 d 3 d 4 00r = (00q + 0d + d 2 ).d 3 d 4 d 5 etc. That is, mutipying by powers of 0 shifts the decima point. Subtracting matching digits: Suppose r and s are rea numbers with matching digits. That is, suppose: Then r has decima expansion q.d d 2 d 3 and s has decima expansion p.d d 2 d 3 r s = q p That is, if the decimas a match, then the difference is an integer. In Proposition.3.2, we proved that every rationa number expands as a repeating decima. Here we prove the converse statement. Proposition.3.5. Every repeating decima is the decima expansion of some rationa number. and Proof: Start with a repeating decima r = q.d d 2...d k d k+ d n. Then: 0 k r = (0 k q + 0 k d + + d k ).d k+ d n 0 n r = (0 n q + 0 n d + + d n ).d k+...d n and these are matching decimas, so we can subtract them to get: 0 n r 0 k r = (0 n q d n ) (0 k q d k ) and dividing both sides by 0 n 0 k, we see that r is rationa: r = (0n q d n ) (0 k q d k ) 0 n 0 k

9 32 CHAPTER. NUMBERS Exampe: To find the rationa number that expands to:.2 we take: = = Finay, from the irrationaity of 2 (see.2) we get an interesting: Coroary: The decima expansion of 2 doesn t repeat!.3. Rea Number Exercises 3- (a) Find the first 5 decimas in the expansion of 3 by squaring. (b) Find the first 5 decimas in the expansion of 3 2 by cubing. 3-2 Find the decima expansions for each of the foowing: (a) 3 (b) 2 3 (c) 3 3 (d) 4 3 (e) 5 3 (f) 6 3 (g) 7 3 Do you see a pattern? 3-3 Convert each repeating decima to a fraction in owest terms. (a) 0.27 (b) (c) (d) (e) Which rationa numbers correspond to terminating decimas? Infinite decimas are not the most efficient way to express a positive rea number as a imit of rationa numbers. The continued fraction expansion actuay works much better. Continued Fraction Expansion: Given a positive rea number r, define its continued fraction by induction: (i) q is chosen so that q r < q + (q is the integer part of r) and s = r q (s is the fractiona part of r, with 0 s < ). (ii) Once s n is defined, then if s n = 0, STOP! Otherwise, q n+ is defined to be the integer part of /s n and: s n+ is defined to be the fractiona part of /s n. This gives a sequence of rationa numbers converging to r: q, q + q 2, q + q 2 +, q + q 3 q 2 +, q 3+ q 4

10 .3. THE REAL NUMBERS. 33 Exampe: Expand 25/7 = as a continued fraction. q = and s = 8/7 (so /s = 7/8), q 2 = 2 and s 2 = /8 (so /s 2 = 8), q 3 = 8 and s 3 = 0. STOP! This gives the sequence:, + 2 = 3 2, = 25 7 Continued fraction expansions of rationa numbers aways terminate. (Why?) 3-5 Expand each of the foowing as continued fractions and write the sequences (as in the exampe above): (a) (b) (c) (d) Another Exampe: Expand 2 =.44 as a continued fraction. q = and s = 2 Next step. Simpify: s = 2 = ( 2 )( 2 + ) = 2 = 2 + = 2.44 q 2 = 2 and s 2 = = 2 (same as s ) q 3 = 2 and s 3 = 2 (same as q 2 and s 2 ) so 2 = q 2 = q 3 = q 4 =, giving the sequence:, + 2 = 3 2, = 7 5, = 7 2, In particuar, the continued fraction for 2 doesn t terminate (this is another proof that 2 isn t rationa!). But the q s do repeat. In fact, continued fraction expansions of soutions to the quadratic formua: b + b 2 4ac 2a with a, b, c Z (see 2.3) aways repeat. (Why?) 3-6 Expand the foowing two numbers as continued fractions, indicating where the repeat in the q s occurs, and write out the first four terms of the sequence, as in the exampe. (a) 3 (b) the goden mean: + 5 2

11 34 CHAPTER. NUMBERS Hint: The goden mean satisfies the coo property: = Cacuator exercise. Use a cacuator to find the first 5 vaues q, q 2,..., q 5 in the continued fraction expansion of π and then find the rationa number: q + q 2 + q 3+ q 4 + q 5 which is an exceent (much better than 3.459) approximation of π.