Exam 1 studyChapter 1

1) Let X={1,2,3,4,5 } and Y={−1,1,3,5,7 }. Find XY.XY= {-1,1,2,3,4,5,7}

2) What is Z∪Z?Z∪Z=Z

3) Find XY for the following:a. X={ x∈R : 0≤ x<6 } and Y= {x∈R :−π≤ x ≤7 }

XY=Xb. X={0,2,4,6,8 } and Y={1,3,5,7,9 }

XY=c. X=Q and Y={0,1 , ,5 }

XY={0,1,5}4) Find Z ∩ Z, Z ∩∅ and Z ∩ R

5) Find the union and intersection of { x∈ R : x>7 } and { x∈N : x>5 }

6) Let X={ x∈Z :0 ≤ x ≤10 }, A={0,2,4,6,8,10 } and B={2,3,5,7 }. Find A ∩ B, A∪B, A ¿, B\A, A B, X A, Ac and Bc.

7) Find the union and intersections of {x∈R : x2−9 x+14=0 } and { y∈Z :3≤ y<10 }.

8) Suppose A, B and C re subsets of X. Use examples of these sets to investigate the following:

Let A={1,2,3,4}, B = {3,4,5,6}, C = {2,4,6,8}, X = {1,2,3,4,5,6,7,8,9}a. ( A ∩ B )∪ ( A ∩ C ) and A ∩ ( B∪C )

( A ∩ B )∪ ( A ∩ C )= {2,3,4 } and A ∩ ( B∪C )={2,3,4 }b. ( A∪B )∩ ( A∪C ) and A∪ ( B∩ C )

( A∪B )∩ ( A∪C )={1,2,3,4,6 } and A∪ ( B∩ C )={1,2,3,4,6 }c. ( A∪B )c and Ac ∩Bc

( A∪B )c={7,8,9 } and Ac ∩Bc={7,8,9 }d. ( A∪B )c and Ac∪Bc

( A∪B )c={7,8,9 } and Ac∪Bc={1,2,5,6,7,8,9 }e. ( A ∩ B )c and Ac∪Bc

( A ∩ B )c={1,2,5,6,7,8,9 } and Ac∪Bc={1,2,5,6,7,8,9 }f. ( A ∩ B )c and Ac ∩Bc

( A ∩ B )c={1,2,5,6,7,8,9 } and Ac ∩Bc={7,8,9 }The sets in parts a, b, c and e are equal.

9) Suppose A, B and C are subsets of X. – will post pictures later.

a. Draw a Venn diagram for the case that A and B have no intersection.

b. Draw Venn diagrams and shade the sets A∪B, Ac and ( A ∩ B )c.c. Draw three intersectiong circles to represent the sets A, B and C. Shade the

intersection A ∩ B∩ C.d. Draw Venn diagrams to illustrate the sets in problem 8.

Chapter 3You may be given one of the two problems in the book to re-write in a clearer form.

Chapter 5 – Write solutions using 4-step method.

1) Show that a+b

2≥√ab for 0<a ≤b.

a. Understand

Validate the inequality a+b

2≥√ab.

b. PlanWork backwards.

c. Implement

Start with a+b

2≥√ab, multiply by 2 and square both sides. Because both a and b

are positive, these steps won’t change the inequality and are reversible. The result

is (a+b )2 ≥ 4 ab or a2+2 ab+b2 ≥ 4 ab. Subtract 4ab from both sides to obtain

a2−2 ab+b2≥ 0 or (a−b )2≥ 0. Since this last statement is true and all the steps are

reversible we can write a solution to the problem starting with (a−b )2≥ 0 and

working the other way.d. Look back

Working backwards is a good technique. We need ab and a+b to be positive because both x2 and sqrt(x) are increasing for positive numbers – this preserves the inequalities.

2) Show that a2+b2+c2≥ ab+bc+ca for all positive integers a, b and c.a. Understand

Validate the inequality a2+b2+c2≥ ab+bc+cab. Plan

Play around with (a-b)2 (which is positive) and other similar constructs until we find something that produces the correct terms.

c. Implement

(a−b )2+( a−c )2+(b−c )2≥ 0

This expands to: 2 (a2+b2+c2)−2 (ab+ac+bc ) ≥ 0The inequality can be seen by subtracting and dividing by 2.

d. Look backMore than one attempt is necessary before you find the right starting point.

3) Let f ( x )= 11−x

. Define f r ( x )=f (f (f (…f ( f ( x ) ) )) ) ( f composed with itself r times.) Find

f 653(56).a. Understand

Since f r is a composition. We need to learn what happens when we start with 56 and keep applying the function 653 times.

b. PlanLook for a pattern.

c. Implement

f1(56) = -1/55, f2(56)=f(-1/55) =55/56, f3(56)=f(55/56) = 56. So f 3 (56 )=56. This

means that every multiple of 3 will return to 56. 653 = 3(217)+2, so we conclude

that f653 (56 )=f 2 (56 )=55

56.

d. Look backThis depends on the repeating quality of f(x). Other numbers probably have

similar qualities. In fact, f3( p

q )= pq

for any rational number.

4) Show, without using a calculator, that

a. 7√7 !< 8√8 ! and

i. UnderstandWe need to show that the 7th root of 7! Is smaller than the 8th root of 8! 7!=7*6*5*4*3*2*1 and 8!=8*7!

ii. PlanUse properties of roots.

iii. Implement

First we observe that 1< 8√8<2. Also, for positive integers 8th roots will be

smaller than 7th roots. So 7√7 !< 8√8 7√7 !< 8√8 8√7 != 8√8 !iv. Look back

Similar inequalities can be found for other integer pairs. It is necessary that n and n! are both greater than1.

b. √100001−√100000< 12√100000

.

i. UnderstandWe need to find a way to manipulate the inequality without using a calculator.

ii. PlanWork the final inequality backwards to find a true statement, then re-write in reverse order. Use properties of numbers and squaring to get rid of the roots.

iii. Implement

Begin with 100001<100001+ 1400000

But 100001+ 1400000

=100000+1+ 14∗100000

¿100000+1+( 12∗√100000 )

2

¿ (√100000 )2+1+( 12∗√100000 )

2

¿ (√100000 )2+2 (√100000 )∗( 12∗√100000 )+( 1

2∗√100000 )2

¿(√100000+ 12∗√100000 )

2

So 100001<(√100000+ 12∗√100000 )

2

If we take the square root of both sides and subtract √100000 we obtain

√100001−√100000< 1

2√100000 as desired.

iv. Look backThis solution shows the final answer. To understand how to solve you need only read the solution backwards – this is how I decided to use 1/400000 on the first line.

5) Suppose that three friends have a meal for $25. They misread the bill, thinking it says $27, they give the waiter $10 each and ask for a total $3 in change. He puts $25 in the the till, gives each friend $1 and pockets the extra $2. The friends each paid $9 which make a total of $27 and the waiter kept $2 which brings the total to $29. The friends gave a total of $30. What happened to the extra dollar

a. UnderstandThis is a riddle. We need to identify the place where the logic fails.

b. PlanWrite out with careful thinking.

c. Implement3*10-3=27 is OK. The Waiter is right to keep the $2.9*3+2 = 27 is the wrong math. Since the waiter kept $2 he should subtract. Then 3*9-2 = 25 which is the correct amount to put in the register.

d. Look backIt is easy to confuse people with simple math. The three friends were jerks for not leaving a tip.

6) Bottle A contains a liter of milk and bottle B contains a liter of coffee. A spoonful of coffee is poured from B into A and the contents are mixed will. Liquid from A is then poured into B until B once again has one liter of liquid. Is the fraction of coffee in A greater than the fraction of milk in B or is it the other way around?

a. Understand

After the first exchange there is a small amount of coffee in the milk. When the containers are balanced again there will be a small amount of milk in the coffee too. Which amount is greater?

b. PlanUse a variable to calculate the ratios.

c. ImplementLet x be the size of the spoon. So x coffee is removed from container B and placed in Container A. This leaves 1+x liters in A and 1-x liter in B. The concentration of coffee in A is x/(1+x). When x mixture is removed from A and returned to A, the amount of coffee in the spoon is (x)(x/(1+x))= x2/(1+x). This means that there will be x/(1+x) milk in the spoon. Simple algebra will show that there is now x-x/(1+x)= x/1+x coffee in the milk. Since the only milk in container B came from the spoon we see that the amount of coffee in A is the same as the amount of milk in B

d. Look backFast way to understand: Suppose the spoon contains a liter. Then we mix both containers and then split them back into two. The concentration is the same in both.

Chapter 61) Which of the following are statements?

a. Life is sweet. – yesb. Is 2 prime? – noc. Prove that 2 is prime. – nod. The president of the united states in 1789 was a man. – yes e. The president of the united states in 2089 will be a woman. - yes

2) Construct truth tables for the following:a. not(A and B)b. not(A or B)c. (not A) or (not B)d. A or (not B)e. (not B) or Bf. (not B) and B

A B not (A and B) not(A or B) (not A) or (not B) A or (not B) (not B) or B (not B) and BT T F F F T T FF T T F T F T FT F T F T T T FF F T T T T T F

3) Negate the following:a. A is true or B is false – A is false and B is trueb. A is false and B is true – A is true or B is false

c. A is true or B is true – A is false and B is falsed. A is true and B is true – A is false or B is false

4) Construct truth tables for the following:a. A and (B or C)b. (A and B) or Cc. (not(A or B)) and C

A B C A and (B or C) (A and B) or C not(A or B) and CT T T T T FT T F T T FT F T T T FT F F F F FF T T F T FF T F F F FF F T F T TF F F F F F

5) Tautology and/or contradiction…….a. Show that “A or (not A)” is a tautology. – A and (not A) have opposite truth

value. One of them is always true.b. Is “Winners don’t quit and quitters don’t win” a tautology? NO both statements

are subjective and might both be false.c. Show that “A and (not A)” is a contradiction. A and (not A) have opposite truth

value. They are never both true at the same time.Chapter 7

1) Suppose A is true and B is false. Which of the following are true?a. A B – false b. B A – true c. (not B) A – true d. A A – true e. A or (not B) f. (not A) A – true g. not(A or B) – false

2) Negate the following:a. If you score 70%, then you have done well in this course

You score 70% and you have not done wellb. If it rains, then I will stay home

It rains and I don’t stay home.c. If x2 + 2x + 1 = 0 then x = -1.

x2 + 2x + 1 = 0 and x -1d. x2 + x – 2 = 0 implies x = 1 or x = -2.

x2 + x – 2 = 0 and neither x = 1 nor x = -2

3) Write the following using “only if”a. If x = -2, then x2 = 4.

x2 = 4 only if x = -2b. If x and y are odd, then xy is odd.

xy is odd only if x and y are odd.c. x2 + x – 2 = 0 x = 1 or x = -2.

x = 1 or x = -2 only if x2 + x – 2 = 0d. x2 + x – 2 = 0 x = 1.

x2 + x – 2 = 0 only if x = 1.4) Construct truth tables for the following:

a. (A B) (not A)b. A or [B (not A)]

A B (A=>B)=>(not A) A or (B=>(not A))T T F TF T T TT F T TF F T T

5) Show the following are tautologies using truth tables:a. A (A or B)

A B A or B A => (A or B)T T T TF T T TT F T TF F F T

b. [(A and B) C] [(not C) {(not A) or (not B)}]

A B C (A and B) => C (not C) => {(not A) or (not B)}{(A and B) => C} =>

[(not C) => {(not A) or (not B)}]T T T T T TF T T T T TT F T T T TF F T T T TT T F F F TF T F T T TT F F T T TF F F T T T

Chapter 81) Rewrite the following as “if..then” statements:

a. A sufficient condition for Peter to win the Championship is that he wins in Brazil.If Peter wind in Brazil, then he will win the championship

b. A necessary condition for Stuart to win the championship is that he beats Peter.If Stuart wins the championship, then he beats Peter.

c. Regular work is sufficient to pass this course.If you do regular work, then you will pass this course.

d. Regular work is not necessary to pass this course.Not(If you pass this course then you did regular work.) i.e. You can pass the class _and_ not do regular work.

e. To be President of the United States of America it is necessary to be born in the USA.

If you are President of the United States then you were born in the USA.f. To be Prime Minister of India it is not necessary to be born in India.

Not(If you are Prime minister of India then you were born in India). i.e. You can be Prime Minister of India _and_ be born in another country.

2) Let A be “x2 – 2x – 3 > 0” and B be “x > 3”. Which of the following are true and which are false?

The solution to the inequality is {x: x > 3 or x < -1}a. A B – false b. B A – true c. A is necessary for B – trued. B is sufficient for A – truee. A is sufficient for B – false f. B is necessary for A – false g. not(B) not(A) – falseh. not(A) not(B) – true

3) Which of the following statements are equivalent? – a and da. If my team lost the last game, then they most have lost the championship.b. If my team lost the last game, then your team won the championship.c. If my team lost the last game, then they won the championship.d. If my team won the championship, then they won the last game.e. If my team won the last game, then they won the championship.f. If my team lost the championship, then they must have lost the last game.

4) What is the contrapositive of A(BC)?“Not(B=>C) => (not A)” – or – If (B and (not C)) then (not A)

Chapter 91) Write the converse of the following statements:

a. If x > 5, then x is red.If x is red then x>5.

b. An integer can be even or odd but it cannot be both.If n is even or odd but not both, then n is an integer.

c. Eating ice cream is necessary for me to be happy all day.If I am eating ice cream then I am happy all day.

d. Eating ice cream is sufficient for me to be happy all day.If I am happy all day then I am eating ice cream.

e. It is not necessary to understand things to argue about them.It is not sufficient to understand things to argue about them

f. Stop or I will shoot. (Remember PQ is the same as (not P) or Q.)If I shoot then you didn’t stop.

2) Which of the following are equivalent? a and c are equivalent and b and d are equivalent.

a. If A and B are both red, then X is true.b. If A and B are both not red, then X is true.c. If X is false, then A and B are not both red.d. If A or B are not red, then X is true.

3) Create truth tables for:a. A Bb. The converse of “A implies B.”

A B A <=> B B => AT T T TF T F FT F F TF F T T

4) Give the converse of each of the following statements:a. If ab = 0, then a = 0 or b = 0.

If a = 0 or b = 0, then ab = 0.b. If 2x2 – 7x + 6 = 0, then x = 2.

If x = 2 then 2x2 – 7x + 6 = 0.c. The product of two odd integers is odd.

If the product is odd then the integers are odd.d. The sum of an odd number and an even number is odd.

If the sum is odd then one number is odd and one is even.e. Suppose that A, B and C are sets. In this case A ∩ ( B∪C )⊆ ( A ∩B )∪ ( A ∩C ).

If A ∩ ( B∪C )⊆ ( A ∩B )∪ ( A ∩C ) then A B and C are sets.f. Suppose that f is a polynomial in the variable x.

If f(a) = 0, then x – a is a factor of f.If x – a is a factor of f then f(a) = 0.

Chapter 101) Rewrite the following using and :

a. For all integers x, x is odd or even.xZ(x is odd or even)

b. There exist two prime numbers such that their sum is prime.(p and q prime)(p+q is prime)

c. There exists a rational number greater than √2.xQ(x>√2)

d. If x is a real number, then x2 is greater than x.xR(x2>x)

e. For all n∈N there exists a prime p such that p > n.n∈Np prime(p > n)

2) Decide whether the following are true or false:a. xy(x2 = y), where both x and y are real numbers. – true b. yx(x2 = y), where both x and y are real numbers. – false c. xy(x2 = y), where both x and y are integers. – true d. yx(x2 = y), where both x and y are integers. – false e. ∀ x∈R∃ y∈ R(x+y=0) – true f. ∃ x∈R ∀ y∈ R(x+y=1) – false g. xP(x) xP(x) – true h. xP(x) xP(x) – false i. ∃n∈N such that n2 n. – true, because n = 1 is a natural number.

Chapter 111) Rewrite the following using the symbolic notation and :

a. If a and b are real numbers with a 0, then ax + b = 0 has a solution.a,bR,a0,x(ax + b = 0)

b. If a and b are real numbers with a 0, then ax + b = 0 has a unique solution.a,bR,a0,!x(ax + b = 0)

2) Negate the following:a. There exists a grey cat.

All cats are not grey.b. For all cats there exists an owner.

There exists a cat with no owner.c. There exists a grey cat for all owners

There exists an owner with no grey cat.d. Every fire engine is red and every ambulance is white.

There exists a non-red fire engine or a non-white ambulance.3) Negate the following:

a. Some students in the class are not here todayAll students are here today.

b. Let x, y, z N. For all x there exists y such that x + y = z.For all x and y, x+yz

c. There exists a unique x such that P(x) is true.(P(x) and P(y) are true for xy) or (P(x) is not true for all x).

d. All mathematics students are hardworking.Some mathematics students are not hardworking

e. Only some of the students in the class are here today.All students are here today.

f. The number √ x is rational if x is an integer.There is some integer x with √ x not rational.

4) Simplify the following:

a. not (∀ y ∃ x (P ( x , y )⟹Q ( x , y )))For for all x there exists y with notP(x,y) and Q(x,y)

b. not (∃ x , y ∀ z not (∀u∃ vP (u , v , x , y , z )))∃ z∀ x , y (∀u∃ v P (u , v , x , y , z ))

c. not(there exist xR and yR such that for all zQ we have xz and z y)For all x and y in R, there exists z in Q with x<z or y>z.

d. not(there exist xR and yR such that xy or for all zQ we have xz and z y)for all x and y in R y<x and for all z in Q x<z or y>z.

5) Show the following:

a. ∃N∈N such that ∀n ≥ N ,1n< 25

37.

Let N = 2 > 37/25.

b. ∃N∈N such that ∀ n ≥ N ,5 n2+2

n2 −5< 11000

Let N = 45.

c. ∀ ε>0 ,∃N∈N such that ∀n ≥ N ,1n<ε

Let N > 1/ε.

d. ∀ ε>0 ,∃N∈N such that ∀ n ≥ N ,5 n2+2

n2 −5<ε

Let N > √2/εChapter 12

1) Find examples of the following:a. A non-constant function f : R → R such that f(x)= 0 for a finite number of x.

f(x) = xb. A non-constant function f : R → R such that f(x)= 0 for an infinite number of x.

f(x) = sin(x)c. A non-polynomial function f : R → R such that f’(x) is always positive.

f(x)= tan(x)d. A non-polynomial function f : R → R such that f’(x) is negative for x < 0 and

positive for x 0.f(x) = |x|

e. A function f : R → R such that f has a maximum at x = -2 and a minimum at x = 7.

f(x) = 2x3 - 15x2 - 84x + 2 (yay calculus!)2) Determine whether or not the following are true: (explain answer)

a. x3 < 0 for all x < -1.True. If x < -1 then x <0 and x3 < 0 too.

b. x3 > 0 for all x > 1000.True. If x >1000 then x > 0. So x3 > 0 too.

c. x3 0 for all x 1.False. Let x = ½. Then x3 = 1/8 > 0.

3) Find counterexamples for the following:

a.1−x

x is not an integer for x∈Z .

If x = 1 the fraction is 0.

b. (a+b )2=a2+b2.

Let a = 1 and b = 2. The left side is 9, while the right side is 5

4) Find a counter example to f’(c)=0 implies that f has a maximum or minimum at x = c. f(x) = x3 satisfies f’(0) = 0, but it does not have a minimum or maximum at x = 0.