For a more detailed treatment of both the \ufb02oor and ceiling see the book

# For a more detailed treatment of both the ﬂoor and

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. For a more detailed treatment of both the ﬂoor and ceiling see the book Concrete Mathemat- ics [5]. According to the definition of x we have x = max { n Z | n x } (4.1) 13
14 CHAPTER 4. THE FLOOR AND CEILING OF A REAL NUMBER Note also that if n is an integer we have: n = x ⇐⇒ n x < n + 1 . (4.2) From this it is clear that x x holds for all x, and x = x ⇐⇒ x Z . We need the following lemma to prove our next theorem. Lemma 4.1. For all x R x 1 < x x. Proof. Let n = x . Then by (4.2) we have n x < n + 1. This gives immediately that x x , as already noted above. It also gives x < n + 1 which implies that x 1 < n , that is, x 1 < x . Exercise 4.1. Sketch the graph of the function f ( x ) = x for 3 x 3. Exercise 4.2. Find π , π , 2 , 2 , π , π , 2 , and 2 . Definition 4.2. Recall that the decimal representation of a positive in- teger a is given by a = a n 1 a n 2 · · · a 1 a 0 where a = a n 1 10 n 1 + a n 2 10 n 2 + · · · + a 1 10 + a 0 (4.3) and the digits a n 1 , a n 2 , . . . , a 1 , a 0 are in the set { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } with a n 1 = 0. In this case we say that the integer a is an n digit number or that a is n digits long. Exercise 4.3. Prove that a N is an n digit number where n = log( a ) +1. Here log means logarithm to base 10. Hint: Show that if ( 4.3 ) holds with a n 1 = 0 then 10 n 1 a < 10 n . Then apply the log to all terms of this inequality. Exercise 4.4. Use the previous exercise to determine the number of digits in the decimal representation of the number 2 3321928 . Recall that log( x y ) = y log( x ) when x and y are positive.
Chapter 5 The Division Algorithm The goal of this section is to prove the following important result. Theorem 5.1 (The Division Algorithm). If a and b are integers and b > 0 then there exist unique integers q and r satisfying the two conditions: a = bq + r and 0 r < b. (5.1) In this situation q is called the quotient and r is called the remainder when a is divided by b . Note that there are two parts to this result. One part is the EXISTENCE of integers q and r satisfying (5.1) and the second part is the UNIQUENESS of the integers q and r satisfying (5.1). Proof. Given b > 0 and any a define q = a b r = a bq Cleary we have a = bq + r . But we need to prove that 0 r < b . By Lemma 4.1 we have a b 1 < a b a b . Now multiply all terms of this inequality by b . Since b is positive, b is negative so the direction of the inequality is reversed, giving us: b a > b a b ≥ − a. 15
16 CHAPTER 5. THE DIVISION ALGORITHM If we add a to all sides of the inequality and replace a/b by q we obtain b > a bq 0 . Since r = a bq this gives us the desired result 0 r < b . We still have to prove that q and r are uniquely determined. To do this we assume that a = bq 1 + r 1 and 0 r 1 < b, and a = bq 2 + r 2 and 0 r 2 < b.

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