HW12 sol.pdf - MATH 2345 section 54\/55 Homework 12 Solutions Must be submitted in a one-page PDF file on D2L Q1 Prove by induction 1 2 \u22ef = Basis step

# HW12 sol.pdf - MATH 2345 section 54/55 Homework 12...

• Homework Help
• 2

This preview shows page 1 out of 2 pages.

Unformatted text preview: MATH 2345 section 54/55 Homework 12 Solutions Must be submitted in a one-page PDF file on D2L )()+,) . Q1. Prove by induction: 1" + 2" + ⋯ + " = ( Basis step: 1" = ( / for all ≥ 1. . ,(,+,) . / which is true since 1=1. . Inductive step: We assume that 1" + 2" + ⋯ + " = ( (2+,)(2+.) . 2" + ⋯ + ( + 1)" = ( 2(2+,) . / is true and we show that 1" + . / is true: . . ( + 1) 1 + 2 + ⋯ + ( + 1) = 1 + 2 + ⋯ + + ( + 1) = 4 5 + ( + 1)" 2 . ( + 1). 4( + 1)" . ( + 1). + 4( + 1)" = + = 4 4 4 ( + 1). [ . + 4( + 1)] ( + 1). [ . + 4 + 4] ( + 1). ( + 2). = = = . 2 4 4 " " " , " " , " , " , ) Q2. Prove by induction: ,×. + .×" + "×; + ⋯ + )()+,) = )+, for all ≥ 1. , , , , Basis step: ,×. = . which is true since . = . . Inductive step: We assume that , , , 2+, , ,×. , , , 2 + .×" + "×; + ⋯ + 2(2+,) = 2+, and we show that + "×; + ⋯ + (2+,)(2+.) = 2+. . .×" 1 1 1 1 + + + ⋯+ ( + 1)( + 2) 1×2 2×3 3×4 1 1 1 1 1 = + + + ⋯+ + 1×2 2×3 3×4 ( + 1) ( + 1)( + 2) 1 ( + 2) 1 = + = + + 1 ( + 1)( + 2) ( + 1)( + 2) ( + 1)( + 2) ( + 1). ( + 2) + 1 . + 2 + 1 +1 = = = = . ( + 1)( + 2) ( + 1)( + 2) ( + 1)( + 2) + 2 Q3. Prove by Induction: For every integer ∈ , we have 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +··· + × ( + 1) = ( + 1)( + 2) . 3 , ,×. + 7 7 7 d. R7 is transitive. 8. a. 2. * 3 b. R 5 is not reflexive because, for example, (2, 2) Basis 1×2= c. R5 step: is symmetric. ,(.)(") " ¢ R5 . which is true since 2 = 2 . d. R5 is transitive. Inductive step: We assume that 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +··· + × ( + 1) = 2(2+,)(2+.) " ()+,)()+.)()+") 10. and S t =we{(0, 0), that (0,3),1 (1,0), (1,3),+( (2, 2), (2,3), (3,+3),2)(3,=O)} show × 2 +(1,2), 2 × 3(2,O), + 3 ×(3,2), 4 + 4(0,2), × 5 +··· + 1) × ( . " * 3(1,2), (2,1), (3, 2), (3,O), (OO), (0,1), (1, 1), (1,3), (3,3), 11. Tt 2. = {(0, 2), (1,0), (2,3), (3,1), (0,3), 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +··· +( + 1) × ( + 2) (2,2), (2,O)} = 1because, × 2 + 2 for × 3example, +3×4+ + × ( + 1) + ( + 1) × ( + 2) b. R 5 is not reflexive (2, 42)×¢5R+··· 5. ( 1)( +X2)for all real numbers x, x C x. By definition of C this 13. C is not reflexive: C is + reflexive c. R5 is symmetric.= ( = (this + 1) + 2) means that for all real numbers x, x2 + + x2 1. × But is false. As a counterexample, take 3 2 d. transitive. x=0.R5 isThenx +x2 ( =02+02=07 + 1)( + 2)1. 3 × ( + 1) × ( + 2) ( + 1)( + 2)[k + 3] = + = . 3 (2,O), 3 (2, x2),and 3 y C x. By is {(0, symmetric: is symmetric X for all (0,2), real numbers y, if(3,x3),C (3, y then 10. SCt = 0), (0,3),C(1,0), (1,2), (3,2), (1,3), (2,3), O)} 2 definition of C this means that for all real numbers x and y, if X2 + y = 1 then y 2 + X2 = 1. 2 2 whether given relation reflexive, of these. 11. Q4. Tt {(0,is2),true (1,0), (2,3),the (3,1), (0,3), (1,2),isof (2,1), (3, 2),symmetric, (1,orall 1),none (1,3), (3,3), But=Determine this because by commutativity addition, x(3,O), + y (OO), = transitive, y 2 (0,1), + X2 for real numbers . . is the circle relation on the set of real numbers: For all , ∈ ℝ, ⇔ + = 1. x and y. (2,2), (2,O)} C is not transitive: C is transitive t# for all real numbers x, y, and z, if x C y and y C z 2 13. C is xnot C is reflexive for allthat realfornumbers x C x.x, By definition then C reflexive: z. By definition of C thisX means all real x, numbers y and z, if X2 of + yC =this 1 2 2 that for all real numbers 2 means x, x + x2 = 1. But this is false. As a counterexample, and y + Z2 = 21 then X2 + Z = 1. But this is false. As a counterexample, take x = 0, y =take 1, x=0. and z =Thenx 0. Then+x2 x2 +=02+02=07 y 2 = 1 because1. 02 + 12 = 1 and y2 + Z2 =1 because 12 + 02 = 1, but 2 92 is + symmetric: Z2 l 1 because +0 2 = 0 5 X1. for all real numbers x and y, if x C y then y C x. By C C is0 symmetric definition of C this means that for all real numbers x and y, if X2 + y2 = 1 then y 2 + X2 = 1. 16. But F is this reflexive: any integer. Since m-nm, =x02 and 0,2 we have that 5 1 (m - m). 25 is trueSuppose because mbyis commutativity of addition, + y = y + X2 for all real numbers Consequently, m F m by definition of F. x and y. F is symmetric: Suppose m and n are any integers such that mr F n. By definition of F this C is not transitive: C is transitive t# for all real numbers x, y, and z, if x C y and y C z means that 5 1 (m - n), and so, by definition of divisibility, m - n = 5k for some integer k. then x C z. By definition of C this means that for all real numbers x, y and z, if X2 + y2 = 1 Now n - m = -(m - n). Hence by substitution, n - m = -(5k) = 5 (-k). It follows that and y 2 + Z2 = 1 then X2 + Z2 = 1. But this is false. As a counterexample, take x = 0, y = 1, 51 (n - m) by definition2 of divisibility (since integer), and thus n F m by definition and z = 0. Then x2 + y = 1 because 02 + 12-k= is1 an and y2 + Z2 =1 because 12 + 02 = 1, but of F. 2 2 92 + Z2 l 1 because 0 +0 = 0 5 1. F is transitive: Suppose m, n and p are any integers such that m F n and n F p. By definition 16. Q5. Fof is reflexive: m(misgiven integer. m-nm, =by0 definition and 5 transitive, 0, of wedivisibility, haveorthat 5m-1of (m -= m). F, this meansSuppose that 5 1the - any n) and 5 1(nSince and so,symmetric, nthese. 5k Determine whether relation is-p), reflexive, none for some integerm k,F and n-p 51 forofsome Consequently, m by definition F. integer 1. Now m-p = (m- n) + (n-p). Hence by is the relation defined on ℤ as follows: For all , ∈ ℤ, ⇔ − is odd.   substitution, mp = 5k + 51 5(k + 1). It follows 1(m -mrp)F by of divisibility F is symmetric: Suppose m and n are any integersthat such5 that n. definition By definition of F this (since k + 1 is an integer), and thus in F p by definition of F. means that 5 1 (m - n), and so, by definition of divisibility, m - n = 5k for some integer k. - 17. 298 Now - reflexive: m = -(m0- isn). Hence -'by for substitution, m 0= m. -(5k) 5 (-k).of It0 follows that 0 is n not reflexive all integers nmn,- m By = definition this means 51 (n m) by definition of divisibility (since -k is an integer), and thus n F m by definition Solutions for Exercises: Relations that for all integers m, m - m is odd. But this is false. As a counterexample, take any integer of m.F.Then m - m = 0, which is even, not odd. Fo isis transitive: and np are are any any integers integers such such that that m m 0F n. n and F p. By of definition symmetric:Suppose Supposem,mnand By n definition 0 this of F, this means 5 1(mand - n)so and 5 1(n -p), ofand m- k.n Now = 5k means that m - that n is odd, by definition oddso, m by - ndefinition = 2k + 1 of fordivisibility, some integer for some integer k, and n-p 51 for some integer 1. Now m-p = (mn) + (n-p). Hence n -m =- (m -n). Hence by substitution, n -m = -(2k + 1) = 2(-k- 1) + 1. It follows thatby substitution, p = 5k + 51 5(k(since+ 1). Itkfollows 5 1(m -and p) by definition of divisibility n -m is odd mby definition of odd -1 is that an integer), thus n 0 m by definition (since of 0. k + 1 is an integer), and thus in F p by definition of F. - not reflexive: transitive:0 0is isreflexive transitive #> for integers and By p, if m 0 n and 0 pmeans then 17. 0o isis not -' for all all integers mn, m, m n, 0 m. definition of 0n this of 0 this means that for all integers m, n, and p, if m -n is odd and m 0 p. By definition that for all integers m, m - m is odd. But this is false. As a counterexample, take any integer n - p is odd then m -p is odd. But this is false. As a counterexample, take m = 1, n = 0, and m. Then m - m = 0, which is even, not odd. 1-= = 1 is odd and n -p = 01 =- 1 is also odd, but m -p = p = 1. Then m -n = 1is not odd. Hence m 0 n and n O p but m q p. 19. A is reflexive: A is reflexive reflexive property of equality. X' for all real numbers x, x 1x1 2 1. But this is true by the A is symmetric: [We must show that for all real numbers x and y, if I x But this is true by the symmetric property of equality. 1=1 y I then yI 1=1I x1- A is transitive: A is transitive -z> for all real numbers x, y, and z, if I x 1=1 y I and I y 1=1 z ...
View Full Document