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# INTER ANALYSIS I MATH 431

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This 11 page Class Notes was uploaded by Sadie Schroeder on Wednesday September 30, 2015. The Class Notes belongs to MATH 431 at Western Kentucky University taught by David Neal in Fall. Since its upload, it has received 10 views. For similar materials see /class/216744/math-431-western-kentucky-university in Applied Mathematics at Western Kentucky University.

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Date Created: 09/30/15

Dr Neal WKU MATH 431431g Euclidean Space For each positive integer n we let Rn denote the set of all ordered n etuples of the form u u1 un where ul 6 SR for 1Sz39 S n Elements in Rquot are called vectors But for n 1 we generally just refer to them as real numbers or scalars and write SR rather than R1 Henceforth let u u1 un and v 21 vn be vectors in Rquot 0 Sum and Scalar Multiplication i We define the sum of vectors u and v by u vu1v1 un vn ii For c 6 SR the scalar product cu is defined by cu c ul c un We can then define the difference v u by v u v 1u v1 u1 vn un which gives the directional vector from u to v With these operations of addition and scalar multiplication Rquot is a vector space over the field of real numbers Properties of Addition 1 lfu v e Rnthen uv ERquot closure 2 uvw uvw for all u v w e Rquot associative 3 u v vu for all u v e Rquot commutative 4 There exists a 0 element in Rquot denoted 6 such that 6 v v v 6 for all v e Rquot additive identity Specifically 6 0 0 which we call the zero vector 5 For every v e Rquot there exists an element v such that v v 6 v v additive inverse Specifically 7v1 vn 421 ivn Properties of Scalar Multiplication 1 If v e Rquot and c is any scalar ie real number then cv e Rquot closure 2 Cdv cdv for all v e Rquot and all c d 6 SR associative 3 cuv cu cv for all u v e Rquot and all c 6 SR scalar distributive 4 cdv cv dv for all v e Rquot and all c d 6 SR vector distributive 5 For the scalar 1 1V v for all v e Rquot multiplicative identity Dr Neal WKU I Norm The norm or length or magnitude of a vector u e Rquot is defined by J L112 un2 In R2 the norm of the vector u x y is simply the length of the segment from the origin 0 0 to the point x y and is given by lxz y2 Properties of Norm i u20andu Oifandonlyifu 6 ii u12 unz iii Icu cgtltu for every scalar c 6 SR x ifoO x ifo039 Moreover ifo 0 then x2 0 thus xSIxI for all x 6 SR Also le2 x2 forall x For x 6 SR the norm is simply the absolute value I x J Ix II Distance The distance duv between vectors u and v is given by the length of the vector from u to v 2 2 duvv uquotv1 u1 vn un Properties of Distance Forallforallu v e Rquot i duv Z 0 ii duu 0 and duv 0 ifand only ifu v iii duv dvu Note For xy e SRwehave dxyy xx y III Dot Product We define the dot product uv also called inner product between vectors u and v ian by uv u1v1unvn Properties of Dot Product a uu u12 un2 Z 0 b lull Juu c uv vu d For any scalar c cuv c X uv and ucv c X uv e For another vector w u v w u w v w Dr Neal WKU Theorem Cauchyischwarz Inequality Let u and v be vectors in Rquot Then Iuvl S lull Ivll Proof Let k be a scalar and consider the vector ku v Then 0 SIIku vl2 ku vku v k2uu kuvkvuvv u2 k2 2uvkv2 This expression defines a quadratic in k which is always noninegative so the i quadratic has zero roots or just one root Thus the discriminant b2 4ac must be less than or equal to 0 where au2 b 2uv and c v2 If the discriminant were positive then the quadratic would have two roots and therefore would have to be negative over some interval Thus 132 4616 4uv2 4u2v2 s 0 which implies that uv2Su2 v2 By taking square roots we obtain v S Iu QED Note In SR the result is actually lfxy GER then xy x y The proof in SR is much easier than Cauchyischwarz in general Simply consider three cases i If x Z 0 andy20then xyZO andthuslxyxyxy ii lfxlt0 and ylt0 then xygt0 and xy xy 1x 1y xIIyI iii If one of xy is negative and the other non negative then xy S 0 thus xy xy xIIyI Theorem Triangle Inequality Let u and v be vectors in Rquot Then uv S lull Ivll Proof We simply expand Iu v2 The first inequality below arises from the fact that the number uv is less than or equal to its absolute value The second inequality is from Cauchyischwarz IuvII2 uvuvIIuI2IVI2 2uv 2 2 Slull v 2luvl Slullzv22uv Ilullv2 By taking square roots we obtain Iu Vquot S Iu QED Dr Neal WKU 0 We frequently will use the triangle inequality applied to real numbers x and y xy Sxyl The proof in SR may be done a little differently as follows xy2xy2 x2 2xyy2 sz 2xyy2 x22xyy2 xy2 By taking square roots we obtain Ix y SI x I I y I Note For vectors u and v in R2 the triangle inequality states that length of the diagonal u v can be no more than the sum of the lengths of the two sides u and v hence the name triangle inequality uV The result can be generalized to the distance V between vectors as shown next Shortest Distance Between Points Theorem Let u and v be vectors in Rquot For any other vector w we have duv duwdwv Proof We use the definition of distance and apply the triangle inequality duv v ul IIv ww ul Sllvwwul dwv duw duwdwv QED The result states that the distance directly from u to v is less than or equal to the sum of the distances from u to w then from w to v We note that if u v and w are collinear with w between u and v then duv will equal duwdwv We shall use the result repeatedly in SR as Ix yx ww yISx ww y Dr Neal WKU Exercises 1 Prove that for all integers n 2 1 x1 x2 xnSx1x2 xn for all real numbers x1 x2 xn 2 For all real numbers x and y prove that lellyIISIXyl 3 Let u v e Rquot Expand and simplify Iu v2 7u v2 Dr Neal WKU MATH 431431 g Proofs This entire course requires you to write proper mathematical proofs All proofs should be written elegantly in a formal mathematical style Complete sentences of explanation are required Do not simply write an equation you must explain what the equation is giving and or why it is being used Moreover all equations must be properly aligned with no scratch outs Always give a conclusion We will begin by reviewing several standard methods of proof using the following basic definitions and using the facts that the sum and product of integers are integers Divisib ity We say that a divides b if there exists an integer k such that b ka For example 6 divides 18 because 18 3 X 6 where k 3 is an integer Even Number An integer n is said to be even if it has the form n 2k for some integer k That is n is even if and only if n divisible by 2 Odd Number An integer n is called odd if it has the form n 2k 1 for some integer k Prime Number A natural number n gt 1 is said to be prime if its only positive divisors are 1 and n If n has other positive divisors then n is called composite Rational Number A real number x is called rational if x can be written as a fraction m n where m and n are integers with n i 0 Otherwise x is called irrational Direct Proof Consider the statements S1 E If p then q and S2 E Vx px At times we may try to prove that these types of statements are true To prove S1 directly we assume p is true and then argue that q must also be true To prove S2 is true we pick an arbitrary x and argue that property px must hold Example 1 Prove the following results directly a If n is an odd integer then n2 1 is even b If a divides b and b divides c then a divides c c For all rational numbers x and y the product x y is also a rational number Proof a Assume n is an odd integer Then n 2k1 for some integer k We then have n2 1 2k12 1 4k2 4k 2 22k2 2k 1 21 where 1 2k2 2k 1 is an integer because k is an integer Thus n2 1 has the form of an even number and so n2 1 is even QED Dr Neal WKU b Assume a divides b and b divides c Then b ka for some integer k and c fl for some other integer j Thus 0 jb jkajka Because the product of integers jk is still an integer we have that a divides c QED c Let x and y be rational numbers Then x m n and y j k for some integers m n j and kwith n i 0 and k i 0 Then a X 1 k xy 7 The products of integers m j and nk are still integers and nk cannot be 0 because neither n nor k is 0 Thus x y is in the form of a rational number and therefore x y is rational QED Indirect Proofs Given an implication S E p gt q its contrapositive is N q gt N p which is logically equivalent to the original implication In order to prove that S is true it might be easier to prove that the contrapositive is true That is we can assume that q is not true and then argue that p is not true Another common method used to prove p gt q is a proof by contradiction In this case we assume that p is true but that q is not true We then argue that a mathematical contradiction occurs We conclude that q in fact must be true Proof by Contrapositive Proof by Contradiction To prove p gt q To prove p gt q assume that q is not true assume that p is true but q is not true then argue that p is not true Then argue that a mathematical contradiction occurs Notes i A logical statement is often a conjunction aA b a and b or a disjunction a v b a or b We apply DeMorgan s Laws to obtain the negations of these statements as follows NabENaVNb NavbENaNb ii The negation of one or the other but not both is given by both or neither iii A logical statement may be in terms of a quantifier such as for every x property px holds The negation is there exists an x for which px does not hold Statement Vx px Negation El x N px Dr Neal WKU Example 2 Prove the following results a If c is a nonzero rational number then c T is irrational b If m3 is even then m is even c For all real numbers x and y if xy gt 0 then xgt 0 or ygt 0 d Let n be an integer If n2 is divisible by 3 then n is divisible by 3 e Let x bea real number lfxlt e forall s gt 0 then x 0 a Proof By contradiction Let c be a nonzero rational number and suppose that c T is rational Then on m n where m n are integers and n i 0 Moreover C j k where j k are integers and k i 0 but also j i 0 because c is nonizero We then have 1 m k m km TE gtlt gtlt c n j n jn The products of integers km and jn are still integers and jn cannot be 0 because neither n nor j is 0 Thus 11 is in the form of a rational number which is a contradiction because 11 is irrational Ergo c 11 must be irrational QED b Claim If m3 is even then m is even We shall prove the contrapositive instead we shall assume that m is not even and show that m3 is not even That is we shall assume that m is odd and show that m3 is odd Proof Assume that m is odd Then m 2k 1 for some integer k Then m3 2k13 8k3 12k2 6k1 24k3 6k2 3k1 211 where l is the integer 4k3 6k2 3k Thus m3 is odd because it is written in the form of an odd integer By contrapositive if m3 is even then m is even QED c Proof Let x and y be real numbers Assume that xS 0 and yS 0 Then by adding we have xyS 00 0 That is xyS 0 Hence xS 0 and yS 0 implies that xyS 0 By contrapositive if xy gt 0 then xgt 0 or ygt 0 Dr Neal WKU d Claim For an integer n if n2 is divisible by 3 then n is divisible by 3 Proof Assume n is not divisible by 3 We then can apply the Fundamental Theorem of Arithmetic to write n uniquely as a product of powers of primes as follows nplrl Xp2r2 gtlt kark where rl are natural numbers and the pi are distinct primes Because n is not divisible by 3 then p1 3 for 1 Si S k By squaring we obtain 2 2 n plrl Xp2r2 Xkark p12r1gtltp2272 gtlt X ka 7k which is a prime factorization of n2 By uniqueness of prime factorization the above p1 for 1 Si S k are the only prime factors of n2 Because none of the p equals 3 then n2 is not divisible by 3 By contrapositive if n2 is divisible by 3 then n must also be divisible by 3 QED Note The preceding result can be generalized as follows Let p be prime and suppose k 2 2 lf nk is divisible by p then n is divisible by p In the proof of d simply replace 3 with p and replace the exponent 2 with k e Claim lfIxIlt s forall a gt 0 then x 0 Proof Suppose Ix I lt e for all s gt 0 We must show that x 0 So suppose that x i 0 Then xlt0 or xgt0 In either case we have Ixgt0 Now let a Ix2gt0 For this a we then have Ix I gtI xI 2 e which contradicts the assumption Thus we must have that x 0 QED Note We also could have argued by contrapositive Assume x i 0 Then we showed that there exists an s gt 0 for which Ix I lt e fails So if x i 0 then it is not true that Ix I lt e for all e gt 0 By contrapositve if it is true that Ix I lt s for all e gt 0 then x 0 Double Implication A double implication is of the form p gt q which is read p if and only if q This statement means p gt q and q gt p and both implications must be proven Example 3 Let n be an integer Then n is even if and only if n2 is even Dr Neal WKU Proof Suppose first that n is even Then n 2k for some integer k So then n2 2k2 4k2 22k2 Because 2k2 is also an integer we see that n2 must be even Next we must prove that if n2 is even then n is even But we shall prove the contrapositive instead So assume that n is not even ie that n is odd Then n 2k 1 for some integer k We then have n22k124k24k122k22k12j1 where j 2k2 2k is an integer Hence n2 is odd that is n2 is not even We have proven that if n is not even then n2 is not even By contrapositive if n2 is even then n is even From both directions we now have that n is even if and only if n2 is even QED Note The second direction also follows from the generalization of Part d using the prime p 2 In other words if 2 divides n2 then 2 divides n Theorem J5 is irrational Proof Assume J5 is rational Then we can write J5 mn where m and n are integers with n i 0 Furthermore we can assume that the fraction is reduced so that m and n have no common divisors Then by squaring the fraction we have 2 f or 2n2m2 n Thus m2 is even because it is divisible by 2 It follows that the integer m must be even by Example 3 So we may write m 2k for some integer k Then 2 2 2 2k 2n m2 2k2 4k2 22k2 which gives 71 Thus n2 is even because it is divisible by 2 It follows that the integer n also must be even So both m and n are even and therefore both are divisible by 2 But this fact contradicts the assumption that we have chosen m and n to have no common divisors This contradiction leads us to conclude that 5 cannot be written as a fraction Thus J5 is irrational QED Corollary For any prime p J is irrational Dr Neal WKU Exercises Prove the following results in a formal elegantly written mathematical style a For all integers m and n if m is even and n is odd then m n is odd b Let a be an integer If a divides b then b is an integer and a divides b2 0 If x and y are rational numbers then x y is a rational number d Let m and n be integers If m n is even then m is even or n is even e Let m and n be integers If m n is odd then m is odd or n is odd but not both f Let x and y be real numbers Then x y if and only iflx y lt 1 for every 2 gt 0 g For any prime p J is irrational

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