Download mae 301/501 course notes for 2/25/08

MAE 301/501 COURSE NOTES FOR 2/25/08
Written by Adam Esposito
First, there are some remarks concerning Homework # 3.
1. a) To prove the function f :N  Z defined by:
n
2
if n is even
1 n
2
if n is odd
is injective, we must show that the statement “if f (x) = f (y), then
x = y” is true in three cases:
(i)
x and y are both even,
(ii) x and y are both odd, AND
(iii) x and y are not the same parity
Proof
Case 1: If x and y are both even, f (x) = f (y) implies
x=y
Case 2: If x and y are both odd, f (x) = f (y) implies
x y

, and therefore
2 2
1 x 1 y

, and
2
2
therefore x = y
x
1 y
and f (y) =
. Since x and y
2
2
are natural numbers, f (x) > 0 and f (y)  0 , so when x and y are not of the
same parity the condition f (x) = f (y) does not hold. Since f (x)  f (y) , the
Case 3: If x is even and y is odd, f (x) =
statement “if f (x) = f (y), then x = y” is true, proving injectivity in this case.
(Note: if y is even and x is odd, we would yield the same result.)




b) To prove f (n) is surjective, we must show for every w  Z , there exists
t  N such that f (t) = w . It is not enough to say “
“
1 n
always maps to
2
Z”
n
always maps to
2
Z ” and
Proof
If w > 0 , let t = 2w, then t is an even natural number, and f (t) =
2w
w
2
If w  0 , let t = 1 – 2w , then t is an odd natural number, and
f (t) =
1  (1  2w)
w
2



2. Question (2b) on Homework # 3 should remind us of a certain rule or
theorem.
First, notice that if n = pk , then n has k + 1 divisors.
For example: 16 = 24 , and 16 has 5 divisors: 1,2,4,8, and 16
where 1 = 20, 2 = 21, 4 = 22, 8 = 23, and 16 = 24
Second, notice that if n = p1e p2e , n has (e1 + 1)(e2 + 1) divisors.
1
2
For example: 36 = (22)(32) , and has 9 divisors: 1,2,3,4,6,9,12,18,36
This is a direct result of the Fundamental Principle of Counting (FPC)
which states: If A can be done in m ways, and B can be done in n ways, the
total number of ways A and B can be done together is mn ways. So, since p1e
can be divided e1 + 1 ways, and p2e can be divided e2 + 1 ways, p1e p2e can
be divided (e1 + 1)(e2 + 1) ways.
1
2
1
2
We can extend the FPC to say: Given A1 , A2 , A3 , …… , Ak , if Ai can be
done ni ways, the total number of ways all A’s can be done together is:
k
ni = n1n2…….nk
i=1
This principle is related to our main topic of exploration:
Enumerative Combinatorics
There are 10 homogeneous cubic monomials in 3 variables:
x3 , y3 , z3 , x2y , x2z , y2x , y2z , z2x , z2y , xyz
How many homogeneous monomials are there of degree n in k variables?
Note: Our definition of homogeneous is “all of the same kind”, so a set of
“homogeneous monomials” have the same sum of exponents. In the context
of linear algebra, a “homogeneous system” of linear equations is one where
all equations in the system equal zero. These two definitions are compatible!
A set of monomials is said to be homogeneous if all elements have an
exponential sum equal to one another.
A system of linear equations is said to be homogeneous if all equations equal
zero, and therefore the terms of the polynomials in the system have the same
exponential sum. For example, if a system included the equation
ax + by + cz = 5 the system is not homogeneous, since:
ax + by + cz = 5  ax1 + by1 + cz1 – 5t0 = 0
Back to our exploration…
For each of the following questions we should make the connection between
how we use our problem solving skills to find solutions, and our use of
factorials, binomial coefficients, and permutations. This way we could easily
explain our methods to high school students:
1. a) How many integers are between 100 and 999 with distinct digits?
9  9  8 = 648
The 1st 9 is derived from all digits 1 through 9.
The 2nd 9 is derived from all digits 0 through 9 excluding the digit in
the 100th place.
The 8 is derived from all digits 0 through 9 excluding the first 2 digits.
b) How many of these are odd?
8  8  5 = 320
We use 5 for the last digit, since there are 5 odd numbers we can use:
1,3,5,7, and 9.
The 1st 8 is derived from all numbers 1 through 9 excluding the odd
number in the last digit.
The 2nd 8 is derived from all numbers 0 through 9 excluding the other
two digits.
2. There are 4 graduate students, 3 undergraduate students, and 3 professors
at a conference.
a) How many ways can they line up in the order: undergrad, grad, prof?
(3  2  1)(4  3  2  1)(3  2  1) = 3!  4!  3! = 864
b) How many ways can they line up if they remain grouped, but not
necessarily in any order?
3!  4!  3!  3! = 5184
There are 3! distinct arrangements of undergraduate students.
There are 4! distinct arrangements of graduate students.
There are 3! distinct arrangements of professors.
There are 3! distinct arrangements of the 3 groups.
3. How many 9 digit numbers, using 1 through 9, are possible where 1 and 2
precede 3 and 4, and no digit is used twice in the same number?
(e.g. 928173645)
We can use the calculation: 7  6  7  6  5  4  3  2  1 to tell us
how many 9 digit numbers there are where 3 and 4 do not come first, but
this is not what we’re looking for. Why doesn’t this work? And, is this
number less than or greater than the real answer?
4. Lisa Berger has 7 students in her MAE 301/501 class, of which 3 are
boys and 4 are girls. They all line up in front of her office to ask her a
question about the homework due the next day. Find the number of ways
they can line up where:
a) the 3 boys are first in line
b) the 3 boys are together in line
c) the 4 girls are together in line
d) No 2 girls are together