Math 104, Solution to Homework 2 Instructor: Guoliang Wu June 21, 2009
Ross, K. A., Elementary Analysis: The theory of calculus : Page 18: 3.3, 3.4, 3.6, 3.7, 3.8 Page 25: 4.1 – 4.4 (only (b), (j), (n), (v)), 4.6, 4.7, 4.11, 4.15
0 3.3 Prove Prove that that (iv) (iv) (−a)(−b) = ab for all a, b; (v) ac = bc and c = imply a = b. Proof. (iv) Using result (iii) in Theorem 3.1 (Ross), we have
(−a)(−b) = −a(−b) = −(−b)a = −(−ba) = ba = ab. (v) ac = bc ⇒ ac + (−bc) = bc + (−bc) = 0 ⇒ ac + (−b)c = 0 (by (iii)). 0 , c− 1 Then by DL and M2, we obtain c(a + (−b)) = 0. Since c = exists, and by M4 and (ii), a + (−b) = c−1 c(a + (−b)) = c−1 · 0 = 0 .
Finally, add b to both sides and use A1, a + ((−b) + b) = 0 + b ⇒ a=b 3.4 Prove that that (v) 0 < 1; (vii) if 0 < a < b, then 0 < b−1 < a−1 . Proof. (v) 1 = 1 · 1, so 0 ≤ 1 by (iv) of Theorem 3.2 (Ross). How 1, so 0 < 1. ever, 0 =
(vii) 0 < a implies that 0 < a−1 by (vi) (vi).. Also Also 0 < a < b implies 1 − 0 < b hence 0 < b again by (vi). Starting Starting from a < b, we have 1 1 1 − − − a·a < b·a and then (a · a ) · b−1 < (b · a−1 ) · b−1 ⇒ b−1 < b · (a−1 · b−1 ) = b · (b−1 · a−1 ) = (b · b−1 ) · a−1 = 1 · a−1 = a−1 . So 0 < b−1 < a−1 . 3.6 3.6
(a) (a) Prove Prove that that |a + b + c| ≤ |a| + |b| + |c| for all a,b,c ∈ R. 1
(b) Use induction to prove
|a1 + a2 + · · · + a | ≤ |a1 | + |a2 | + · · · + |a | n
n
for n numbers a1 , a2 , · · · , a . n
Proof. (a) Using Triangle inequality twice,
|a + b + c| = |(a + b) + c| ≤ |a + b| + |c| ≤ |a| + |b| + |c|. (b) (1) When n = 1 , the inequality holds since |a1 | ≤ |a1 |. (2) Suppose the inequality holds for n numbers, i.e.,
|a1 + a2 + · · · + a | ≤ |a1 | + |a2 | + · · · + |a |. n
n
We want to show that
|a1 + a2 + · · · + a + a n
+1
n
| ≤ |a1 | + |a2 | + · · · + |a | + |a n
+1
n
|.
By Triangle Inequality and the above assumption,
|a1 + a2 + · · · + a + a +1 | = |(a1 + a2 + · · · + a ) + a ≤||a1 + a2 + · · · + a | + |a +1 | ≤|a1 | + |a2 | + · · · + |a | + |a +1 |. n
n
n
n
+1
n
|
n
n
n
By induction, we proved the claim. 3.7
(a) Show that |b| < a if and only if −a < b < a. (b) Show that |a − b| < c if and only if b − c < a < b + c. (c) Show that |a − b| ≤ c if and only if b − c ≤ a ≤ b + c. To prove a statement like “ A if and only if B ” (or, “A and B are equivalent”), we need to show that (1) A implies B , and (2) B implies A. Proof. It is clear that
−|x| ≤ x ≤ |x| for any real number x since |x| is either x, or −x. (You can use this fact later on.)
2
(a) “⇒”: From the above inequality, −|b| ≤ b ≤ |b|. So b ≤ |b| < a which implies b < a by O3. (b = a cannot happen since otherwise, |b| = a.) On the other hand, −|b| ≤ b ⇒ −b ≤ |b| < a. Thus −b < a. So −a < b. We proved that
−a < b < a. “⇐”: If b ≥ 0, then |b| = b. Thus |b| < a since b < a. If b < 0, then |b| = −b < a since −a < b. (b) By what we have proved in (a),
|a − b| < c if and only if − c < a − b < c. Since −c < a − b < c if and only if b − c < a < b + c, we have
|a − b| < c if and only if b − c < a < b + c. (c) The same argument as in (a) shows that |b| ≤ a if and only if −a ≤ b ≤ a. (This is Exercise 3.5 (a).) “⇒”: Again, we use the obvious fact that −|b| ≤ b ≤ |b|. So b ≤ |b| ≤ a which implies b ≤ a by O3. On the other hand, −|b| ≤ b ⇒ −b ≤ |b| ≤ a. Thus −b ≤ a. So −a ≤ b. We proved that −a ≤ b ≤ a. “⇐”: If b ≥ 0, then |b| = b. Thus |b| ≤ a since b ≤ a. If b < 0, then |b| = −b ≤ a since −a ≤ b. Therefore,
|a − b| ≤ c if and only if − c ≤ a − b ≤ c. Since −c ≤ a − b ≤ c if and only if b − c ≤ a ≤ b + c, we have
|a − b| ≤ c if and only if b − c ≤ a ≤ b + c. 3.8 Let a, b ∈ R. Show that if a ≤ b1 for every b1 > b, then a ≤ b. Proof. We prove it by contradiction. Suppose b < a, then we take 1 b1 = (a + b).
2 We check that b1 > b since a + b > 2b. However, b1 < a since a + b < 2a. We obtain a contradiction. This proves that a ≤ b. 3
4.1 – 4.4 (b) (0, 1) is bounded above: it has lots of upper bounds. For instance, 1, 2, 3. It has lower bounds 0, −1, −2 (or any nonpositive number.)
sup(0, 1) = 1,
inf(0, 1) = 0.
, · · · }. Upper bounds: 1, 2, 3; lower (j) {1 − 31 : n ∈ N} = { 23 , 89 , 26 27 2 bounds: 3 , 0, −1. n
sup{1 −
1 : n ∈ N} = 1 , 3 n
inf {1 −
1 2 : n ∈ N} = . 3 3 n
(n) {r ∈ Q : r < 2}. Upper bounds: 2, 3, 4. The set is NOT bounded below.
sup{r ∈ Q : r < 2} = 2. (We will learn the notation −∞. This set is not bounded below, so it does not have a lower bound/infimum. But sometimes we write inf {r ∈ Q : r < 2} = −∞ to mean that it is not bound below.)
(v) {cos 3 : n ∈ N}. Note that this is in fact a finite set since cos is a cyclic function: nπ
{cos
nπ 3
1 2
1 1 : n ∈ N} = { , − , −1, 1}, 2 2 2π 3
= , cos = − since cos = , cos = 1 , · · · − , cos π
1 2
5π 3
3
1 2
6π 3
1 2
, cos
3π 3
= −1, cos
4π 3
=
Upper bounds: 1, 2, 3; lower bounds −1, −2, −3.
sup{cos
nπ 3
: n ∈ N} = 1,
inf {cos
nπ 3
: n ∈ N} = −1.
4.6 Let S be a nonempty bounded subset of R. (a) Prove that inf S ≤ sup S . (b) What can you say about S if inf S = sup S . Proof. (a) Take any element s ∈ S (nonempty). Note that sup S is an upper bound of S and inf S is a lower bound, thus s ≤ sup S and inf S ≤ s. Therefore
inf S ≤ sup S. 4
(b) If inf S = sup S , then the set S has only one element inf S (same as sup S ). This is because, as argued in (a), for any element s ∈ S , inf S ≤ s ≤ sup S = inf S. Then it must be that
inf S = s = sup S = inf S. Otherwise, there is a contradiction inf S < inf S . Any element s equals inf S means that the set S has only one element. 4.7 Let S and T be nonempty bounded subsets of R. (a) Prove that if S ⊂ T , then inf T ≤ inf S ≤ sup S ≤ sup T . (b) Prove that sup(S ∪ T ) = max{sup S, sup T }. Proof. (a) In 4.6 (a), we proved that inf S ≤ sup S for any subset S of R. So we need to prove that inf T ≤ inf S and sup S ≤ sup T .
To show that inf T ≤ inf S , it suffices to show that inf T is a lower bound of S , then it follows that inf T ≤ inf S since inf S is the greatest lower bound (greater than or equal to any other lower bound of S ). For any s ∈ S , it belongs to T since S ⊂ T . Thus inf T ≤ s since inf T is a lower bound of T and s ∈ T . Then, by definition, inf T is a lower bound of S and therefore inf T ≤ inf S . Similarly, to show sup S ≤ sup T , we need only to show that sup T is an upper bound of S . For any s ∈ S , it is also in T because S ⊂ T . Thus s ≤ sup T . Therefore sup T is an upper bound of S . (b) Note that S ⊂ S ∪ T and T ⊂ S ∪ T . We have
sup S ≤ sup(S ∪ T );
sup T ≤ sup(S ∪ T ).
So
max{sup S, sup T } ≤ sup(S ∪ T ). On the other hand, for any element s ∈ S ∪ T , either (i) s ∈ S or (ii) s ∈ T . If (i) happens (s ∈ S ), then s ≤ sup S ≤ max{sup S, sup T }. 5
Otherwise (ii) happens (s ∈ T ), which implies that s ≤ sup T ≤ max{sup S, sup T }. In either case s ≤ max{sup S, sup T }.
Since s is taken arbitrarily from S ∪T , we see that max{sup S, sup T } is an upper bound of S ∪ T . Thus
sup(S ∪ T ) ≤ max{sup S, sup T }. Finally, we conclude that
sup(S ∪ T ) = max{sup S, sup T }.
4.11 Consider a, b ∈ R where a < b. Use denseness of Q to show that there are infinitely many rationals between a and b. Proof. See textbook (P313).
4.15 Let a, b ∈ R. Show that if a ≤ b +
1 n
for all n ∈ N, then a ≤ b.
Proof. Suppose not, i.e., b < a. Then a − b > 0. By Archimedean Property, there exists an integer n such that n(a − b) > 1, which implies that a − b > 1 and a > b + 1 . This is a contradiction to what we have: a ≤ b + 1 for all n ∈ N. Therefore a ≤ b. n
n
n
6