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MATH 135, Fall 2010

Midterm SolutionProblem 1.(8pts)

(a) (6pts) Construct a truth table for the statementP =

(NOTQ) R

.

(b) (2pts) State the contrapositive of the statement ifn is prime, then n = 2 or n is odd.

Solution.

(a) (6pts) The truth table is

P Q R (NOT Q) (NOT Q) R P =(NOT QR)T T T F F FT T F F T TT F T T T TT F F T F FF T T F F TF T F F T TF F T T T TF F F T F T

(b) (2pts) Ifn = 2 and nis even, then n is not a prime.

Problem 2.(12pts) Consider the following statement for any two numbers x and y, ifx > y,then x2 > y.

(a) (4pts) Write down this statement using mathematical quantifiers.(b) (4pts) Write down and simplify the NOT of this statement. Your answer should not have

the word NOT in it.(c) (4pts) Is the statement in (a) true if the universe of discourse is the set of positive real

numbers? Justify your answer.Solution.

(a) (4pts)

x, y, (x > y = x2 > y)

(b) (4pts) Note that NOT (P = Q) is equivalent to P AND (NOT Q), and NOT is NOT. Thus, NOT

x, y, (x > y = x2 > y)

is equivalent to

x, y(x > y ANDx2 y).

(c) (4pts) Over the positive real numbers, the statement is FALSE. The following is a coun-terexample. Let x = 1/2 and y = 1/3. Then x2 = 1/4< 1/3 =y.

Problem 3.(9pts) A sequence {yn} is defined recursively by y1

= 16, y2

= 44, and

yn= 2yn1+ 3yn2, n 3.

Prove that for all n P, yn= (1)n+1 + 5 3n.

Solution. We prove the statement by Principle of Strong Induction.Ifn= 1, (1)2 + 5 31 = 1 + 15 = 16 =y1. Thus the statement is true for n= 1.

Ifn = 2, (1)3 + 5 32 =1 + 45 = 44 =y2. Thus the statement is true forn= 2.

Suppose there is a positive integer k 2 so that for 1 r k,

yr = (1)r+1 + 5 3r

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Considern= k + 1. We have

yk+1 = 2yk+ 3yk1 From the recursive definition of{yn} andk+ 1 3

= 2[(1)k+1 + 5 3k] + 3[(1)(k1)+1 + 5 3k1] From the assumption,yk = (1)k+1 + 5 3k

and yk1= (1)k

+ 5 3k1

= (1)k[2(1) + 3] + 5 3k1[2 31 + 3]

= (1)k + 5 3k1[9]

= (1)k+2 + 5 3k+1 Since (1)k = (1)k(1)2 = (1)k+2

Thus the statement is true for n = k+ 1 when it is true for 1 r k. By Principle of StrongInduction,

yn= (1)n+1 + 5 3n, n P.

Problem 4.(11pts)

(a) (3pts) Determine the coefficient ofx3 in the expansion of

2x2 1

x6

.

(b) (8pts) Prove that all n, k Pwith n > k , we have

n 1

k

n 1

k 1

=

n 2k

n

n

k

.

Solution.

(a) (3pts) The general term is

6

r

(2x2)r

1

x

6rWe want 2r+r 6 = 3 r = 3.

So the coefficient is

6

3

(2)3(1)63 = (20)(8)(1) =160

(b) (8pts)n 1k

n 1k 1

= (n 1)!

k!(n 1 k)!

(n 1)!

(k 1)!(n 1 (k 1))!

= (n 1)!

k!(n 1 k)!

(n 1)!

(k 1)!(n k)!

= (n 1)!(n k)

k!(n k 1)!(n k)

(n 1)!k

k(k 1)!(n k)!

=(n 1)!(n k) (n 1)!k

k!(n k)!

=(n 1)!(

n

k

k)

k!(n k)!

= n(n 1)!(n 2k)

k!(n k)!n

= n!

k!(n k)!

n 2k

n

= n 2k

n

n

k

Problem 5.(7pts) Find x, y Z such that gcd(46134, 2244) = 46134x+ 2244y.

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Solution.

x y 46134x+ 2244y Euclidean Algorithm q q1 0 461340 1 2244 46134 = 20 2244 + 1254 20 20

(20) row 2 + row 1 1 -20 1254 2244 = 1 1254 + 990 1 1(1) row 3 + row 2 1 21 990 1254 = 1 990 + 264 1 1

(1) row 4 + row 3 2 41 264 990 = 3 264 + 198 3 3(3) row 5 + row 4 7 144 198 264 = 1 198 + 66 1 1(1) row 6 + row 5 9 185 66 gcd 198 = 3 66 + 0

Thus we have66 = 9 46134 185 2244,

and x = 9, y= 185 is a solution.

Problem 6.(8pts) Leta, b Z, not both zero. Prove that gcd(a, b) = gcd(3a+ 2b, a+b).

Solution. Letd= gcd(a, b) and e= gcd(3a+ 2b, a+b). We want to prove that d = e.Sinced|aand d|b, by Proposition 2.11(ii),d|(3a + 2b) andd|(a + b). Sinced|(3a + 2b),d|(a + b) ande= gcd(3a+ 2b, a+b), we have d e.

Sincee|(3a + 2b) ande|(a + b), by Proposition 2.11(ii), e|3(a + b) (3a + 2b), i.e.,e|b. Sincee|(a + b)ande|b, by Proposition 2.11,e|(a + b) b, i.e.,e|a. Sincee|a,e|b, andd = gcd(a, b), we have e d.Since d e and e d, we have d= e, i.e., gcd(a, b) = gcd(3a+ 2b, a+b).

Problem 7.(5pts) Leta, b, c Z with c|ab and gcd(a, c) = 1. Prove that c|b.

Solution. Since c|ab, there exists q Zsuch that ab = qc.Since gcd(a, c) = 1, by the Euclidean algorithm, there exist x, y Z, such that

1 =ax+cy.

Multiplying by b, we getb= abx+bcy= qcx+bcy= c(qx+by).

Since qx+by Z, we have c|b.