1 Section 2.2, Absolute value

Transcription

1 .Math 0450 Hoors itro to aalysis Sprig, 2009 Notes #6 1 Sectio 2.2, Absolute value It is importat to uderstad iequalities ivolvig absolute value. I class we cosidered the iequality jx 1j < jxj ; ad discussed a graphical method of solutio. A secod method is to square both sides. For ay a 2 R; jaj 2 = a 2 ;by Theorem 2.2.2(a).Hece, jx 1j < jxj, (x 1) 2 < x 2 by Example (a). Note that this uses the fact that jx 1j 0: Therefore, jx 1j < jxj, x 2 2x + 1 < x 2 (by expadig (x 1) 2 ), 2x + 1 < 0, 1 < 2x, 1 2 < x: Note that i i some steps of this proof, I used results from sectio 2.1 without explaatio. We will assume these results from ow o, except for homework from 2.1. However, it would be easy to make a mistake o the rst step, ad try to square a iequality without kowig that both terms are positive. Hece, I cited a reaso there. I had to adjust a little to get everythig ito coheret seteces. This is the same aswer as we got graphically. If a problem says :\Fid all x such that jx 1j < jxj.", the you ca d your aswer by ay method you wat, but for full credit you the have to prove that your aswer is correct usig algebra. Here is a secod example: (Corollary 2.2.4(a)) Prove that jjaj jbjj ja bj : 1

2 First method: \cases". Oe method for hadlig absolute value iequalities is to cosider various cases. For this problem, these are: a > 0; b > 0 a < 0; b > 0 a > 0; b < 0 a < 0; b < 0: (i) If a 0 ad b 0; the the iequality becomes which is certaily true. (ii) If a > 0 ad b < 0; the ja bj ja bj ; jjaj jbjj = j a bj = ja + bj jaj + jbj while a b < 0; so ja bj = b a = jbj + jaj because a < 0: Combiig the last two lies gives jjaj jbjj ja bj : (iii) If a > 0; b < 0; the use (ii), substitutig b for a ad a for b: (iv) If a < 0; b < 0, the jjaj jbjj = jj aj j bjj ad use (i) d method { square both sides Method 1 was a fair amout of work. did earlier. We get Istead, we ca square the iequality, as we j jaj jbj j 2 = jaj 2 2 jaj jbj + jbj 2 = a 2 + b 2 2 jabj a 2 + b 2 2ab = (a b) 2 : Here we have use the priciple that jxj x 2

3 for ay x 2 R; with x = ab. Obviously this is a lot quicker. It is good to look for ways to hadle several cases at oce, but sometimes, cosiderig each case separately is uavoidable or at least more straightforward. The text uses still aother method to prove this result. { The idea of a "-eighborhood is very importat. etire secod half of page 33. We will use this ofte. Be sure you uderstad the 2 Completeess property Read rst: 2.2.3, Recall that earlier we showed that there is o ratioal umber r = m such that r 2 = 2. We ca ask whether there is ay real umber r such that r 2 = 2: I other words, \Does p 2 exist?" Ca we deduce the existece of p 2 from the axioms we gave earlier (A1-A4,M1-M4,D, plus order axioms (i), (ii), (iii)? The aswer is o. This is obvious because the set Q of ratioal umbers obeys all of those axioms, ad yet withi Q; p 2 does ot exist. So, we eed aother axiom for R: This is called the \completeess" property. Several deitios are eeded to state this property Be sure you read them o page 35. We eed rst the terms \upper boud", \bouded above", \supremum". The latter is abbreviated \sup". Here are some examples: Example 1 Let A = m : m 2 N; 2 N; m <. The 2 is a upper boud for A; so A is bouded above. However 2 is ot the supremum of A. Istead, sup A = 1. I this case, sup A =2 A. Example 2 A = fx 2 R : 0 < x 2:5g. The 147 is a upper boud for A. However (obviously!), 147 is ot the supremum of A. Istead, sup A = 2:5: I this case, sup A 2 A. Example 3 A = fx 2 R : 0 < x < 3g : I this case, sup A = 3 =2 A. 3

4 A questio you probably would ot thik to ask is: Is N bouded? Surprisigly, we ca't aswer that questio based o the axioms we have had so far. We will eed the completeess axiom, which I will ow state: Completeess axiom for R : Every subset of R which is bouded above has a supremum i R: Notice that Q does ot satisfy that property. Let A = fr 2 Q : r 2 < 2g. The 13 is a upper boud for A (which is i Q), but A does ot have a supremum i Q 2 sice, as we will see, a supremum would have to satisfy r 2 = 2. Theorem 4 N is a ubouded subset i R. Proof. Suppose that N is bouded. The by the completeess axiom, N has a supremum, say u. The by deitio of supremum, u 1 is ot a upper boud for N : There must be a iteger 2 N such that > u 1: By 2.1.7(b), + 1 > u: Hece, u is ot a upper boud for N ; a cotradictio. This theorem is ofte called the \Archimedea property" : is a x 2 N such that x > x. If x 2 R; the there Corollary (2.4.4) If S = x : x = 1 for some 2 N, the if S = 0. (see the text for the deitio of if.) Proof: 0 1 for each 2 N, so 0 is a lower boud for S. By the completeess property, S has a imum. Say the imum is p. The p 0 because 0 is a lower boud. Suppose that p > 0: The 1 2 R; so there is a 2 N with > 1: The p p < p ad p is ot a lower boud. Hece, p = 0: 1 I the text, these properties are used to prove that p 2 exists. The proof is surprisigly complicated, ad I may ot have time to go over it i class. Homework, due Thursday, Jauary 29. Remember to use complete seteces. Do't give a strig of equatios or iequalities that are ot part of a logical setece, usig correct grammar. pg. 30, # 16 (a,d), (Aswers i back. Prove these aswers, givig a reaso for each step (each ew = ad each ew iequality. pg. 34, # 13(d), 15.(give a specic ", i terms of a ad b. Prove your " works.) pg. 38, # 4 (brief reaso, ot a proof) 4

5 pg. 38, # 5. (Cite a specic reaso for ay step that requires a result from sectio 2.3. You ca omit the citatio whe the reaso is somethig from before sectio 2.3.). pg. 43, # 2 (aswer i back; explai it), 3, 8(a).9(a) (aswer i back; explai it.) 5