Chapter 5 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you've probably known and used for a long time are true. It also offers some exposure to generalization, that is, of taking some specific examples that are true and finding a general statement that includes these as specific cases. 5.1 Floors and Ceilings For x 2 R, the floor of x is the largest integer that is less than or equal to x, and is denoted by bxc. If x is an integer, then bxc = x, and otherwise it is the nearest integer to the left of x on the number line. Correspondingly, for x 2 R, the ceiling of x is the smallest integer that is greater than or equal to x, and is denoted by dxe. If x is an integer, then dxe = x, and otherwise it is the nearest integer to the right of x on the number line. Proposition 5.1.1 Let x be a real number. Then • x − 1 < bxc ≤ x; and • x ≤ dxe < x + 1. Proof. We prove the first statement. The proof of the second statement is similar. By definition, bxc ≤ x. If x is an integer, then bxc = x, so 1 2 CHAPTER 5. NUMBER THEORY x − 1 < bxc ≤ x. If x is not an integer, then there is an integer n so that x belongs to the open interval (n; n+1). In this case x−n < 1. Rearranging this inequality and using the definition of the floor of x gives x−1 < n = bxc ≤ x. This section concludes with an example of proving a statement about the ceiling operation. Proposition 5.1.2 Let x; y 2 R. Then dxe + dye ≥ dx + ye: Proof. Since dxe ≥ x and dye ≥ y, addition of inequalities gives dxe + dye ≥ x + y. Thus, dxe + dye is an integer that is greater than or equal to x + y. By definition dx + ye is the smallest integer that is greater than or equal to x + y. Consequently, dx + ye ≤ dxe + dye. 5.2 The Division Algorithm Something everyone learns in elementary school is that when one integer is divided by another there is a unique quotient and a unique remainder. For example, when 65 is divided by 17 the quotient is 3 and the remainder is 14. That is, 65 = 3 × 17 + 14. What about when 65 is divided by −17? We have 65 = (−3) × (−17) + 14, and we also have 65 = (−4) × (−17) − 3. Should the remainder be 14 or -3? The convention is that the remainder is always non-negative when dividing by a negative number. The fact that there is a unique quotient and a unique remainder is a theorem. It bears the name \The Division Algorithm" because the proof tells you how to find the quotient and the remainder when an integer a is divided by an integer b { keep subtracting multiples of b from a until what's left is a number r between 0 and jbj−1 (inclusive). The total number of times b was subtracted from a is the quotient, and the number r is the remainder. That is, a = bq + r; 0 ≤ r < jbj. Theorem 5.2.1 The Division Algorithm Let a; b 2 Z, with b 6= 0. Then there exist unique integers q and r so that a = bq + r and 0 ≤ r < jbj. 5.2. THE DIVISION ALGORITHM 3 The integers q and r in The Division Algorithm are the quotient and remainder when a is divided by b, respectively. The integer b is the divisor, and (for completeness we will note that) the integer a is the dividend. Instead of proving the Division Algorithm, we illustrate the proof with the example below below where 2024 is divided by 75. First, ten 75s are subtracted, leaving 1274. Then, ten more 75s are subtracted, leaving 524. From this number five 75s are subtracted, leaving 149. And finally, one 75 is subtracted leaving 74 (the remainder). The quotient is the total number of 75s subtracted, which is 26. Thus 2024 = 26 × 75 + 74. 75 ) 2024 10 −750 1274 10 −750 524 5 −375 149 1 −75 74 26 Suppose, for the moment, that a and b are both positive. The quotient when a is divided by b is the largest integer q such that bq < a. This is the floor of a=b: if a = bq + r with 0 ≤ r < b then ba=bc = b(bq + r)=bc = bq + (r=b)c = q. The same thing happens when a is negative (notice that the quotient is a negative number). Now suppose a is positive and b is negative. Again, the quotient when a is divided by b is the largest integer q such that bq > a. (Such a q is negative!) This is the ceiling of a=b: if a = bq + r with 0 ≤ r < jbj then da=be = d(bq + r)=be = dq + (r=b)e = q, where the last equality follows because b is negative and so r=b is in the interval (−1; 0]. The same thing happens when a is negative. We state the observations just made formally, and also indicate a different proof. Proposition 5.2.2 Let a; b 2 Z, with b 6= 0. If q and r are integers such 4 CHAPTER 5. NUMBER THEORY that a = bq + r; 0 ≤ r < jbj, then ( ba=bc if b > 0; and q = da=be if b < 0: Proof. We give the proof when b > 0. The proof when b < 0 is similar. By definition of the floor function, (a=b) − 1 < ba=bc ≤ a=b. Multiplying through by the positive number b and simplifying gives a − b < bba=bc ≤ a. Let r = a − bba=bc. Then a = bba=bc + r. Rearranging the right hand inequality gives, 0 ≤ a − bba=bc = r. Rearranging the left hand inequality gives r = a − bba=bc < b. Therefore, by The Division Algorithm, r is the remainder when a is divided by b, and ba=bc is the quotient. 5.3 Base b representations When we use the symbol 5374 to represent the integer five thousand, three hundred and seventy four, we understand the notation to mean 5 × 103 + 3 × 102 + 7 × 101 + 4 × 100. Every place in the notation has a value that is a power of ten, the base of the system. The number 5 is in the thousands place, 3 is in the hundreds place, 7 is in the tens place and 4 is in the ones place. (Aside: if there were a decimal point, then the places to the right of it would be the tenths, hundredths, and so on because 1=10 = 10−1, 1=100 = 10−2, etc.) If the digits 0; 1;::: 9 are used in the various places, then every integer can be uniquely represented (in base 10). It turns out that there is nothing special about 10, any integer b > 1 can be used in its place, as can integers b < −1, though we will not consider this case. Let b > 1 be an integer. A base b digit is one of the numbers 0; 1; : : : ; b−1. If dk; dk−1; : : : ; d0 are all base b digits, then the notation (dkdk−1 : : : d1d0)b is k k−1 1 0 shorthand for dk × b + dk−1 × b + ··· + d1 × b + d0 × b . Notice that every place in the notation has a value that is a power of the base: the value of the i-place from the right is bi−1 (the value of the rightmost place is b0). k k−1 1 0 For an integer n ≥ 0, if n = dk ×b +dk−1 ×b +···+d1 ×b +d0 ×b and dk 6= 0, then (dkdk−1 : : : d1d0)b is called the base b representation of n. By analogy with what happens in base 10, if n ≥ 0 then the base b representation of −n is −(dkdk−1 : : : d1d0)b. 5.3. BASE B REPRESENTATIONS 5 2 For example, (234)5 is the base 5 representation of the integer 2 × 5 + 3 × 5 + 4 = 69 (in base 10). Every integer has a unique base b representation. We illustrate the proof that a representation exists with an example. Suppose we want to find the base 5 representation of 69, that is, base 5 digits d0; d1;::: so that 69 = k k−1 1 0 dk × 5 + dk−1 × 5 + ··· + d1 × 5 + d0 × 5 . Rearranging the right hand k−1 k−2 expression gives 69 = (dk × 5 + dk−1 × 5 + ··· + d1) × 5 + d0, so that the ones digit, d0 is the remainder when 69 is divided by 5, that is, 4. Furthermore, the quotient is the number (69 − 4)=5 = 13, which equals k−1 k−2 dk × 5 + dk−1 × 5 + ··· + d1), and hence has base 5 representation (dkdk−1 : : : d1)5. (Notice the absence of d0 in this representation.) Repeating what was just done, the ones digit of this number, that is d1, is the remainder on dividing the 13 by 5. Hence d1 = 3 and, continuing in this way, d2 is the remainder on dividing the new quotient of (13 − 3)=5 = 2 by 5 (so d2 = 2). Repeating again gives dj = 0 for all j > 2. Hence 69 = (234)5, as above. Why is the representation unique? Because of The Division Algorithm. It says that the quotient and remainder are unique. Since the base 5 digits are remainders on division of a uniquely determined number by 5, there can be only one representation.