MAT 230 EXAM ONE
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Jacqueline Amoah
1
then x is
1
, not real and false.
x
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 exists at least one x for all the values of y where x
∗
y
>
0. If y
=
0,
0
(b)
Translate each of the following English statements into logical expressions.
(i)
There are two numbers whose ratio is less than 1.
∃
x
∃
y (
y
<
1)
(ii)
The reciprocal of every positive number is also positive.
∀
x (x
>
0
→
1
>
0)
P
ROBLEM
2
≤
· ≤
∗ ≤
∗ ≤
≥ ≥
Prove
i
the
i
following
i
using
i
the
i
specified
i
technique:
(a)
Let
i
x
i
and
i
y
i
be
i
two
i
real
i
numbers
i
such
i
that
i
x
i
+
i
y
i
is
i
rational.
i
Prove
i
by
i
contrapositive
i
that
i
if
i
x
is
i
irrational,
i
then
i
x
i
−
i
y
i
is
i
irrational.
Contrapositive:
i
If
i
x
i
-
i
y
i
is
i
rational
i
then
i
x
i
is
i
rational.
i
x
i
+
i
y
i
is
i
rational
i
(given)
x
i
-
i
y
i
is
i
rational
i
(premise)
Consider,
i
(x
i
+
i
y)
i
+
i
(x
i
-
i
y)
i
=
i
x
i
+
i
x
i
+
i
y
i
-
i
y
i
=
i
2x
i
Addition
i
of
i
rational
i
numbers
i
is
i
rational.
=
i
2x
i
is
i
rational
=
i
x
i
is
i
rational
i
Proved
(b)
Prove
i
by
i
contradiction
i
that
i
for
i
any
i
positive
i
two
i
real
i
numbers,
i
x
i
and
i
y,
i
if
i
x
i i
y
50,
i
then
i
either
i
x
i
<
i
8
i
or
i
y
i
<
i
8.
Contradiction:
i
x 8
i
and
i
y 8.
Consider,
i
x y 50
i
substituting
i
values
i
of
i
x
i
and
i
y
i
(8) (8) 50
64 50
Therefore,
i
the
i
assumption
i
is
i
false
i
and
i
the
i
original
i
statement
i
is
i
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
=
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
1
+
kx for k
>
1
To show that (1
+ x)
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
i
the
i
following
i
problems:
(a)
Howi manyi waysi cani ai storei manageri arrangei ai groupi ofi 1i teami leaderi andi 3i teami
workersi fromi hisi 25i employees?
=>Total
i
number
i
of
i
employees
i
=
i
25
i
employees.
25
=>Number
i
of
i
ways
i
of
i
selecting
i
1
i
team
i
leader
i
from
i
25
i
employees
i
=
i
1
i
C
24
=>Number
i
of
i
ways
i
of
i
selecting
i
3
i
team
i
workers
i
from
i
rest
i
24
i
employees
i
=
i
3
i
C
25 24
=>Total
i
number
i
of
i
ways
i
of
i
selecting
i
a
i
group
i
of
i
1
i
team
i
leader
i
and
i
3
i
team
i
workers
i
=
i
1
i
C
i
∗
i i
3
i
C
=>Total
i
number
i
of
i
ways
i
of
i
selecting
i
a
i
group
i
of
i
1
i
team
i
leader
i
and
i
3
i
team
i
workers
i
=
i
25*2024
=>Total
i
number
i
of
i
ways
i
of
i
selecting
i
a
i
group
i
of
i
1
i
team
i
leader
i
and
i
3
i
team
i
workers
i
=
i
50,600
i
ways
(b)
A
i
state’s
i
license
i
plate
i
has
i
7
i
characters.
i
Each
i
character
i
can
i
be
i
a
i
capital
i
letter
i
(A
i
Z),
i
or
i
a
i
non-
zero
i
digit
i
(1
i
9).
i
How
i
many
i
license
i
plates
i
start
i
with
i
3
i
capital
i
letters
i
and
i
end
i
with
i
4
i
digits
i
with
i
no
i
letter
i
or
i
digit
i
repeated?
=>State’s
i
licence
i
plate
i
has
i
7
i
characters.
=>Number
i
of
i
ways
i
of
i
placing
i
capital
i
letter
i
at
i
the
i
first
i
three
i
places
i
such
i
that
i
no
i
letter
i
re-
i
peated
i
=
i
26*25*24
=>Number
i
of
i
ways
i
of
i
placing
i
digits
i
at
i
the
i
last
i
places
i
such
i
that
i
no
i
digit
i
repeated
i
=
i
9*8*7*6
=>Number
i
of
i
plates
i
starts
i
with
i
3
i
capital
i
letters
i
and
i
end
i
with
i
4
i
digits
i
with
i
no
i
letter
i
or
i
digit
i
repeated
i
=
i
26*25*24*9*8*7*6
=>Number
i
of
i
plates
i
starts
i
with
i
3
i
capital
i
letters
i
and
i
end
i
with
i
4
i
digits
i
with
i
no
i
letter
i
or
i
digit
i
repeated
i
=
i
47,174,400
i
plates
(c)
How
i
many
i
binary
i
strings
i
of
i
length
i
5
i
have
i
at
i
least
i
2
i
adjacent
i
bits
i
that
i
are
i
the
i
same
i
(“00”
i
or
i
“11”)
i
somewhere
i
in
i
the
i
string?
=>Strings
i
should
i
be
i
of
i
length
i
5
i
and
i
at
i
least
i
2
i
adjacent
i
bits
i
are
i
same
i
either
i
00
i
or
i
11
=>Total
i
number
i
of
i
strings
i
of
i
length
i
5
i
=
i
2
5
=>Total
i
number
i
of
i
strings
i
of
i
length
i
5
i
=
i
32
i
strings
=>Strings
i
01010
i
and
i
10101
i
only
i
are
i
violating
i
the
i
given
i
condition
i
means
i
no
i
00
i
or
i
11
i
adjacent
i
combination
=>Hence
i
total
i
number
i
of
i
favourable
i
ways
i
=
i
total
i
number
i
of
i
string
i
-
i
violating
i
cases
=>Hence
i
total
i
number
i
of
i
favourable
i
ways
i
=
i
32