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By DAVID ALEXANDER BRANNAN

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2 1. Solve the following inequalities: pffiffiffiffiffiffiffiffiffiffiffiffiffi xþ1 (a) xxÀ1 (b) 4x À 3 > x; 2 þ4 < x2 À4;   (c) 17 À 2x4  15; (d) jx þ 1j þ jx À 1j < 4. 3 1. Use the Triangle Inequality to prove that jaj 1 ) ja À 3j ! 2: 2. Prove that (a2 þ b2)(c2 þ d2) ! (ac þ ad)2, for any a, b, c, d 2 R. 3. Prove the inequality 3n ! 2n2 þ 1, for n ¼ 1, 2, . : 35 1: Numbers 36 (a) by using the Binomial Theorem, applied to (1 þ x)n with x ¼ 2; (b) by using the Principle of Mathematical Induction. 4. Use the Principle of Mathematical Induction to prove that, for n ¼ 1, 2, .

We can express this formally as follows: A sequence {an} is null if and only if the corresponding sequence {janj} is null. 2. A null sequence {an} remains null if we add, delete or alter a finite number of terms in the sequence. Similarly, a non-null sequence remains non-null if we add, delete or alter a finite number of terms. 3. If one number serves as a suitable value of X for the inequality in (1) to hold, then any larger number will also serve as a suitable X. Hence, for simplicity in some proofs, we may assume if we wish that our initial choice of X in (1) is a positive integer.

In this case, Theorem 1 asserts the existence of a real number b such that b2 ¼ 2. In other pffiffiffi words, it asserts the existence of a decimal b which can be used to define 2 precisely. Here is a direct proof of Theorem 1 in this special case. 414, . . 3. 5 Manipulating real numbers 33 12 < 2 < 22 ð1:4Þ2 < 2 < ð1:5Þ2 ð1:41Þ2 < 2 < ð1:42Þ2 ð1:414Þ2 < 2 < ð1:415Þ2 .. (1) This process gives an infinite decimal b ¼ 1:414 . ; and we claim that b2 ¼ ð1:414 . 414 .. 414 .. 999396 .. Notice that b ¼ 1:414 .

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