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Unformatted text preview: Math 25: Advanced Calculus Fall 2010 Homework # 3 – Solutions 1.3.7: Which of the field axioms does Z 6 fail to satisfy? Solution: By observation, Z 6 satisfies A1,A2,M1,M2, and AM1. Furthermore, Z 6 satisfies A3 because 0 + a = a for any a ∈ Z 6 , and satisfies M3 because 1 · a = a for any a ∈ Z 6 . We can check that Z 6 satisfies A4 because: 0 + 0 = 0 = ⇒  0 = 0 1 + 5 = 0 = ⇒  1 = 5 2 + 4 = 0 = ⇒  2 = 4 3 + 3 = 0 = ⇒  3 = 3 4 + 2 = 0 = ⇒  4 = 2 5 + 1 = 0 = ⇒  5 = 1 . However, Z 6 does not satisfy axiom M4 because, for example, 2 does not have a multiplicative inverse: 2 · 0 = 0 2 · 1 = 2 2 · 2 = 4 2 · 3 = 0 2 · 4 = 2 2 · 5 = 4 . 1 1.4.1: Using just the axioms, prove that ad + bc < ac + bd for all a < b and c < d . Proof: Since c < d , axiom O3 implies that 0 < d c . Letting z = d c , x = a , and y = b in axiom O4 tells us that a ( d c ) < b ( d c ) . Applying axiom AM1 and adding ac + bc to both sides of this inequality (which we can do by O3) gives ad ac + ac + bc < bd bc + ac + bc. By commutativity, associativity, and A4, we obtain the desired result. 2 1.4.3: Using just the axioms, prove that for any a > 0 and b > 0 in R , √ ab < a + b 2 . Proof: By axiom O1, we know that either a > b , b > a , or a = b . If a = b , then √ ab = a = a + b 2 and the result is verified. Otherwise, we may assume without loss of generality that a > b . First, we claim that if a > b , then √ a > √ b . We prove this claim by contraposition. Suppose that √ a ≤ √ b . Multiplying both sides of this equation by √ a tells us that a ≤ √ a √ b, by axiom O4. Similarly, multiplyinb both sides of the inequality √ a < √ b by √ b tells us that √ a √ b ≤ b, again by O4. By transitivity, this implies that a ≤ b , which establishes the claim. Next, since √ a > √ b , axiom O3 implies that √ a √ b > 0. Using problem 3.1(I), we compute < ( √ a √ b ) 2 = a 2 √ a √ b + b. By axiom O3, we can add 2 √ a √ b to both sides of this equation to obtain 2 √ a √ b < a + b. ( * ) First of all, we notice that √ a √ b = √ ab . Second of all, by Problem 3.1(II), we know that 1 > 0, and hence 2 = 1 + 1 > 0 + 1 = 1 by O3. Therefore 2 > 0 as well. By Problem 3.1(III), it follows that 1 2 > 0 also. Therefore, by axiom O4, we can multiply both sides of equation (*) by 1 2 to obtain the desired result....
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This note was uploaded on 03/18/2012 for the course MAT MAT 25 taught by Professor Stevenklee during the Fall '10 term at UC Davis.
 Fall '10
 StevenKlee
 Calculus

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