Math 104: Homework 2 solutions
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1 Math 04: Homework solutios. A (0, ): Sice this is a ope iterval, the miimum is udefied, ad sice the set is ot bouded above, the maximum is also udefied. if A 0 ad sup A. B { m + : m, N}: This set does ot have a miimum, sice for ay elemet m +, there is a smaller elemet m+ +. The maximum elemet is, which is attaied for m. Hece max B. B is a bouded below by 0. However, for ɛ > 0, there exists c N such that /c < ɛ, by the Archimedea property. Thus by puttig m c, we see that there exists b B such that b < ɛ. Hece, ɛ is ot a lower boud. Hece 0 is the greatest lower boud, ad thus if B 0. C {x x : x R}: By completig the square, this ca be writte as {(x ) 5 4 x R}. We kow that (x ) 0 by Theorem 3.(iv). Hece mi C 5 4 which is attaied for x. Sice the values of C are ot bouded above, the maximum does ot exist. Hece if C 5 4 ad sup C. D [0, ] [, 3]: Sice this set is composed of closed itervals, we have mi D 0 ad max D 3. Hece if D 0 ad max D 3, ad thus if D 0 ad sup D 3. E [, + ]: The first iterval i this uio is [, 3], ad there are a ifiite umber of cosecutive itervals i the positive directio. Hece mi E, but the maximum does ot exist. Thus if E ad sup E. F (, + ): We begi by showig that F {}. Choose ay x >. The x + ɛ for ɛ > 0. Hece, by the Archimedea property, there exists N such that < ɛ. Hece x / (, + ), ad thus x / F. Similarly if x <, the x ɛ, ad ad there exists a such that x / (, + ). However, (, + ) for all N. Thus F {}, ad hece mi F max F if F sup F.. (a) To begi, we show that sup A + sup B is a upper boud for S. Ay elemet i S ca be writte as a + b for a A, ad b B. However, sice sup A is a upper boud for A, the a sup A. Similiarly, b sup B, ad thus a + b sup A + sup B. We ow wish to show that sup A + sup B is the least upper boud for S. Assume that t is a upper boud for S, but that t < sup A + sup B. The for some ɛ > 0, t sup A + sup B ɛ. Now, sice sup A is the supremum of A, there exists a A such that a > sup A ɛ. (If this was ot the case, the sup A ɛ would be a upper boud for A.) Similarly, there exists b B such that b > sup B ɛ. But a + b S, ad ( a + b > sup A ɛ ) ( + sup B ɛ ) t.
2 Hece t is ot a upper boud, which is a cotradictio. Thus if t is a upper boud, it must satisfy t sup A + sup B. sup A + sup B is a upper boud for S, ad it is the least upper boud. Hece sup S sup A + sup B. (b) This could be proved by repeatig the above argumet but with lower bouds istead of upper bouds. However, a alterative method is to defie egated sets A { a a A}, B { b b B}, ad S { s s S}. We see that S ca be costructed as the set of sums a + b where a A ad b B. Thus, by applyig the above result, we kow that sup( S) sup( A) + sup( B). However, by Corollary 4.5, for ay set C, sup( C) if C. Hece if S if A if B ad thus if S if A + if B. 3. This result is ot true. As a couterexample, choose A B {, }. The sup A sup B, ad hece sup A sup B. However M {,, 4} ad hece sup M 4 which is ot equal to. Note that the couterexample relies o havig two egative terms that multiply together to give a large positive term. If we restrict A ad B to be subsets of the positive real lie, (0, ), the the result sup M sup A sup B would hold, ad could be proved followig similar logic to the previous exercise. 4. (a) By dividig through by, we obtai ( ) ( ) ad sice 0 as, we see that ( ) ( ) 3 9. (b) By makig use of Example i Sectio, we ca write as (+) / + + /
3 (c) We first write ( ) a b a + b ba ( ) c + ba + c where c b/a. Sice a > b > 0, we kow that > c > 0. Thus c 0 as by Theorem 9.7(b), ad hece (a b )/(a + b ) as. (d) Although rapidly becomes much bigger that, we must be careful to show this rigorously. Oe method is to use the biomial theorem to expad for 3 accordig to ( + ) + ( ) + + ad hece, by eglectig all but oe term, ( )( ) 6 ( )( ) >. 6 Now, for 3, we kow that ( ) > ad ( ) > 4, ad hece > ad therefore > 3 /4. Thus for 3, 0 < / < 4/, ad thus / 0 as by the Squeezig Lemma. (e) This ca be carried out by itroducig a factor that completes the square: ( ) ( + )( + + ) + + ( + ) Sice as, we must have + 0 as. 5. (a) Let s for all N. We kow that s is irratioal, sice if s p/q for some itegers p ad q, the p/(q), but has bee show to be irratioal. Now cosider a ɛ > 0. We see that s 0 ad thus if > ɛ, the s 0 < ɛ. Hece s 0 as. 3
4 (b) There are may ways this could be achieved, such as defiig s as the first digits of π, so that the first few terms would be 3, 3., 3.4, 3.4, 3.45, However, here a method is preseted which shows explicitly how to costruct all the umbers i a sequece, ad show that they coverge to a irratioal. Defie s p /q ad put p ad q. Now, defie the rest of the sequece recursively by puttig p + p + q, q + p + q. It is straightforward to see that if p > 0 ad q > 0, the p + > 0 ad q + > 0, so by mathematical iductio s > 0 ad q 0 for all. The first few terms are, 3, 7 5, 7, 4 9, 99 70, 39 69, The last of these is which differs from by Here, we prove that s does ideed coverge to. Suppose that s differs from by a amout, so that p q. () The cosider how much s + differs from : + p + q + p + q p + q p q + p q ( + + ) + + ( ) + +. Sice p /q is positive, we kow from Eq. that + > 0. We also kow that < < 3/ sice < ad (3/) 9/4 >. Hece / < 4
5 < 0. Usig these iequalities, Hece by mathematical iductio, (/), ad thus by Theorem 9.7(b), 0 as. Hece Eq. shows that s p /q as. 6. We ca rewrite a term i the sequece as a product of fractios, ( ) ( ) ( ) s.... Each of these fractios is less tha or equal to oe, ad the first is equal to. Thus s. Now choose ɛ > 0. We see that / s 0 s ad thus we see that for all > ɛ, s 0 < ɛ. Hece lim s A arbitrary polyomial ca be writte as a sum p(x) k a j x j j0 where a j R ad a k 0. To begi, we show by iductio that if s s as 0, the (s ) j s j for all j N {0}. Cosider the base case whe j 0. Sice (s ) 0 for all, this is a costat sequece, ad thus coverges to, which is equal to s 0. Now assume the result is true for j ad cosider the case for j +. We ca defie (s ) j+ (s ) j s ad thus by Theorem 9.4. we kow that (s ) j s s j s s j+ as. Hece the iductio step holds, ad by mathematical iductio (s ) j s j for all j N {0}. Now, if a j is a costat, the we kow that a j (s ) j a j s j by Theorem 9.. Fially, by applyig Theorem 9.3, we see that p(s ) p(s) as. 5
6 8. We begi by provig that s f () which is defied accordig to Cosider the case whe : f () ( t). f () ( t) ( t) t ad thus s f (). Now assume that the result is true for ad cosider the case for + : s + + s ( t) + ( t) (+). ad thus s + f ( + ). Hece by mathematical iductio, s N. f () for all Now, by Theorem 9.7(b), we kow that a 0 if a <. Hece, by usig Theorems 9. ad 9.3 about the scalig ad additio of sequeces we kow that s ( t) 0. 6
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