Decimal Expansion Representation of Real Numbers Fall 2016 Do˘ganC¸¨omez Contents: 1. Introduction 2. Decimal Expansion 3. The Connection with the Ten-fold Map 4. The n-fold Map Tn 5. The Full n-shift Space Σn 1. Introduction A real number can be represented in many ways. The most popular representation is by decimal expansion; i.e., expressing a real number as a geometric series with base 10. A modification of this representation is expressing a real number as a geometric series with base n for any positive integer n ≥ 2; called as the n-ary expansion. The decimal expansion of a real number provides a convenient method of representing rational and irrational numbers as well as approxima- tions of irrational numbers by rational numbers. Furthermore, the process of obtaining the decimal expansion of a real number has close connections with dynamical systems and probability theory, which en- ables one investigate its properties in more depth. In this presentation I will briefly introduce the decimal expansion representation of real numbers, obtain some important properties, and then focus on the dy- namical properties of the \n-ary" map and some of its consequences. The concept of decimal expansion goes back to 8th century when Muhammad ibn-Musa Khwarizmi (c. 780-c. 850) introduced the ara- bic number system in his book who on algebra. In its numerous trans- lations into latin during 12th century, one encounters occasional use of decimal notation. The regular use of the decimal point and decimal notation appears to have been introduced about 1585 by Flemish sci- entist Simon Stevinus (c. 1548-1620). It was Scottish mathematician John Napier (1550-1617), the inventor of logarithms in 1614, who first used and then popularized the decimal point to separate the whole part from the fractional part of a number. P1 k When jrj < 1, the geometric series k=0 r converges with the sum 1 1 1−r : If r = 10 and ai 2 f0; 1;:::; 8; 9g; for all i ≥ 0; then the series 1 X ak 10k k=0 converges to a real number by comparison test. The idea of decimal expansion is the process of obtaining the converse of this fact; namely, given a real number x 2 [0; 1), determining the sequence of integers P1 ak 0 ≤ ai ≤ 9 such that x = k=0 10k : There is nothing special about the 1 2 number 10; indeed, one can develop expansion of real numbers with respect to any other base n 2 N as well. In that case, the expansion is called as the n-ary expansion. 2. Decimal Expansion Representation Let's look at the decimal expansion process closely. Given a real number x ≥ 0: (i) Let a0 = [x]; where [] is the greatest integer function. So, x = a0 + p0; where 0 ≤ p0 < 1: Therefore, a0 ≤ x < a0 + 1; and hence 10a0 ≤ 10x ≤ 10a0 + 9: Also, 0 ≤ 10p0 < 10: (ii) Now, 10p0 = a1 + p1; where a1 = [10p0] is an integer and 0 ≤ p0 < 1: Thus, 0 ≤ a1 ≤ 9; and a1 ≤ 10p0 < a1 + 1 () a1 ≤ 10(x − a0) < a1 + 1; which implies that a a 1 a + 1 ≤ x < a + 1 + : 0 10 0 10 10 (iii) Similarly, 10p1 = a2 + p2; where a2 = [10p1] is an integer and 0 ≤ p1 < 1: Thus, 0 ≤ a2 ≤ 9; and a2 ≤ 10p1 < a2 + 1 () a2 ≤ 10(10p0 − a1) < a2 + 1 2 () a2 ≤ 10 p0 − 10a1 < a2 + 1 2 () a2 ≤ 10 (x − a0) − 10a1 < a2 + 1; which implies that a a a 1 a a + 1 + 2 ≤ x < a + 1 + + 2 : 0 10 102 0 10 10 102 Continuing in this process and letting 10pk−1 = ak + pk; where ak = [10pk−1] and 0 ≤ pk < 1; we obtain integers ai 2 f0; 1;:::; 9g; 0 ≤ i ≤ k; such that a a a a 1 a + 1 + ··· + k ≤ x < a + 1 + ··· + k + : 0 10 10k 0 10 10k 10 Observe that if an = 0 for some n; then ai = 0 for all i ≥ n; hence the process terminates. Otherwise the process continues indefinitely. In the case that the process terminates at the nth step, we have a a (1) x = a + 1 + ··· + n ; 0 10 10n and when the process continues indefinitely, we have 1 a1 an X ak (2) x = a + + ··· + + ··· = : 0 10 10n 10k k=0 3 These expressions (1) and (2) are called the decimal expansions of the real number x; where (1) is finite and (2) is infinite, and are denoted by x = a0:a1a2 : : : an−1an; or x = a0:a1a2 : : : an :::; respectively: Remarks. 1. In the case that the process continues indefinitely, the decimal expansion process outlined above shows that, at any step n ≥ 1; we have a a a a 1 a + 1 + ··· + n ≤ x < a + 1 + ··· + n + : 0 10 10n 0 10 10n 10 Hence, we obtain two sequences of finite decimals each converging to x; one increasing from below and the other decreasing from above. By repeating this procedure sufficiently many times, the decimal expansion of x can be obtained to any desired degree of accuracy. 2. In the case that the process continues indefinitely, there are two possibilities: either the sequence of terms faig has no repetition, or it repeats, in the sense that, for some positive integers k and m; ak = ak+m; hence, the decimal expansion has the form a0:a1a2 : : : akak+1ak+2 : : : ak+m−1akak+1 :::; which we denote, for convenience, as a0:a1a2 : : : ak−1akak+1ak+2 : : : ak+m−1; where akak+1ak+2 : : : ak+m−1 denotes that the string of terms ak; ak+1; : : : ; ak+m−1 repeats indefinitely. 3. If a0:a1a2 : : : ak−1ak is a terminating decimal expansion, then it can also be rewritten as as repeating decimal expansion as a0:a1a2 : : : ak−1ak000 ··· = a0:a1a2 : : : ak0: However, in order to distinguish such decimals from the decimals with nonzero repeating parts, we will call them as finite decimal expansion. Pk ai Observe that if x = a0:a1a2 : : : ak0 = i=0 10i ; hence x 2 Q: If x = a0:a1a2 : : : ak−1akak+1ak+2 : : : ak+m−1; then letting y = a0:a1a2 : : : ak−1 and z = 0:akak+1ak+2 : : : ak+m−1; we see that y; z 2 Q: Then, 10m a a a : : : a = z; k k+1 k+2 k+m−1 10m − 1 10m and hence x = y + 10m−1 z 2 Q: Thus, any finite or repeating decimal expansion represents a rational number. The converse of this statement is also valid. Fact.1 If x 2 Q; then it has either finite or repeating decimal expan- sion. Proof. For convenience, we will assume that x 2 (0; 1); hence, we r + have a0 = [x] = 0: Let x = s ; r; s 2 Z ; and recalling the process of 4 r obtaining decimal expansion of numbers above, we observe that p0 = s ; and r r r 10( ) = a + 1 ; where 0 ≤ 1 < 1; s 1 s s r r r 10( 1 ) = a + 2 ; where 0 ≤ 2 < 1; s 2 s s ::: r r r 10( k−1 ) = a + k ; where 0 ≤ k < 1: s k s s Multiplying both sides by s at each step above, we see that this process is exactly the division algorithm: 10r = a1s + r1; where 0 ≤ r1 < s; 10r1 = a2s + r2; where 0 ≤ r2 < s; ::: 10rk−1 = ak + rk; where 0 ≤ rk < s: Notice that if one of rk = 0; then ri = 0 for all i ≥ k; i.e., the process stops and we have finite decimal expansion. For otherwise, since ri 2 f1; 2;:::; 9g; there will be integers k < m such that rm = rk: Then, by division algorithm, we will have rm+i = rk+i for 0 ≤ i ≤ m − k: In turn, we also have that am+i = ak+i for 0 ≤ i ≤ m − k; hence the block akak+1 : : : am−1 repats, which yields to repeating decimal expansion. Corollary. A real number x is irrational if and only if it has a non- repeating decimal expansion. Remark. Observe that, if x = a0:a1a2 : : : an; i.e., x has finite decimal expansion, then, by the division algorithm, we also have that x = a0:a1a2 : : : an−1(an − 1)999 ··· = a0:a1a2 : : : an−1(an − 1)9: Thus, decimal expansion of a rational numbers need not be unique! This is an unpleasant situation. Question. Can we circumvent this problem? It turns out that the remedy is bringing some tolls of dynamical systems into the process. Hence, dynamics comes to the rescue! Next we will discuss this in length and at the same time study an interesting type of dynamical systems. 3. The Connection with the Ten-fold Map Let's look at the process of finding the (decimal) digits in decimal expansion process in an alternative way. Again, we will focus on x 2 [0; 1): i i+1 Partition [0; 1) into 10 equal subintervals [ 10 ; 10 ); 0 ≤ i ≤ 9; and, if i i+1 i i+1 x 2 [ 10 ; 10 ); let a1 = i: Then divide [ 10 ; 10 ) into 10 equal subintervals 5 i j i+1 j i j i+1 j [ 10 + 102 ; 10 + 102 ); 0 ≤ j ≤ 9; and, if x 2 [ 10 + 102 ; 10 + 102 ); let a2 = j: Continuing in this manner, we obtain all the digits ak such that x = 0:a1a2a3 : : : an :::: At each step, this process is the same as multiplication of x by 10 and taking mod 1: Namely, repeated the action of the ten-fold map T on [0; 1); where 8 1 > > 10x − 0 if 0 ≤ x < > 10 > 1 2 <> 10x − 1 if ≤ x < T (x) = 10 10 > ::: > > 9 > 10x − 9 if ≤ x < 1: : 10 Therefore, i1 i1+1 a1 = i1 if 10 ≤ x < 10 () T x = 10x − i1 i2 i2+1 2 a2 = i2 if 10 ≤ T x < 10 () T x = 10(T x) − i2 ::: ik ik+1 k k−1 ak = ik if 10 ≤ x < 10 () T x = 10(T x) − ik; and consequently, a 1 x = 1 + T x 10 10 a 1 a 1 = 1 + ( 2 + T 2x) 10 10 10 10 ::: a a a a = 1 + 2 + 3 + ··· + k T kx: 10 102 103 10k k−1 Remark.