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Question

1. Suppose w, x, y, and z are positive integers such that w+z = x+y. Which of the following must be true?

a. w+z b. x+y< z
c. w = x
d. z < y
e. w - x = y - z

2. If a, b are odd numbers and c is even, which of the following is an even number.

a. ab + c
b. a(b+c)
c. (a+b) + (b+c)
d. (a+b) - c
e. a+bc

If possible, can you show how you got it and explain it. Thanks so much. =]

Answer 1

1/e

2/d

Answer 2

1. True or False? Give your reason. There is a largest rational number less than 1.



Solution FALSE. For any rational number x less than 1, (1+x)/2 is another rational number less than 1 which is greater than x.



2. True or False? Give your reason. (-\sqrt(3)+1)/(\sqrt(3)-1) is a rational number.



Solution This number is in fact -1, which is rational.



3. Verify that x=1+i\sqrt(3) satisfies the equation x^2-2x+4=0.



Solution x^2=1+2i\sqrt(3)+3i^2=-2+2i\sqrt(3), -2x=-2-2i\sqrt(3), so x^2-2x+4=0.



4. Prove that the cube root of 5 is irrational.



Solution Suppose that the cube root of 5 does equal p/q, for integers p and q. Cubing this equation we have 5=p^3/q^3 or 5*q^3=p^3. Now consider how many 5's are in the prime factorization of the right side. We must a multiple of 3 number of 5's. The left side must have one more than a multiple of three number of fives, and this contradicts the fundamental theorem of arithmetic.



5. Suppose that w is complex number such that w^6=-1. Can you find w^15?



Solution w^15=w^6*w^6*w^3=(-1)*(-1)*w^3=w^3. Also w^3 squared is -1, so w^3 is either i or -i.



6. Let p(x)= x^3+10*x^2+x. Show that there are three distinct real values values of x such that p(x)=0.



Solution p(x)=x(x^2+10x+1), so x=0 solves p(x)=0. There are two other real solutions. To see this either write them out using the quadratic formula, (-10\pm \sqrt(100-4))/2, or use the intermediate value theorem x^2+10x+1 is 1 at x=-10, -8 at x=-1 (so there is zero between -10 and -1), and is 1 at x=0 (so there is zero between -1 and 0).



Math 3118 Practice Exam 4-2



1. Let x=(1/2+i)/(2/3-i). Find a rational number r such that x-ir is a real number.



Solution x=(1/2+i)*(2/3+i)/(13/9)=(-2/3+7/6i)/(13/9) so r=7/6*9/13.



2. True or False? Give your reason. The cube root of a positive integer m is irrational unless m is perfect cube.



Solution TRUE. Could m=p^3/q^3? Then we would have m*q^3=p^3. If m is not a perfect cube, then some prime r in the prime factorization of m occurs either 1 mod 3 times or 2 mod 3 times. This contradicts m*q^3=p^3. (see prob 1-4).



3. Prove that the number x=(\sqrt(3)-\sqrt(5))/4 is irrational.



Solution 4x=(\sqrt(3)-\sqrt(5)) so 16x^2=3+5-2*\sqrt(15), or \sqrt(15)=-8x^2+4. If x were rational then the right side of this equation would be rational, so \sqrt(15) would be rational which we know is wrong. So x cannot be rational.



4. Suppose that x^6=1. Find rational numbers a,b,c,d,e,f such that x^1000=a+b*x+c*x^2+d*x^3+e*x^4+f*x^5.



Solution 1000=6*166+4, so x^1000=x^4 and we can take a=b=c=d=f=0, d=1.



5. Show that there is some number x between 1 and 3 such that x^7+x-3=0.



SolutionIf x=0 the polynomial is -3, if x=2 the polynomial is 128+2-3>0, so the IVT applies.



Math 3118 Practice Exam 4-3



1. True or False? Give your reason. There is positive real number x such that x^3+3x^2-100*x-4=0



Solution Let p(x)=x^3+3x^2-100*x-4. Since p(0)=-4 and p(1000)>0, the IVT implies there is some x between 0 and 1000 such that p(x)=0. TRUE.



2. Give an example of two irrational numbers x,y such that x/y is rational.



Solution x=\sqrt(3)=y, x/y=1.



3. True or false: If x and y are any two real numbers, then there is some rational number r between x and y.



Solution TRUE. We did this in class. Find the first place where the decimal representations of x and y disagree, and choose a digit for r in between. Then add all zeros.



4. The infinite decimal which consists of the consecutive integers 0.1234567891011121314... is an irrational number.



Solution TRUE. This infinite decimal does not terminate. Could it be repeating? Suppose the repeating part has some length, say 17. Eventually we come to the integer 1111111111111111 which appears in our given number, so our repeating part would have to be 1111111111111111. But the next number in our expansion would be 11111111111111112, so it does not repeat!



5. Let z=(2-i)/i. Find the a+bi form, a and b rational, for z^2.



Solution z^2=(2-i)^2/i^2=(4-4i+i^2)/(-1)=-3+4i, so a=-3, b=4.



6. Let z=1-2*\sqrt(5). Find the a+b\sqrt(5) form, a and b rational, for z^2.



Solution z^2=1-4*\sqrt(5)+20=21-4*\sqrt(5), so a=21, b=-4.



Math 3118 Practice Exam 4-4



1. Prove that the real number (2-\sqrt(2))/\sqrt(2) is irrational.



Solution x=(2-\sqrt(2))/\sqrt(2)=\sqrt(2)-1, so x+1=\sqrt(2). If x were rational, the left side would be rational, and then the right would be rational. But we know that \sqrt(2) is irrational.



2. Let z be the complex number z=(1+i)/\sqrt(2). Prove that z^2=i.



Solution z^2=(1+2i+i^2)/2=i.



3.Suppose that x is a complex number such that x^2+x+1=0. Prove that x^3=1.



Solution x^3=x*x^2=x*(-x-1)=-x^2-x=(x+1)-x=1.



4. True or False: The average of two irrational numbers is always irrational.



Solution FALSE. Try \sqrt(2) and -\sqrt(2) whose average is 0.



5. True or False: The square of an irrational number is rational.



Solution FALSE The square of the cube root of 2, is t

Answer 3

1



a == F example w = 1 , z = 3 , x = 2 , y = 2 clearly 4 < 2 is False

b == F remember each variable is a positive integer

c == F see above values

d == F again see above

e == T try the values above



2



a,b odd => a and b can be written as a = 2n-1 and b = 2m-1 where m is some integer



c is even => c can be written as c = 2p where p is some integer



let a = 3 b = 1 c =0



a == odd

b == odd

c == odd

d == even

e == odd

Answer 4

For number 1;

A. can be eliminated because we know that W+Z = X+Y therefore the value of W+Z can NOT be less than Y.

B. can be eliminated for the same reason.

C. Is not true because if Z and Y are different the given equation would not hold true.

D. Same reason as for A.

E. Is the answer because of algebra:

W+Z-Z = X+Y-Z

W = Y+X-Z

W-X = Y+X-Z-X

W-X = Y-Z



number 2;

any odd number multipled by an odd number will still yield an odd number. And any odd number plus and even number will always yield a odd number



therefore a and b can quickly be eliminated. C on the other hand is not right because A+B (is even) + (B+C) (odd) = odd.



D is the right answer because a even number - and even number is still even.



E is wrong because an odd number multiplied by an even number yields is always even, and an even number plus and odd number is odd.

Answer 5

1. the correct answer is e. w - x = y - z



a. w+z


c. w = x can be right but it does not 'have' to be right. Lets take the example of 2(w) + 12(z) = 5(x) +9(y). In this equation, the sum of both sides equal to 14, but w is not equal to x.



d. z < y can be right but it does not 'have' to be correct. you can see in my above given example that z is not less than y.



Now, the last choice we have is e. w - x = y - z that is absolutely correct. w+z = x +y, move x to the left and it will become negative, move z to the right and it will become negative as well. Finally you will get, w - x = y - z



The correct answer to the second question is d. (a+b) - c



Option a. ab + c is wrong because the product of two odd numbers is always odd and the sum of an odd and an even number is odd as well.



Option b. a(b+c) is wrong since the sum of an even and an odd number is always an odd number and the product of two odd numbers is also odd.



Option c. (a+b) + (b+c) is wrong because the sum of two odd numbers is even, the sum of an even and an odd number is odd and finally, the sum of even and an odd is odd as well.



Option d. (a+b) - c is correct since the sum of two odd numbers is even, and when you subtract an even number from an even number, you get an even number.



Option e. a+bc is wrong since the product of odd and even is even and the sum of odd and even is odd.

I hope you understand my explanation:)

Answer 6

1. e - I'm not really sure how to explain it, but if you plug some numbers in, it works out...(for example: 8 + 2 = 6 + 4 or 15 + 1 = 7 + 9)



2. d - Two odd numbers added together always make an even number, so a + b would be even....then, an even number minus an even number will always be even....

Answer 7

1/e

w-x=y-z try to make the negative signs positive by moving them to the other side of the equal sign. see what you will get:

*w+z=y+x*



2/d

for this question I suggest you to chose numbers which represents odd or even numbers and then plug them in the equations and see which one gives you an even number.

for example my numbers are:

a=3 odd

b=5 odd

c=2 even



now plug in:

ab+c=(3)(5)+2=17 odd

a(b+c)=3(5+2)=21 odd

(a+b) + (b+c)= (3+5)+(5+2)=15 odd

(a+b) - c = (3+5)-2=6 even*

(odd#+odd#=even# even# - even#=even#)

a+bc= 3+5.2=3+10=13 odd

Answer 8

1 e



.

Answer 9

1. w+z = x+y

subtract z from both sides, subtract x from both sides

you get w-x = y-z. E.



2. odd + odd = even, odd * odd = odd. odd + even = even. odd * even = even. even anything with even = even.



so D is right. (odd + odd = even, even - even = even)