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Unformatted text preview: c) add 3 to both sides to get x 2 4 = 0 , use distributivity to factor and conclude like in b). d) Same: apply theorem 2.1.3 ( a.b = 0 implies a = 0 or b = 0 ). square Exercise 3. Page 30, #8. Proof. a) Clearly a b + c d = a.d + b.c b.d ∈ Q and a b . c d = a.c b.d ∈ Q . b) If x is rational and y irrational, then x + y can’t be rational because y = ( x + y ) + ( x ) would also be rational. Now if in addition x , then x.y can’t be rational because y = ( x.y ) . (1/ x ) would also be rational from a). square Exercise 4. Page 30, #18. 1 Proof. By contradiction: assume that a > b . Then take ε = a b 2 . One should have a b lessorequalslant ( a b 2 ) which is absurd. square Exercise 5. Page 30, #23. Proof. Let us prove that if a, b > then a < b if and only if a n <b n for any n ∈ N . Clearly the right hand side implies the left one. Let’s prove the rest by induction: 1. The property is true for n = 1 because a < b ⇒ a 1 < b 1 . 2. Assume that one has a n <b n . Then a.a n < a.b n <b.b n so we are done. square Exercise 6. Page 34, #1. Proof. a) Clearly vextendsingle vextendsingle a  2 = a 2 , and since vextendsingle vextendsingle a vextendsingle vextendsingle...
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 Fall '10
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 Irrational number

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