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By Mischa Cotlar

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1). 3. Prove that n j= A(n) : j=1 n(n + 1) , n ≥ 1. e. e. A(1) holds. Now suppose that A(n) holds for arbitrary but fixed n ∈ N. We want to show that then A(n + 1) holds too. Indeed we have n+1 n j= j=1 j + (n + 1), j=1 and this is already the crucial step since it allows us to use statement A(n), namely n+1 n j= j=1 j + (n + 1) j=1 n(n + 1) + 2(n + 1) n(n + 1) + (n + 1) = 2 2 (n + 1)(n + 2) , = 2 which is A(n + 1). 4. For x = 1 the statement n xn+1 − 1 , n≥0 x−1 xj = A(n) : j=0 holds. Recall that x0 = 1, thus we have n = 0 the statement A(0) is correct: 0 xj = x0 = 1 and j=0 n j j=0 x = 1 + x + x2 + · · · + xn .

5in reduction˙9625 2 THE ABSOLUTE VALUE, INEQUALITIES AND INTERVALS We may also look at unions of intervals which is less problematic since we do not need to solve inequalities however we might have to combine them. 20 Note that in the case of closed or half-open intervals we may meet some new possibilities(compared with open intervals). The two intervals (a, b] and (b, c), for example, do not intersect (a, b] ∩ (b, c) = {x ∈ R | a < x ≤ b and b < x < c} = ∅, however (a, b] ∪ (b, c) = {x ∈ R | a < x ≤ b or b < x < c} = {x ∈ R | a < x < c} = (a, c).

83) x ≤ y implies a · x ≤ a · y. 84) a > b > 0 and x > y > 0 imply a · x > b · y. 87) x ≤ y implies a · x ≥ a · y. 88) In the next section we will often make use of these rules. Here are some simple examples: i) 7 3 7 7 3 ≤ , hence 4 · = 3 ≤ = 4 · , 4 8 4 2 8 however (−4) · 3 7 7 = −3 ≥ − = (−4) · . 4 2 8 ii) 3 + x > 2 + y implies 1 + x > y or y − x < 1. 5in reduction˙9625 1 NUMBERS - REVISION iii) Consider 7x−5 > 21x+30. This inequality is equivalent to 7x > 21x+35, which is again equivalent to x > 3x + 5, or −5 > 2x, implying x < − 52 .

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