OUTLINING PROOFS IN CALCULUS. Andrew Wohlgemuth

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1 OUTLINING PROOFS IN CALCULUS Andre Wohlgemuth

2 ii OUTLINING PROOFS IN CALCULUS Copyright 1998, 001, 00 Andre Wohlgemuth This text may be freely donloaded and copied for personal or class use

3 iii Contents Section Page Introduction v 1 Limits 1 Using Definitions and Theorems13 3 Limit Theorems 19 4 Continuity31 5 Derivatives 39 Index45

4 iv Logical Deduction is the one and only true poerhouse of mathematical thinking Jean Dieudonne Conjecturing and demonstrating the logical validity of conjectures are the essence of the creative act of doing mathematics National Council of Teachers of Mathematics

5 v Introduction Consider a point p and a line l in some plane, ith p not on l: Ho many lines are there in the plane that pass through point p and that are parallel to line l? It seems clear, by hat e mean by point, line, and plane, that there is just one such line This assertion is logically equivalent to Euclid's 5th, or parallel, postulate (in the context of his other postulates) In fact, this as seen as so obvious by everyone, mathematicians included, that for to thousand years mathematicians attempted to prove it After all, if it as true, then it should be provable One approach, taken independently by Carl F Gauss, Janos Bolyai, and Nikolai Lobachevsky, as to assume the negation of the obvious truth and attempt to arrive at a logical contradiction But they did not arrive at a contradiction Instead, the logical consequences ent on and on They included theorems, all provable from the assumptions In fact, they discovered hat Bolyai called a strange ne universe, and hat e today call non-euclidean geometry The strange thing is that the real, physical universe turns out to be non-euclidean Einstein's special theory of relativity uses a geometry developed by H Minkoski, and his general theory, a geometry of Gauss and G F B Riemann 1 Mathematics is about an imaginative universe a orld of ideas, but the imagination is constrained by logic The basic idea behind proof in mathematics is that everything is exactly hat its definition says it is A proof that something has a property is a demonstration that the property follos logically from the definition alone On the intuitive level, definitions serve to lead our imagination In a formal proof, hoever, e are not alloed to use attributes of our imaginative ideas that don't follo logically solely from definitions and axioms relating undefined terms This vie of proof, articulated by David Hilbert, is accepted today by the mathematical community, and is the basis for research mathematics and graduate and upper-level undergraduate mathematics courses There is also a very satisfying aspect of proof that comes from our ability to picture situations and to dra inferences from the pictures In calculus, many, but not all, theorems have satisfactory picture proofs Picture proofs are satisfying, because they enable us to see the truth of the theorem Rigorous proof in the sense of Hilbert has an advantage, not shared by picture proofs, that proof outlines are suggested by the very language in hich theorems are expressed Thus both picture proofs and rigorous proofs have advantages Calculus is best seen using both types of proof A picture is an example of a situation covered by a theorem The theorem, of course, is not true about the actual picture the molecules of ink stain on the molecules of paper It is true about the idealization that e intuit from the picture When e see that the proof of a theorem follos from a picture, e see that the picture is in some sense completely general that e can't dra another picture for hich the theorem is false To those mathematicians that are satisfied only by rigorous mathematics, such a situation ould merely represent proof by lack of imagination: e 1 See Marvin Greenberg, Euclidean and Non-Euclidean Geometries, W H Freeman, San Fransisco, 1980 The ord theorem literally means object of a vision

6 vi Outlining Proofs in Calculus can't imagine any situation essentially different from the situation represented by the picture, and e conclude that because e can't imagine it that it doesn't exist Non-Euclidean geometry is an example of a situation here possibilities not imagined are, nevertheless, not logically excluded Gauss 3 kept his investigations in this area secret for years, because he anted to avoid controversy, and because he thought that the Boeotians 4 ould not understand It is surely a profound thing that the universe, hile it may not be picturable to us, is nevertheless logical, and that folloing the logical but unpicturable has unlocked deep truths about the universe In mathematics, unpicturable but logical results are sometimes called counterintuitive Rigorous and picture proofs are both necessary to a good course in calculus and both ithin grasp 5 To insist on a rigorous proof, here a picture has made everything transparent, is deadening Mathematicians ith refined intuition kno that they could, if pressed, supply such a proof and it therefore becomes unnecessary to actually do it We therefore focus the method of outlining proofs on those theorems for hich there is no satisfactory picture proof Our purpose is not to cover all theorems of calculus, but to do enough to enable a student to catch on to the method 6 A detailed formal exposition of the method can be found in Introduction to Proof in Abstract Mathematics and Deductive Mathematics an introduction to proof and discovery for mathematicsd education In calculus texts, examples are given that illustrate computational techniques, the use of certain ideas, or the solution of certain problems In these notes, e give examples that illustrate the basic features of the method of discovering a proof outline 3 One logical consequence of there being more than one line through p parallel to l is that the sum of the number of degrees in the angles of a triangle is less than 180 Gauss made measurements of the angles formed by three prominent points, but his measurements ere inconclusive We kno today that the difference beteen the Euclidean 180 and that predicted for his triangle by relativity ould be too small to be picked up by his instruments 4 A term of derision Today's Boeotians have derisive terms of their on claiming courses that depend on rigorous proof have rigor-mortis 5 The fundamentals of discovering proof outlines can be picked up in a relatively very short time compared to the years of study of descriptive mathematics prerequisite to calculus 6 In the American Math Monthly 10, May 1995, page 401, Charles Wells states: A colleague of mine in computer science ho majored in mathematics as an undergraduate has described ho as a student he suddenly caught on that he could do at least B ork in most math courses by merely reriting the definitions of the terms involved in the questions and making a fe deductions The reason this orked for Prof Well's colleague is that definitions are basic for deductive mathematics and upper-level math courses are taught from a deductive perspective Our method for discovering proof steps is merely a systematic ay of making the deductions a ay that can be taught

7 Section 1 1 Limits Consider the it of a function 0 as the independent variable B approaches some value - If 0 has such a it, it is denoted by 0ÐBÑ This is read the it of 0 of B as B approaches - Þ Here is a picture of a function and its graph that expresses a situation here the it is equal to a number P: We see from the picture that as B goes to, or approaches -, that, in turn, 0ÐBÑ goes to, or approaches P This is the intuitive idea of the it of a function It is necessary to capture this idea in a formal definition Theorems and definitions in formal mathematics are expressed in a language that employs only 7 different types of sentences 7 The types could be listed as: for all (or for every) there exists if, then or and not iff The formal definition of the it of a function uses sentences of these types Suppose that denotes the set of real numbers and H is a subset of on hich a function 0 is defined We have: Definition For a function 0ÀH Ä and real numbers - and P, define 0ÐBÑœP if for every %,!, there BÄexists $,! such that for all B, if! / lb 1 -l / $, then l0ðbñ 1 Pl / % 7 in addition to the sentences expressing relationships beteen the objects e are studying such as numerical equality or set membership

8 Outlining Proofs in Calculus In a definition, e make the thing defined logically equivalent to the defining condition Thus, 0ÐBÑœP is logically equivalent to its defining condition for every %,!, there exists $,! such that for all B, if! / lb 1 -l / $, then l0ðbñ 1 Pl / % Logical equivalence is generally denoted by the ords if and only if abbreviated iff Thus the ording of the definition should technically have been 0ÐBÑœP if and only if for every %,!, there exists BÄá In a definition hoever, the ord if, hen used to relate the thing defined to its defining condition, alays means iff At this point it is imperative that you see that the intuitive idea of it (conveyed by the picture and the concept of movement ) is captured exactly by the formal definition; that is, the formal condition (hich begins ith the ords for every %,! ) is true in situations here the it exists according to our intuition, and the it exists according to our intuition hen the formal definition is true You can't prove this connection beteen the intuitive idea and the formal definition That is not ithin the realm of proof But you must see the connection The hole idea behind formal proof in mathematics is that anything that is proved true about the it of a function must be shon to be a consequence of the formal defining condition The intuitive idea of it should inform all e find to be true about its It should motivate our investigations It can be used to convince ourselves or our sympathetic listeners of the truth of our investigations But it cannot be used in a formal proof Right in the middle of a formal proof e can't appeal, for example, to the physical idea of motion A formal proof depends on the precise meaning (given by accepted rules of inference) of sentences of the seven types listed above Theorems are expressed in terms of this formal language as are definitions and a theorem about the it of a function, for example, must be shon to be true (in a formal proof) solely as a logical consequence of the definition of it As e said in the introduction, hat might appear to be a burden the need to be formal actually turns out to be a help in learning to do one's on proofs: the language in hich formal mathematics is expressed provides the key to the structure of a proof Here is the first example that illustrates the method of developing an outline of a proof structure Example 1 If 0ÐBÑ œ $B4#, then 0ÐBÑ œ "( Most mathematical propositions or theorems assert that something is true (the conclusion) under certain conditions (the hypotheses) The first step in developing a proof of such a theorem or proposition is to identify, in a sentence or to, the hypotheses and conclusion We ill develop proofs themselves in a shaded area that ill differentiate proofs from standard text here e ill talk about proofs The first entry in the shaded area identifies the hypotheses and conclusion: it tells the reader hat you are assuming and hat you ill sho Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( The statement 0ÐBÑ œ $B4# serves to define the function 0 Function definitions are accomplished by giving a rule for hat the function does to any element in its domain some subset of the real numbers The statement 0ÐBÑ œ $B4# is really a customary abbreviation for the more complete, formal statement for all B, 0ÐBÑœ$B4# For all B is read for all B in Function definitions alays need to specify hat the function does to each element in its domain In general, e rite 1À H Ä to express the fact that the function 1 has domain H (a subset of ) and codomain For our function 0 here, H œ Þ The phrase for all B H is alays implied even hen it's not explicitly ritten in a function definition In our development, e ill alays take H to be either the entire set of real numbers or some open interval This ill keep proofs (at our introductory level) free from complicating special cases (such as one-sided its), hile nonetheless exhibiting the generality of the method

9 Section 1: Limits 3 The next step in developing the proof is to rite the conclusion as the last statement at the bottom of the proof A proof is a sequence of statements that lead to the conclusion, so the last statement is alays the conclusion This leaves a gap in the proof 8, namely, all the statements needed to establish the conclusion The method of outlining a proof ill, in general, fill in many, if not all, of these statements Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( 0ÐBÑ œ "( The next step is to consider hat it means for the conclusion 0ÐBÑ œ "( to be true In a formal proof, this meaning can come only from the definition in this case, the definition of it Here again is the definition: Definition For a function 0ÀH Ä and real numbers - and P, define 0ÐBÑœP if for every %,!, there BÄexists $,! such that for all B, if! / lb 1 -l / $, then l0ðbñ 1 Pl / % The next step in the proof outlining process is to rite hat 0ÐBÑ œ "( means by definition, as the next-to-last statement in the proof In a definition, the thing defined is logically equivalent to the defining condition Thus, 0ÐBÑ œ "( is logically equivalent to its defining condition for every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "( In order to prove 0ÐBÑ œ "(, e prove (in the preceding statement) the defining condition for 0ÐBÑ œ "( This is the condition given in the definition applied to the function and it in Example 1: Pœ"(, Hœ, and -œ& ÞThere is no other choice, at this stage, since e have no theorems about its Once e have proved the defining condition, the conclusion ill follo logically from it (by definition) since for any mathematical statement, e can substitute a logically equivalent one Our task is no to prove the statement beginning ith for every %,!, á this has the form for every %,!, (loer-level statement involving %) 8 At this point the gap is the entire proof

10 4 Outlining Proofs in Calculus Mathematics, not just calculus, but all areas of mathematics, can be formalized in terms of the 7 types of statements: for all there exists if, then or and not iff The method of proof outlining depends on rules for proving and using statements of the first 4 types For-all statements have the form for all B E, (loer-level statement involving B), here 4 4 E is some set In particular, E can be the set of positive real numbers The statement B means exactly the same as the statement B,!, so that for all B 4 means the same as for all B,! This, in turn, means the same as for every B 4 or for every B,!, since, if a statement is true for all elements in a set, then it is true for every one of them and vice versa The defining condition for it is therefore one form of a for-all statement Its form for every %,!, (loer-level statement involving %) is equivalent to for all %,!, (loer-level statement involving %) We use calligraphic letters to represent statements Thus c and d may denote statements We may rite cðbñ, for example, if e ish to emphasize that the statement c involves the symbol B Thus, a typical for-all statement is of the form for all B E, cðbñ, here E is some set and cðbñ is a statement involving B The top-level for-all statement contains the loer-level statement cðbñ The ay that mathematicians prove for-all statements is somehat subtle In order to prove a statement of the form for all B E, cðbñ, select some element, say C, from the set E, and then sho cðcñ is true The idea is to prove cðcñ ith no information about C, except that it is in EÞ Such an element of E is called an arbitrary or given element of E If cðcñ is true for an arbitrary element C of E, then it is true for all elements of E The statement for all B E, cðbñ is therefore proved by this procedure In order to prove the statement for all B E, cðbñ that appears at the bottom of a gap in a proof, e insert to ne statements in the gap First, rite let B be arbitrarily chosen in E, let B Ebe arbitrary, let B Ebe given, or some such phrase at the top of the gap This defines the symbol B Second, rite cðbñ at the bottom of the gap It is then necessary to prove c ÐBÑ to fill the remaining gap In our developing proof, e have inserted let %,! be given at the top of the gap, and the statement there exists $,! such that á at the bottom: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "( Our task is no to prove the ne statement at the bottom of the gap, hich has the form there exists $,! such that (loer-level statement involving $ ) In order to prove such statements (called there-exists statements) it is necessary for us to define $, to sho $,!(if this is not obvious), and to sho the loer-level statement involving $ Thus a statement that begins let $ œá needs to be inserted somehere in the gap (e'll leave space above and belo the

11 Section 1: Limits 5 definition of $ ) 9, and the loer-level piece of the there-exists statement needs to be put at the bottom of the gap: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œá For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "( The statement let $ œá is like an assignment statement in computer science It ill be necessary, as the proof develops, for us to define $ in terms of symbols that already exist in the proof (in the hypotheses or preceding statements) At this stage e don't kno ho to define $, but e kno here e ill need to do it So e rough in the ords let $ œá Our task is no to prove the statement for all B, if!/lb1"l/$, then l0ðbñ 1 #l / % It is a for-all statement, at the top level Our rule for proving such statements dictates that e insert a line such as let B be arbitrary at the top of the gap, and the line if! / lb 1 "l / $, then l0ðbñ 1 #l / % at the bottom: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œá Let B be arbitrary If! / lb 1 &l / $, then l0ðbñ 1 "(l / % For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "( The next step is to prove the ne statement at the bottom of the gap hich is of the form if c, then d and is called an if-then statement or implication In order to prove statements of this 9 The proof outlining method depends upon inserting statements that are dictated by the form of the statement that is at the bottom of the gap For the purposes of this method, the ne gap ill be belo the statement let $ œá The space abov e this statement ill be merely set aside for any ork needed to define $ (say, by using the hypotheses) Although let $ œá appears at the bottom of this space, it does not dictate the insertion of other statements

12 6 Outlining Proofs in Calculus form, assume that c is true (at the top of the gap) and sho that d is true under this assumption (at the bottom of the gap) In our proof e have: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œá Let B be arbitrary Assume!/lB1&l/$ l0ðbñ 1 "(l / % If! / lb 1 &l / $, then l0ðbñ 1 "(l / % For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "( Proof outlining is done by orking up from the bottom, beginning ith the conclusion At each stage, statements are added to the outline as they are dictated by the top-level form of the statement at the bottom of the gap hat it means for such a statement to be true This may be by definition, or because the statement has a certain logical form ( there exists, for all, and so on) In the latter case, an inference rule (such as those on page 8) ill dictate statements to be put in the proof gap When the ne statement at the bottom of the gap is not any one of the logical forms at the top level, and has not been defined, e can proceed ith this program no further, When this stage is reached (and not before), it is time to use the hypotheses to help bridge the remaining gap We are no at this stage in our developing proof At the top level, the statement at the bottom of the gap, l0ðbñ 1 "(l / %, is of the form number / number hich has not been defined in our development Thus e use the hypothesis (hich defines the function 0) We do this by substituting $B 4 # for 0ÐBÑ as e ork up from the bottom Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œá Let B be arbitrary Assume!/lB1&l/$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) If! / lb 1 &l / $, then l0ðbñ 1 "(l / % For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / %

13 Section 1: Limits 7 For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "( In a proof, one should alays note here the hypotheses are used, so e have noted in parentheses that l0ðbñ 1 "(l / % follos from lð$b 4 #Ñ 1 "(l / % by hypothesis and the definition of 0 This brings us to the end of the proof outlining process The remaining tasks have been identified: e need to sho that lð$b 4 #Ñ 1 "( l / % follos from! / lb 1 & l / $, and e are alloed to define $ any ay e ish, in order to do this It seems clear that e can multiply lb 1 & l / $ by l$l to get l$l lb 1 & l / l$l $ œ $ $ Thus e get Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) If! / lb 1 &l / $, then l0ðbñ 1 "(l / % It's clear that e need to define $ œ % Î$, so that $ $ œ % hich finishes the proof: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œ % Î$ Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) If! / lb 1 &l / $, then l0ðbñ 1 "(l / % For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "(! The symbol! signals the end of a complete proof By using the same symbol, B, in to different ays, e have made the proof look a little simpler than it really is Recall that the statement 0ÐBÑ œ $B4# serves to define the function 0 It

14 8 Outlining Proofs in Calculus is a customary abbreviation (in a function definition) for the more complete, formal statement for all B, 0ÐBÑœ$B4# The symbol B in the statement for all B, 0ÐBÑ œ $B4# is called a local variable It has no meaning outside the statement; that is, for all B, 0ÐBÑœ$B4# is about 0, not about B Any other letter (not already in use) could serve as the needed local variable Thus for all B, 0ÐBÑœ$B4# and for all >, 0Ð>Ñœ$>4# have exactly the same meaning Part ay don in our proof e have the statement Let B be arbitrary This serves to define the global variable B for all the steps up to, but not including, the statement For all B, if!/lb1& l/ $, then l0ðbñ1"( l/ % This global variable Bis a particular, fixed element of It is not the same B used in the hypothesis to define the function 0 The hypothesis stated that 0ÐBÑ œ $B4#, but this as not (and could not be) a statement about the particular B defined later in the proof In order to avoid this confusion, e could have ritten the hypothesis of the proposition as for all >, 0Ð>Ñœ$>4# Using this hypothesis, e can conclude 0ÐBÑ œ $B4# about the B defined in the proof: since 0Ð>Ñ œ $>4# is true for all >, it is true for the B defined later in the proof steps We have implicitly invoked an inference rule for using for-all statements: Inference Rule Using for-all statements: If the statement for all > E, cð>ñ is true, and if B E, then cðbñis true In our proof development, e have also used the folloing inference rules: Inference Rule Proving for-all statements: In order to prove the statement for all B E, cðbñ, let B be an arbitrarily chosen element of E, then prove that cðbñis true Inference Rule Proving if-then statements: In order to prove the statement if c, then d, assume that c is true and sho that d is true Inference Rule Proving there-exists statements: In order to prove the statement there exists B E cðbñ, define the symbol B, and prove B E and cðbñ for the B you have defined such that In the proof e have developed, the same phrases are ritten over and over again Although this as done to expose the logic of the outlining project, proofs ritten in texts don't have this redundancy In order to make our proof look more like a text proof, e can abbreviate it by omitting those lines consisting of formal statements at the top level; that is, for-all, there-exists, ifthen, and or statements We ill do this one line at a time, in order to see hy Consider our complete proof:

15 Section 1: Limits 9 Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œ % Î$ Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) If! / lb 1 &l / $, then l0ðbñ 1 "(l / % For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "(! The formal if-then statement just belo the lines ith hite background is proved by precisely those lines The if-then statement itself ould not be stated in an informal or narrative proof, since it seems only to say all over again that hich has just been said implicitly (in the hite background lines preceding it) That is, e kno that the if-then statement is true, as soon as e have read the lines above it, so it looks as if the if-then statement is a mere recapitulation In an abbreviated proof, it is omitted When e omit it from our proof, e have the folloing (the area to hich e ill refer next is put in the hite background ahead of time) : Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œ % Î$ Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) For all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "(!

16 10 Outlining Proofs in Calculus Again, the formal for-all statement for all B, if! / lb 1 "l / $, then l0ðbñ 1 #l / % just after the hite background lines seems only to recapitulate hat e kno from these lines It is customarily omitted in an abbreviated proof If e omit it, e get: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œ % Î$ Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) There exists $,! such that for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "(! Similarly, the formal there-exists statement there exists $,! such that for all B, if! / lb 1 "l / $, then l0ðbñ 1 #l / % just after the hite background lines recapitulates hat e kno from these lines, and e delete it to get: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œ % Î$ Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) For every %,!, there exists $,! such that for all B, if!/lb1&l/ $, then l0ðbñ 1 "(l / % 0ÐBÑ œ "(! Finally, the formal for-all statement for every %,!, there exists á is deleted There are to reasons that justify this deletion First, as e have seen before, it seems only to repeat hat e kno to be true by the hite background lines preceding it Second, by the definition of it, it is equivalent to the statement 0ÐBÑ œ "( just belo it In our minds, e identify the statement 0ÐBÑ œ "( ith its defining condition We prove the defining condition, but e think of this as

17 Section 1: Limits 11 proving the it statement Thus the definition of it is not ritten out explicitly as a proof step, but used implicitly We justify the last step by mentioning this definition This gives: Assume 0ÐBÑ œ $B4# We ill sho 0ÐBÑ œ "( Let %,! be given Let $ œ % Î$ Let B be arbitrary Assume!/lB1&l/$ l$l lb 1 &l / l$l $ œ $ $ l$b 1 "&l / $$ lð$b 4 #Ñ 1 "(l / % l0ðbñ 1 "(l / % (by hypothesis and the definition of 0) Therefore 0ÐBÑ œ "( by the definition of it! Thus all statements that ere introduced by the proof outlining procedure and that ere of one of the forms or, for-all, there-exists, or if-then have been deleted to abbreviate the proof These statements dictated the form of the proof and once they have done this, need not be retained We no change the series of lines to paragraph style, add a fe ords to soften the abruptness, and explain anything e think the reader might not see: Assume for all B, 0ÐBÑœ$B4# We ill sho 0ÐBÑœ"( Let %,! be given Let $ œ % Î$ and let B be arbitrary Then if e assume! / lb1&l / $, e get l$l lb 1 &l / l$l $ œ $ $, so that l$b 1 "&l / $ $ ; that is, lð$b 4 #Ñ 1 "(l / % Therefore l0ðbñ 1 "(l / % by definition of 0 Therefore 0ÐBÑ œ "( by the definition of it! In this final version of the proof, e neglect to mention that e are using the hypothesis hen e say by definition of 0 The hypothesis being the place here 0 is defined, e necessarily must be using it so e needn't say so Although the final version of the proof is ritten in the ay that proofs are most often ritten, it is no more correct than the original, long-inded version It merely involves many things implicitly that in the original version ere explicit Even in the original, long-inded version, the inference rules ere used implicitly They ere not mentioned in the proof itself, but they dictated the form of the proof In the author's texts mentioned in the introduction, everything is done explicitly at first Furthermore, the statement defining it in the present development is a statement of numerical inequality contained in an if-then statement contained in a for-all statement contained in a there- exists statement contained in a for-all statement We therefore begin our development ith a definition that involves 5 levels of nested logical statements In the more basic texts mentioned in the introduction, on the other hand, statements build sloly in complexity The early material in these texts hile still developing the proof outlining method is thus accessible to students ith less maturity than that necessary for folloing our development here In turn, our development here demands much less maturity than the riting of proofs ithout the benefit of a method hich is beyond most students' grasp

18 1 Outlining Proofs in Calculus A proof is a sequence of statements leading to the conclusion Each of these statements must either define ne symbols or involve only symbols that have been previously defined either in the hypotheses, or earlier in the proof statements Consider the variables in the sentences in the proof above The first sentence Assume for all B, 0ÐBÑœ$B4# merely copies the hypothesis The symbol B is a local variable hose scope is restricted to the statement itself It is free to be used again in other ays in the proof, and is used again as a local variable in the next statement e ill sho 0ÐBÑ œ "( The statement let B be given, hoever, defines the symbol B as a global variable hose scope is the set of statements ending ith l0ðbñ 1 "(l / % This is the set of statements needed to prove the top-level for-all statement for all B, if! / lb 1 &l / $, then l0ðbñ 1 "(l / % This use of B as a global variable ceases ith the statement l0ðbñ 1 "(l / % just before the for-all statement It is therefore free to be used as a local variable in the for-all statement (if the for-all statement is included in the proof) It is also used as a local variable in the conclusion therefore 0ÐBÑ œ "( The symbol B is thus used over and over again in the proof in different ays Although this is customary, it is absolutely essential that one symbol not represent to different things at the same time In particular, e can't use a currently defined global variable also as a local variable: e cant talk about all > 's, if > has already been defined and is still in use as some particular number Note that in the final, abbreviated proof, the symbols % and $ each have a single use as a global variable, folloing their definitions in the proof itself In order to illustrate the general breakdon of a proposition into hypotheses and conclusion, e orded Example 1 to have a function definition in the hypotheses This need not be done explicitly Þ In the folloing example, e are asked merely to sho that Ð$B 4 #Ñ œ "( Example Sho that Ð$B 4 #Ñ œ "( It is implicit that $B 4 # is a function (and not a number, say), since the only things that have its as B goes to & are functions of B We can change our proof to one based on this phrasing of the proposition (no hypothesis all conclusion) merely by leaving out the hypothesis and riting $B 4 # in place of 0ÐBÑ We also don't bother to say e ill sho Ð$B 4 #Ñ œ "(, since that is obvious With these changes, e have Let %,! be given Let $ œ% Î$ and let B be given Then if e assume!/lb1&l/ $, e get l$l lb 1 &l / l$l $ œ $ $, so that l$b 1 "&l / $ $ ; that is, lð$b 4 #Ñ 1 "(l / % Therefore Ð$B 4 #Ñ œ "( by the definition of it! EXERCISE 1 Let 0ÐBÑ œ &B1$ Prove 0ÐBÑ œ (ÞDon't do this exercise by copying steps in the proofs given BÄ# above Instead, copy the procedure used to get the steps

19 Section 13 Using Definitions and Theorems In developing a proof outline for the example in the previous section, a statement of the form number / number presented itself at the bottom of the gap Since the relation / had not been defined, and since the statement as not one of the basic logic forms (there exists, for all, and so on), the outlining process came to an end at that point In this section, e ill define the relation /, in order to make a fe important points about proof Our aim is not to seek to justify algebraic manipulations or computations logically 11 Generally, our employment of logical reasoning ill be restricted to the development of a proof outline Familiar (and, perhaps, not-so-familiar) algebraic computations ill be accepted as part of a proof on the same basis that they have been accepted in mathematics prerequisite for the calculus Today, the generally accepted foundation for mathematics is given in terms of sets Thus everything, including number, is ultimately defined in terms of sets Our definition of the relation / ill be in terms of the set 4 of positive real numbers We ill assume as an axiom about this set that it is closed under addition and multiplication That is: 4 4 Axiom For all elements + and, in, the sum +4, and product +, are also in Definition For numbers + and, in the set, define + /, iff there exists 4 such that +4 œ, Assume +/, and,/- for +ß,ß- Þ We ill sho +/-Þ The next step in the outlining process is to put the definition of immediately preceding this statement: Proposition 1 (Transitivity of /) For real numbers +ß,ßand -, if +/, and,/-, then +/- +/- +/- as the line 11 All mathematics is based on logical principles The familiar algebraic computations are logically based on properties of the set of real numbers the axioms for its being a complete, totally-ordered field Some of these axioms, notably the commutative, associative, and distributive properties of addition and multiplication, are normally shon to be the basis of algebraic manipulation in school mathematics

20 14 Outlining Proofs in Calculus Assume +/, and,/- for +ß,ß- Þ We ill sho +/-Þ There exists 4 such that +4œ- +/- The rule for proving there-exists statements dictates that e need to define the element, (to sho here this is done, e rough in the line let œá ), and then sho for this that 4 and that +4œ-sho: This leaves to gaps in the proof above the to things e need to Assume +/, and,/- for +ß,ß- Þ We ill sho +/-Þ Let œá Þ 4 Þ +4œ- There exists 4 such that +4œ- +/- We can ork up from the bottom no longer, so e no use the hypotheses By the first hypothesis, +/,, e can add the defining condition there exists 4 such that +4 œ, to the proof as the second line This there-exists statement ill serve to define the element for the rest of the proof Also, from this statement e may infer the statements 4 and +4 œ, about this Formally, e say that the rule for using there-exists statements allos us to do this Informally, e merely take the there-exists statement at its ord Either ay, is defined by the there-exists statement Hoever, this may not be the same that e ant later in the proof, for the lines that follo from let œá (the statement e have roughed in) In the proof, e therefore must use a symbol other than in the condition defining the relation + /, We'll use Similarly, e need to use some symbol such as in the defining condition for the second hypothesis: Assume +/, and,/- for +ß,ß- Þ We ill sho +/-Þ There exists 4 such that +4 œ, (by the first hypothesis) There exists 4 such that,4 œ - (by the second hypothesis) Let œá Þ 4 Þ +4œ-

21 Section : Using Definitions and Theorems 15 There exists 4 such that +4œ- +/- This ends the proof outlining process, and leaves us ith a ell-defined (computational) task We need to sho +4œ- from the equations +4 œ, and,4 œ -, and e're free to define any ay e ant, in order to do this If e add to both sides of +4 œ,, e get +4 4 œ,4 Since,4 œ-, e get +4 4 œ- It seems clear that e need to define œ 4 Since 4 is closed under addition, e have 4 This completes the proof: Assume +/, and,/- for +ß,ß- Þ We ill sho +/-Þ 4 There exists such that +4 œ, (by the first hypothesis) There exists 4 such that,4 œ - (by the second hypothesis) Let œ Then, since is closed under addition Adding to the equation in line gives: +4 4 œ,4 Substituting line 3 gives: +4 4 œ- Therefore +4œ-, by the definition of There exists 4 such that +4œ- +/-! The next-to-last line can be deleted, since it recapitulates the lines before it (starting ith let œ 4 ) Although lines and 3 are expressed in formal language, they can't be deleted, since they serve to define the numbers and This gives: Assume +/, and,/- for +ß,ß- Þ We ill sho +/-Þ There exists 4 such that +4 œ, (by the first hypothesis) There exists 4 such that,4 œ - (by the second hypothesis) Let œ Then, since is closed under addition Adding to the equation in line gives: +4 4 œ,4 Substituting line 3 gives: +4 4 œ- Therefore +4œ-, by the definition of Therefore +/-, by the definition of /! The notation +,, means exactly the same as,/+ The use of one or the other might depend, not on mathematics, but on its linguistic context For example, e might ish to make + the subject and is greater than, the predicate, or e might ish to make, the subject and is

22 16 Outlining Proofs in Calculus less than + the predicate In formal mathematics, hoever, everything 1 is defined in terms of previously defined things A formal definition of the relation,, in terms of the previously defined relation / ill serve to make a point Definition For +ß,, define +,, iff, / + Proposition (Transitivity of,) For real numbers ;, <, and =, if ;,< and <,=, then ;,= A proof of Proposition begins, as usual, ith the identification of hypotheses and conclusion, and the conclusion as the last line: Assume ;,< and <,= for ;ß<ß= We ill sho ;,= ;,= At the top level, the conclusion is of the form number, number The meaning of this is given by the definition of,, hich gives us the next-to-last line: Assume ;,< and <,= for ;ß<ß= We ill sho ;,= =/; ;,= No the statement at the bottom of the gap is of the form number / number The normal development of the proof outline method ould have us use the definition of / to rite the preceding line Hoever, e no also have a theorem (Proposition 1) about /, hich e might use Supposedly not knoing hich is best at this stage, e could abandon the bottom-up development and use the information from the hypotheses: Assume ;,< and <,= for ;ß<ß= We ill sho ;,= </; (by the first hypothesis and the definition of,) =/< (by the second hypothesis and the definition of,) =/; ;,= 1 except the feest possible undefined terms, ith properties given by axioms

23 Section : Using Definitions and Theorems 17 It's clear no that the gap can be closed merely by appealing to Proposition 1 The next-tolast line =/; follos from the second and third lines, by Proposition 1 This gives the complete proof: Assume ;,< and <,= for ;ß<ß= We ill sho ;,= </; (by the first hypothesis and the definition of,) =/< (by the second hypothesis and the definition of,) =/; (by Proposition 1) ;,=(by the definition of,)! Consider the stage of the proof just before e employed Proposition 1: Assume ;,< and <,= for ;ß<ß= We ill sho ;,= </; (by the first hypothesis and the definition of,) =/< (by the second hypothesis and the definition of,) =/; ;,= The second and third lines are the hypotheses of Proposition 1 The statement at the bottom of the gap is the conclusion of Proposition 1 If e had had failed to use Proposition 1, but continued the proof outlining process using the definition of /, e ould merely have repeated the proof of Proposition 1, no ithin the proof of Proposition In principle, it's not necessary to use theorems to prove other theorems Everything must come from definitions Practically, hoever, e couldn't possibly rite out complete proofs, if e didn't appeal to theorems The use of theorems to prove other theorems is an essential shortcut But it is, nevertheless, a shortcut Proof at its most basic level depends on definitions The easiest proofs the ones a mathematician ould call straightforard are the proofs for hich the outlining method produces a complete proof While the outlining method doesn't reveal creative steps need to fill a gap in a proof, it at least gets one going Furthermore, as the footnote on page vi attests, it is poerful enough to enable students to do very ell in many advanced courses The material in this section has illustrated the folloing formal rule: Inference Rule Using there-exists statements: The statement there exists B E such that cðbñ defines the symbol Bas a global variable and allos us to use the statements B E and cðbñ about B

24 18 Outlining Proofs in Calculus EXERCISE Use the proof outlining method and the definition of / to sho that one can add the same number to each side of an inequality That is, prove that for +ß,ß-, if +/,, then +4-/,4- Note that adding the same number to both sides of an equality can be justified by substitution: if +œ,, then substituting, for + in +4-œ+4- produces +4-œ,4-, but this doesn't ork for inequality

25 Section 3 19 Limit Theorems With the proof of Theorem 1, e ill begin the process of abbreviation from the start of the development of a proof rather than developing a long-inded version to be abbreviated later The theorem assumes that the functions 0 and 1 have a common domain H Definition If 0À H Ä and 1À H Ä, define the function 0 4 1À H Ä by Ð0 4 1ÑÐBÑ œ 0ÐBÑ 4 1ÐBÑ Theorem 1 If 0ÐBÑ œ P and 1ÐBÑ œ Q, then Ð0 4 1ÑÐBÑ œ P 4 Q We first state the hypotheses and conclusion at the top of the proof, and then rite the conclusion as the last line Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Ð0 4 1ÑÐBÑ œ P 4 Q The condition equivalent to Ð0 4 1ÑÐBÑ œ P 4 Q, by the definition of it, is no BÄput as the statement immediately preceding the conclusion: Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Þ For every %,!, there exists $,! such that for all B H, if!/lb1-l/ $, then lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q The rule for proving for-all statements dictates that the folloing statements (ith hite background) be added to the proof:

26 0 Outlining Proofs in Calculus Assume 0ÐBÑœPand 1ÐBÑœQ We ill sho Ð0 4 1ÑÐBÑ œp4q Let %,! be given There exists $,! such that for all B H, if!/lb1-l/ $, then lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % For every %,!, there exists $,! such that for all B H, if!/lb1-l/ $, then lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q After adding the statements ith hite background, e can delete the formal for-all statement for every %,!á just belo them Note that this has the effect of replacing the forall statement ith the to statements one at the top and one at the bottom of the gap If one ere developing the proof on scrap paper, one could merely erase for every %,!, from the for-all statement and put let %,! be given at the top of the gap hich ould leave the there-exists statement at the bottom of the gap No need to capitalize the t It ill be erased in the next step Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Let %,! be given There exists $,! such that for all B H, if!/lb1-l/ $, then lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q The statement at the bottom of the gap is a there-exists statement The rule for proving such statements says that e must define $ in the proof steps, and prove the loer-level statement for all B H, á that involves this $ We therefore add the statement define $ œá in the gap, and erase the ords there exists $,! such that to get the ne statement at the bottom of the gap: Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Let %,! be given Define $ œá For all B H, if!/lb1-l/ $, then lð0 4 1ÑÐBÑ 1ÒP4QÓl/ % Ð0 4 1ÑÐBÑ œ P 4 Q

27 Section 3: Limit Theorems 1 The for-all statement at the bottom of the gap dictates that the line let B Hbe arbitrary be added at the top of the gap, and that for all B H be erased to leave an if-then statement at the bottom: Assume 0ÐBÑœPand 1ÐBÑœQ We ill sho Ð0 4 1ÑÐBÑ œp4q Let %,! be given Define $ œá Let B Hbe arbitrary If! / lb 1 -l / $, then lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q The if-then statement dictates that e add a ssume!/lb1-l/$ at the top of the gap an erase the ords if!/lb1-l/$, then from the statement at the bottom This gives: Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Let %,! be given Define $ œá Let B Hbe arbitrary Assume!/lB1-l/$ lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q It is common to contract the to lines just above the gap into a single line: assume!/lb1-l/ $ for an arbitrary B H Thus instead of applying the rules for proving for-all statements and if-then statements one at a time as e have done above, e do it all at once ith a contracted rule for proving statements of the very common form for-all-if-then This formal rule is analogous to the ay e prove a theorem after deciding on its informal hypotheses/conclusion interpretation The rule, illustrated by statements in our proof, allos us to infer the for-all-if-then statement belo from the to statements above it:

28 Outlining Proofs in Calculus Assume!/lB1-l/ $ for an arbitrary B H lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % For all B H, if!/lb1-l/ $, then lð0 4 1ÑÐBÑ 1ÒP4QÓl/ % In general, e have the folloing rule: Inference Rule Proving for-all-if-then statements: In order to prove the statement for all > E, if cð>ñ, then dð>ñ, assume cð>ñ for an arbitrary > E, then prove that dð>ñ is true Incorporating this in our outline, e have: Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Let %,! be given Define $ œá Assume!/lB1-l/ $ for an arbitrary B H lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q We can continue to ork up from the bottom of the gap, if e use the definition of the function 041: Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Let %,! be given Define $ œá Assume!/lB1-l/ $ for an arbitrary B H lð0ðbñ 4 1ÐBÑ 1 ÒP 4 QÓl / % lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q It is time to use the hypotheses From the first hypothesis, 0ÐBÑ œ P, e infer the BÄdefining condition: for every %,!, there exists $,! such that for all B H, if!/lb1-l/ $, then l0ðbñ 1 Pl / % This statement could be ritten at the top of the first gap just belo the

29 Section 3: Limit Theorems 3 statement Let %,! be given This latter statement, hoever, defines % as a global variable for use in the set of statements up to the next-to-last line of the proof, so e can't use the same symbol % as a local variable ithin this set of statements Since % has been defined as a specific number by the statement let %,! be given, e can't talk about all % or every % in the very next statement We therefore use some other symbol such as % in the defining condition Also, since the $ given by the defining condition may not be the same $ e need to define later (it on't be) e use the symbol $ in the defining condition We thus get the folloing: Assume 0ÐBÑ œ P and 1ÐBÑ œ Q We ill sho Ð0 4 1ÑÐBÑ œ P 4 Q Let %,! be given By the first hypothesis, e have for all %,!, there exists $,! such that for all B H, if! / lb 1 -l / $, then l0ðbñ 1 Pl / % By the second hypothesis, e have for all %,!, there exists $,! such that for all B H, if! / lb 1 -l / $, then l1ðbñ 1 Ql / % Define $ œá Assume!/lB1-l/ $ for an arbitrary B H lð0ðbñ 4 1ÐBÑ 1 ÒP 4 QÓl / % lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q At this point of the proof outline, it remains to define $, and for that $ to sho that lò0ðbñ 4 1ÐBÑÓ 1 ÒP 4 QÓl / % follos from! / lb 1 -l / $ We can use the validity of to forall statements: for all %,!, á and for all %,!, á In order to do this, e need to apply the statement to specific values of the variables % and % (by the rule for using for-all statements) That is, % and % must be defined in terms of the only other (non-local) variable e have in the proof at that stage, namely % In order to see ho to do this, e ork up from the bottom no using computations: Ò0ÐBÑ 4 1ÐBÑÓ 1 ÒP 4 QÓ œ Ò0ÐBÑ 1 PÓ 4 Ò1ÐBÑ 1 QÓ By the proof lines obtained from the hypotheses, e can make Ò0ÐBÑ 1 PÓ and Ò1ÐBÑ 1 QÓ as small as e ant: less than % and %, respectively, here e can choose % and % any ay e ant If e make 0ÐBÑ1P less than % Î# and also 1ÐBÑ 1 Q less than % Î#, then certainly their sum ill be less than % Actually, e need to make the absolute value of their sum less than % The ay to handle absolute values is ith the triangle inequality, hich states that for all numbers + and,, l+ 4,l Ÿ l+l 4 l,l Continuing to ork up from the bottom, and applying the triangle inequality, e get: Assume!/lB1-l/ $ for an arbitrary B H lò0ðbñ 4 1ÐBÑÓ 1 ÒP 4 QÓl œ lò0ðbñ 1 PÓ 4 Ò1ÐBÑ 1 QÓl Ÿ l0ðbñ 1 Pl 4 l1ðbñ 1 Ql / % Î# 4 % Î# œ % Therefore lð0 4 1ÑÐBÑ 1 ÒP 4 QÓl / % Ð0 4 1ÑÐBÑ œ P 4 Q