266 CHAPTER 5 Number Theory and the Real Number System Therefore, = = − 2 2 2 2 . 5 2 5 2 3 Quotient Rule for Exponents = ≠ − a a a a , 0 m n m n Example 2 Using the Quotient Rule for Exponents Use the quotient rule to simplify. a) 2 2 7 2 b) 4 4 8 6 Now try Exercise 9 Solution a) = = = − 2 2 2 2 32 7 2 7 2 5 b) = = = − 4 4 4 4 16 8 6 8 6 2 7 Consider ÷ 2 2 . 3 3 The quotient rule gives = = − 2 2 2 2 3 3 3 3 0 But = = 2 2 8 8 1. 3 3 Therefore, 20 must equal 1. This example illustrates the zero exponent rule. Did You Know? Catalan’s Conjecture Consider all the squares of positive integers: 1 1, 2 4, 3 9, 2 2 2 = = = 4 16, 2 = and so on Now consider all the cubes of positive integers: 1 1, 2 8, 3 27, 3 3 3 = = = 4 64, 3 = and so on Next, consider all the fourth powers of positive integers: 1 1, 2 16, 3 81, 4 4 4 = = = 4 256, 4 = and so on If we were to put all powers of the positive integers greater than one, into one set, the set would begin as follows: {1, 4, 8, 9, 16, 25, 27, 32, 36, … 49, 64, }. In 1844, Belgian mathematician Eugène Catalan proposed that of all the numbers in this infinite set, only 8 and 9 are consecutive integers. Although this conjecture was easily stated, the formal proof of it eluded mathematicians for more than 150 years. In April 2002, Preda Mihailescu, a Romanian-born mathematician, proved Catalan’s Conjecture. The study of mathematics involves many conjectures that are often easy to pose, but difficult to prove. For another example of a now-proven longstanding conjecture, see the Did You Know box on page 480 regarding Fermat’s Last Theorem. For two more still unproven conjectures see the discussion of Goldbach’s conjecture and the twinprime conjecture in Section 5.1. Zero Exponent Rule = ≠ a a 1, 0 0 Note that 00 is not defined by the zero exponent rule. Example 3 Using the Zero Exponent Rule Use the zero exponent rule to simplify. Assume ≠ x 0. a) 20 b) −( 2)0 c) −20 d) x (5 )0 e) x5 0 Solution a) = 2 1 0 b) − = ( 2) 1 0 c) − = − ⋅ = − ⋅ = − 2 1 2 1 1 1 0 0 Now try Exercise 11 d) = x (5 ) 1 0 e) = ⋅ = ⋅ = x x 5 5 5 1 5 0 0 7 Consider ÷ 2 2 . 3 5 The quotient rule yields = = − − 2 2 2 2 3 5 3 5 2 But = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 2 2 2 2 2 2 2 2 2 2 1 2 . 3 5 2 Since 2 2 3 5 equals both −2 2 and 1 2 , 2 then −2 2 must equal 1 2 . 2 This example illustrates the negative exponent rule.
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