Questions for the exam
1
Theoretical questions
1. Formulate binomial theorem. Right the formula for (x + y )n and formula for the binomial coecients. 2. For two positive integer numbers n, m dene gcd(n, m) and prove that it exists. You can not use prime n
Math 311, Quiz 6
Section:
Name:
Instructions: Clearly answer each of the questions below. Remember to check the back side. Use full
sentences and proper grammar. Show your work and any formulas you employ. Simplify all answers as
far as possible. Box your
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Math 311, Quiz 1
Section:
Name:
Instructions: Clearly answer each of the questions below. Remember to check the back side. Use full
sentences and proper grammar. Show your work and any formulas you employ. Simplify all answers as
far as possible. Box your
Math 311, Quiz 3
Section:
Name:
Instructions: Clearly answer each of the questions below. Remember to check the back side. Use full
sentences and proper grammar. Show your work and any formulas you employ. Simplify all answers as
far as possible. Box your
Math 311, Quiz 1
Section:
Name:
Instructions: Clearly answer each of the questions below. Remember to check the back side. Use full
sentences and proper grammar. Show your work and any formulas you employ. Simplify all answers as
far as possible. Box your
MATH 311W - Midterm Exam II (Practice)
April 1, 2016 (50 minutes)
This exam contains 6 problems. This is a closed book exam. Show all of your computations
and justify all of your answers. Calculators are NOT allowed.
1. (10 points) Complete the following
SOLUTIONS TO MIDTERM II
#1:
(a) Two integers a, b Z are relatively prime if gcd(a, b) = 1.
(b) Let n N. Two integers a, b Z are congruent mod n if n|a b.
#2: Note that 4x + 1 is divisible by 5 and by 13 if and only if
(
4x + 1 0 mod 5,
4x + 1 0 mod 13.
Si
SOLUTIONS TO PRACTICE MIDTERM I
#1:
(a) A function f : X Y is injective if whenever x1 , x2 X are such that
f (x1 ) = f (x2 ), it holds that x1 = x2 .
(b) An equivalence relation on a set A is a relation R which is reflexive,
symmetric, and transitive. Th
SOLUTIONS TO PRACTICE MIDTERM I
#1:
(a) A prime number is a positive integer which has exactly two divisors.
(b) Let n N. An integer a Z is invertible mod n if there exists an integer
b Z such that ab 1 mod n.
#2: We are asked to solve the system
2a + 1 0
SOLUTIONS TO PRACTICE MIDTERM I
#1:
(a) A function f : A B is surjective if for every b B, there exists an a A
such that f (a) = b.
(b) A relation R on a set A is reflexive if for every a A, it holds that (a, a) R.
#2:
Since R is symmetric and weakly-anti
MATH 311W - Midterm Exam I (Practice)
February 26, 2016 (50 minutes)
This exam contains 6 problems. This is a closed book exam. Show all of your computations
and justify all of your answers. Calculators are NOT allowed.
1. (10 points) Complete the followi
HW1 SOLUTIONS
2.1.1: X = Z = W and Y = V .
2.1.3: Recall that X \ Y = cfw_x X | x Y and X Y c = cfw_x X Y | x Y .
Suppose x X \ Y . Then x X and x Y . Since x X, we see that x X Y .
Thus x X Y c . Conversely, suppose x X Y c . Then x X Y and x Y .
Thus x
MATH 311W
Hints and Partial Solutions to Some Homework Problems
Gary L. Mullen
Fall Semester 2006
1.1.1 Part (iii); the other parts are done the same way. 126 = 1(91) + 35
91 = 2(35) + 21
35 = 1(21) + 14
21 = 1(14) + 7
14 = 2(7) + 0
Thus (91, 126) = 7, th
Midterm Exam MATH 311W Section 003
(Solutions Included)
Stephen G. Simpson
Pennsylvania State University
October 17, 2011
There are 6 problems on 6 pages. For each problem, use the space below the
problem to exhibit your work leading to the solution of th
HW6 SOLUTIONS
1.2.3: The Fibonacci sequence is defined by f1 = 1, f2 = 1, and fn = fn1 + fn2
for n 3. We claim that (fn , fn+1 ) = 1 for all n N. Clearly (f1 , f2 ) = 1. Suppose
that (fn , fn+1 ) = 1 for some n N. Then
(fn+1 , fn+2 ) = (fn+1 , fn+1 + fn )
HW3 SOLUTIONS
2.3.7: Since (1, 1), (2, 2), (3, 3), (4, 4) R, we see that R is reflexive. Since (1, 2), (2, 1)
R and (3, 4), (4, 3) R and there are no other pairs of distinct integers in R, we see
by exhaustion that R is symmetric. A similar consideration
HW4 SOLUTIONS
3.2.1(a):
(i) Every Scottish person likes whisky.
(ii) Everyone who likes whisky is Scottish.
(iii) There is a Scottish person who doesnt like whisky.
(iv) Not every Scottish person likes whisky.
(v) Not everyone is Scottish and likes whisky
HW2 SOLUTIONS
Problem 2.2.5:
(i) f : R+ R defined by f (x) = ln x is a bijection from R+ = cfw_x R | x > 0
to R.
(ii) f : (/2, /2) R defined by f (x) = tan x is a bijection.
(iii) f : N Z defined by
(
m,
if n = 2m + 1,
f (n) =
m, if n = 2m
is a bijection
1
2. (G, ) is a group. c is a fixed element of G. New operation on G by a 'b = a c b
Seeing as (G, ) is a group, operation abides by the four conditions, and c is from G and
unchanging. How do we know c-1 is in G?
G(1-4) for (G, )
G1: a b G
G2:
( a b ) c
c)
G = ! * defined by a b = a + b 1
G1) x, y G
x, y !
x=x
y=y
(x) + (y) 1 = x + y 1 G
! + ! 1 = !
G2) x, y, z G
( x y) z = x ( y z)
( x + y 1) + z 1 = x + ( y + z 1) 1
x+ y+z2 = x+ y+z2
G3) x e = x
x + e 1 = x
e=1
G4) y = x 1
xy = e
x + y 1 = 1
x+y=2
y=