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Unformatted text preview: n such that na > b. Proof. Suppose not. ie. na ≤ b for all integers n . Then since a > 0 we have n ≤ b a for all n ∈ Z . Hence b a is an upper bound for Z . By LUB Z must have a least upper bound u . ie. for some u ∈ R n ≤ u for all n ∈ Z . 1 2 8. Prove √ 2 + √ 3 + √ 5 is irrational. HINTS: (i) Suppose not; i.e. suppose √ 2 + √ 3 + √ 5 is rational. (ii) By squaring show that this implies √ 6 + √ 10 + √ 15 is rational. (iii) By squaring again show that this implies 5 √ 6 + 3 √ 10 + 2 √ 15 is rational. (iv) Hence show that this implies that 3 √ 15 + 2 √ 10 is rational. (v) Hence show that this implies √ 150 is rational to obtain a contradiction....
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 Spring '08
 Staff
 prime factorizations, test roman letters, Proof. Suppose not., LUB Z

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