MAT 230 EXAM
ONE
1
0
y
x
Directions:
Type
your
solutions
into
this
document
and
be
sure
to
show
all
steps
for
arriving
at
your
solution.
Just
giving
a
final
number
may
not
receive
full
credit.
P
ROBLEM
1
(a)
The
domain
for
all
variables
in
the
expressions
below
is
the
set
of
real
numbers.
Determine
whether
each
statement
is
true
or
false.
(i)
∀
x
∃
y
(x
+
y
≥
0)
For
all
values
of
x,
there
exists
at
least
one
y
where
x
+
y
≥
0
is
true.
(ii)
∃
x
∀
y
(x
·
y
>
0)
There
e
exists
at
least
one
e
x
e
for
all
e
the
e
values
e
of
y
where
e
x
e
∗
e
y > 0.
If
y
=
0,
then
x
is
1
,
not
real
and
false.
(b)
Translate
each
of
the
following
English
statements
into
logical
expressions.
(i)
There
are
two
numbers
whose
ratio
is
less
than
1.
∃
x
∃
y
(
x
<
1)
(ii)
The
reciprocal
of
every
positive
number
is
also
positive.
∀
x
(x
>
0
→
1
e
>
e
0)
P
ROBLEM
2
Prove
the
following
using
the
specified
technique:
(a)
Let x and y e be two real numbers such that x + y e is rational. Prove e by
contrapositive that
if
x
is
irrational,
then
x
−
y
is
e
irrational.
Contrapositive:
If
x
-
y
is
rational
then
x
is
rational.
x
+
y
is
rational
(given)
x
-
e
y
is
rational
(premise)
Consider,
(x
+
y)
+
(x
-
y)
=
x
+
x
+
y
-
y
=
2x
Addition
of
rational
numbers
is
rational.
=
2x
is
rational
=
x
is
rational
Proved
(b)
Prove
by
contradiction
that
for
any
positive
two
real
numbers,
x
and
y,
if
·
e
x
either
x
<
8
or
y
<
8.
≤
y
e
50,
then
Contradiction:
x
≥
8
and
y
≥
e
8.
Consider,
x
∗
y
≤
50
substituting
values
of
x
and
y
(8)
∗
(8)
≤
50
64
≤
e
50
Therefore,
the
assumption
is
false
and
the
original
statement
is
proved
P
ROBLEM
3
Let
n
≥
1,
x
be
a
real
number,
and
x
≥
−
1.
Prove
the
following
statement
using
mathe-
matical
induction.
(1
+
x)
n
≥
1
+
nx
1)
Base
Case:
The
base
case
is
when
n
=
1.
We
have
(1
+
x)
1
=
1
+
x
e
=
1
+
1x
(1
+
x)
1
=
1
+
x
(1
+
x)
=
1
+
x
So
that
(1
+
x)
n
≥
1
+
nx
holds
for
integer
n
=
1
2)
Inductive
Step:
We
assume
that
(1
+
x)
k
To
show
that
(1
+
x)
k
+
1
≥
1
+
kx
for
k
>
1
≥
1
+
(k
+
1)x,
we
have:
(1
+
x)
k
+
1
=
(1
+
x)
k
(1
+
x)
≥
(1
+
kx)
(1
+
x)
=
1
+
kx
+
x
+
kx
2
=
1
+
(k
+
1)x
+
kx
2
=
1
+
(k
+
1)x
(1
+
x)
k
+
1
≥
1
+
(k
+
1)x
The
statement
holds
true
for
k+1.
Conclusion:
Both
the
base
case
and
the
inductive
step
have
been
proved
as
true,
so
by
mathematical
induction
the
statement
holds
for
every
natural
number
n.
P
ROBLEM
4
Solve
the
following
problems:
(a)
How many ways can a store manager arrange a group e e of 1 team leader
and 3 team workersfrom his 25 employees?
=>Total
number
of
employees
=
25
employees.
25
=>Number
of
ways
of
selecting
1
team
leader
from
25
employees
=
1
C
24
=>Number
of
ways
of
selecting
3
team
workers
from
rest
24
employees
=
3C
25 24
=>Total
number
of
ways
of
selecting
a
group
of
1
team
leader
and
3
team
workers
=
1
C
e
∗
3
C
=>Total
number
of
ways
of
selecting
a
group
of
1
team
leader
and
3
team
workers
=
25*2024
=>Total
number
of
ways
of
selecting
a
group
of
1
team
leader
and
3
team
workers
=
50,600
ways
(b)
A
state’s
license
plate
has
7
characters.
Each
character
can
be
a
capital
letter
(A
Z),
−
or
a
non-
zero
e
digit
e
(1
9).
−
How
e
many
e
license
e e
plates
e
start
e
with
e
3
e
capital
e
letters
e
and
e
end
e
with
e
4
digits
with
no
letter
or
digit
repeated?
=>State’s
licence
plate
has
7
characters.
=>Number
of
ways
of
placing
capital
letter
at
the
first
three
places
such
that
no
letter
re-peated
=
26*25*24
=>Number
of
ways
of
placing
digits
at
the
last
places
such
that
no
digit
repeated
=
9*8*7*6
=>Number
of
plates
starts
with
3
capital
letters
and
end
with
4
digits
with
no
letter
or
digitrepeated
=
26*25*24*9*8*7*6
=>Number
of
plates
starts
with
3
capital
letters
and
end
with
4
digits
with
no
letter
or
digitrepeated
=
47,174,400
plates
(c)
How
many
binary
strings
of
length
5
have
at
least
2
adjacent
bits
that
are
the
same
(“00”
or
“11”)
somewhere
in
the
string?
=>Strings
should
be
of
length
5
and
at
least
2
adjacent
bits
are
same
either
00
or
11
=>Total
number
of
strings
of
length
5
=
2
5
=>Total
number
of
strings
of
length
5
=
32
strings
=>Strings
01010
and
10101
only
are
violating
the
given
condition
means
no
00
or
11
adjacentcombination
=>Hence
total
number
of
favourable
ways
=
total
number
of
string
-
violating
cases
=>Hence
total
number
of
favourable
ways
=
32
-
2
=>Hence
total
number
of
favourable
ways
=
30
ways
=>Hence
30
strings
of
length
5
are
possible
with
the
given
condition
P
ROBLEM
5
A
class
with
n
kids
lines
up
for
recess.
The
order
in
which
the
kids
line
up
is
random
with
each
ordering
being
equally
likely.
There
are
two
kids
in
the
class
named
Betty
and
Mary.
The
use
of
the
word
“or”
in
the
description
of
the
events,
should
be
interpreted
as
the
inclusive
or.
That
is
“A
or
B”
means
that
A
is
true,
B
is
true,
or
both
A
and
B
are
true.
What
e
is
e e
the
e
probability
e e
that
e e
Betty
e
is
e e
first
e
in
e
line
e
or
e
Mary
e e
is
e
last
e e
in
e
line
e e
as
e
a
e
function
of
n?
Sim-
plify
your
final
expression
as
much
as
possible
and
include
an
explanation
of
how
you
calculated
this
probability.
Given
total
number
of
kids
lines
up
=
nLet,
B
=’Betty
is
first
in
line’
M
=’Mary
is
last
in
line’
P(B
or
M)
=
P(B)
+
P(M)
-
P(B
and
M)
P(B)
=
(n-1)!/n!
(fixing
the
position
of
Betty
is
first
in
line,
and
arrange
rest
n-1
kids)
=
1/nP(M)
=
(n-1)!/n!
(fixing
the
position
of
Mary
is
last
in
line,
and
arrange
rest
n-1
kids)
=
1/n
P(B
and
M)
=
(n-2)!/n!
(fixing
the
position
of
Betty
is
first
in
line
and
Mary
is
last
in
line,
andarrange
rest
n-2
kids)
=
1/n(n-1)
P(B
or
M)
=
1/n
+
1/n
-
1/n(n-1)
=((n-1)
+
(n-1)
+
1)/n(n-1)
=
(2n-1)/n(n-1)
P
ROBLEM
6
The
general
manager,
marketing
director,
and
3
other
employees
of
Company
A
are
hosting
a
visitby
the
vice
president
and
2
other
employees
of
Company
B.
The
eight
people
line
up
in
a
randomorder
to
take
a
photo.
Every
way
of
lining
up
the
people
is
equally
likely.
(a)
What
is
the
probability
that
the
general
manager
is
next
to
the
vice
president?
Total
Number
of
people
=
8
Total
Number
of
ways
8
people
are
line
up
=
8!
=
40320
Probability
that
the
general
manager
is
next
to
the
vice
president
=
P(A)
Approach:- e e combine the general manager and vice president as one e e then e
the e e total number ofpeople reduced to 7.
Arrange
the
7
people
in
a
line
=
7!
=
5040
Arrange
the
two-combine
people(general
manager
and
vice
president)
=
2!
=
2
P(A)
=
(5040
*
2
)/(40320)
=
1/4
(b)
What
is
the
probability
that
the
marketing
director
is
in
the
leftmost
position?
Probability
that
the
marketing
director
at
the
leftmost
position
=
P(B)
Approach:-
Fix
the
position
of
marketing
director
at
leftmost
position
and
arrange
rest
7
peo-ple
by
7!
ways
that
is
5040
ways
P(B)
=
5040/40320
=
1/8
(c)
Determine
whether
the
e
two
events
are
independent.
Prove
your
answer
by
showing
that
oneof
the
conditions
for
independence
is
either
true
or
false.
P(A)
→
refers
→
1st
sum
P
(B)
→
refers
→
2nd
sum
They’re
dependent.
