s5_3 - 3. APPLICATIONS OF NUMBER THEORY 163 3. Applications...

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3. APPLICATIONS OF NUMBER THEORY 163 3. Applications of Number Theory 3.1. Representation of Integers. Theorem 3.1.1 . Given an integer b > 1 , every positive integer n can be expresses uniquely as n = a k b k + a k - 1 b k - 1 + ··· + a 1 b + a 0 , where k 0 , 0 a 0 ,a 1 ,a 2 ,...,a k < b , and are all integers. Definition 3.1.1 . Base b expansion of n is ( a k a k - 1 ··· a 1 a 0 ) b if the a i are as described in Theorem 3.1.1. Example 3.1.1 . Here are examples of common expansions other than the more familiar decimal expansion. Binary expansion is the base 2 expansion. Octal expansion is the base 8 expansion. Hexadecimal expansion is base 16 expansion. The symbols A through F are used to represent 10 through 15 in the expansion. Discussion Theorem 3.1.1 asserts that each positive integer n can be expressed uniquely as a linear combination of powers of a fixed integer b > 1. The coefficients in the linear combination must be less than b and must be greater than or equal to zero. These coefficients are, by definition, the digits of the base b expansion of n , n = ( a k a k - 1 ...a 1 a 0 ) b . 3.2. Constructing Base b Expansion of n . Use the division algorithm to get the base b expansion of n : 1. n = bq 1 + a 0 , 0 a 0 < b and q 1 < n . 2. q 1 = bq 2 + a 1 , 0 a 1 < b and q 2 < q 1 . 3. q 2 = bq 3 + a 2 , 0 a 2 < b and q 3 < q 2 . 4. etc. until q i = 0. Then n = ( a k a k - 1 ...a 1 a 0 ) b . Example 3.2.1 . Find the binary expansion of 42. Solution: We can use the division algorithm to get the a i ’s.
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3. APPLICATIONS OF NUMBER THEORY 164 42 = 2(21) + 0 21 = 2(10) + 1 10 = 2(5) + 0 5 = 2(2) + 1 2 = 2(1) + 0 1 = 2(0) + 1 This gives us 42 = (1)(2 5 ) + (0)(2 4 ) + (1)(2 3 ) + (0)(2 2 ) + (1)(2 1 ) + 0 . Thus the binary expansion of 42 is (101010) 2 . Example 3.2.2 . Find the hexadecimal expansion of 42. Solution: This time we use 16 for b . 42 = 16(2) + 10 2 = 16(0) + 2 So the hexadecimal expansion of 42 is (2 A ) 16 (recall we use A = 10 in hexadecimal notation). Example 3.2.3 . Find the decimal notation of the octal representation (1024) 8 . (1024) 8 = 1(8 3 ) + 0(8 2 ) + 2(8 1 ) + 4 = 532 3.3. Cancellation in Congruences. Theorem 3.3.1 . Suppose GCD( c,m ) = 1 and ac bc ( mod m ) . Then a b ( mod m ) . Proof. Suppose GCD( c,m ) = 1 and ac bc (mod m ). (We may assume m > 1 so that c 6 = 0.) Then ac - bc = c ( a - b ) = km for some integer k . This implies c | km. Since GCD(
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s5_3 - 3. APPLICATIONS OF NUMBER THEORY 163 3. Applications...

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