hw2[1]

# hw2[1] - m such that m> 1 a ± b But then a ± b< 1 m...

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HOMEWORK 2: SOLUTIONS/HINTS 3.3 By (A4), (M2) and (DL) we get that ( ± a ) b + ab = ( ± a + a ) b = 0 ² b which is zero by Theorem 3.1 (ii), hence ( ± a ) b = ± ab . For part (iv), note that ( ± a )( ± b ) = ± a ( ± b ) by part (iii). Now a ( ± b ) + ab = a ( ± b + b ) = 0 by (DL) and part (ii) again, so ( ± a )( ± b ) = ± a ( ± b ) = ab 3.5 Part (a) is easily checked by distinguishing the cases b ³ 0 and b < 0. For part (b), the triangle inequality gives us j b j ´ j b ± a j + j a j as well as j a j ´ j a ± b j + j b j . Thus ±j a ± b j ´ j a j ± j b j ´ j a ± b j and therefore we have jj a j ± j b jj ´ j a ± b j by part (a). 3.6 We prove part (b) by induction on n . The case n = 1 is clear and the case n = 2 is the triangle inequality. So let n ³ 3, then j a 1 + ::: + a n j ´ j a 1 j + ::: + j a n ± 2 j + j a n ± 1 + a n j ´ j a 1 j + ::: + j a n j by induction respectively the triangle inequality. 4.6 Let x 2 S , then we have by de±nition inf S ´ x ´ sup S: For (b) we then get that inf S ´ x ´ sup S = inf S and thus x = inf S = sup S . As x 2 S was arbitrary, it follows that S is just one point. 4.7 See the solution in the book, p. 313. 4.11 See the solution in the book, p. 313. 4.15 Suppose a > b . Then a ± b > 0 and so 1 a ± b > 0 : By the Archimedean Property there is a natural number

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Unformatted text preview: m such that m > 1 a ± b . But then a ± b < 1 m and therefore we get the contradiction a > b + 1 m : 7.4 (a) Let x n := p 2 n for n ³ 1, then x n is irrational since p 2 is so and lim n !1 x n = 0 ; which is rational. (b) De±ne x n = (1 + 1 n ) n for n ³ 1. Then each x n is rational and lim n !1 x n = e; which is irrational. 1 2 HOMEWORK 2: SOLUTIONS/HINTS 7.2 In the following let ± > 0. (a) Let N > 1 ± , then j a n j = ± ± ± ± n n 2 + 1 ± ± ± ± ± ± ± ± n n 2 ± ± ± = 1 n < ± for n ² N and hence lim a n = 0 : (d) Let N > 16 25 ± ± , then we get ± ± ± ± d n ³ 2 5 ± ± ± ± = ± ± ± ± 16 5(5 n + 2) ± ± ± ± < ± ± ± ± 16 25 n ± ± ± ± ± ± ± ± ± 16 25 N ± ± ± ± < ± for any n ² N and therefore the limit is just 2 5 : (e) Pick N > 1 ± , then j s n j = 1 n ± 1 N < ± for n ² N and thus lim n !1 s n = 0...
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