# 9ourr~ns— I EXAII

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9ourr~ns— I EXAII
```9ourr~ns—
Math 3283W
Exam 1
-
EXAII I
Page 2 of 7
02/19/10
~~sure to answer a total of FIVE QUESTIONS: Complete all questions 1 through ~, a1J
then either 5(a) or 5~. Answer the questions in the space provided on the question sheets.
If you need extra space, write on the other side of the page, in this case please clearly indicate
j~~t your work is continued on the. other side.
1. (20 points) Determine if the following pairs of statements are equivalent. In each part, provide
proof if the statements are equivalent, or if they are not, provide a case where they have different
truth values.
--
(a)
-
(FAQ) VR~versusFA(QvR)
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(b) (FAQ) ~ R versis (F ~ R) V
(Q
R).
a-se
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Math 3283W
2.
(20
Exam 1
-
Page 3 of 7
02/19/10
points) Quantify the following statements using the mathematical expression
P(z,y): x<y.
Your responses should contain no use of the symbols “<“or”>.”
(a) Given any two real numbers, a and b with a < b; their average,
less than b.
is greater than a but
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(b) Given any positive number less than 1, its square is less than itself.
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Math 3283W
Exam 1
-
Page 4 of 7
02/19/10
p
3. (20 points) Consider the following statement:
(*) If A and B are subsets of IR such that A fl B is infinite, then both A and B are infinite.
(a) Write the contrapositive of (*).
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(b) Prove that (*) is true.
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(c) Write the converse of (*).
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(d.) Is the converse of (*) true or false? Justi~’ your answer with either a proof or a counterex
ample.
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S B N]
MATH3283W EXAM1 SOLUTIONS
5.
/
(a) Consider the set
A={~1InEN}
n+1
i Show that A is bounded above abd bounded below.
= fl+1~2 = 1—41< lforanyn EN. SoAisbounded
above by 1.
On the other hand,
n—1\$o,n+l>o~Th—l\$o
71+1~
So A is bounded below by 0.
ii Find L = inf A and M = sup A (no proof is necessary in
this part). Determine whether or not L c A. Determine
whether or not M e A.
Since i-4~ is an increasing sequence, the ihfimum is the
first term and the supremum is the limit of the sequence.
L=infA=
1—1
=0,OEA
1+1.
-
M=supA= lim ~1=l,lØA
n—fcc TI + 1
iii Prove the value M
= sup A yoft found in (ii) is in fact the
least upper bound of A.
We know that 1 is an upper bound of A from (1). We also
have to show:
For any c >, Bx e A such that 1 e < x % 1
—
1
2
MATH3283W EXAM1 SOLUTIONS~
i.e. we want to find n E N such that
n—i
2
=1—
n+1
n+1
2
—e<—
n+1
2
6>
n+1
i—s<
*
*
*
n+1>-
By Archimedean Property we can find n C N such that
n>~.Then
2
n—i
n+1>n> -*i—s<
E
n+1
So 1 is the least upper bound of A.
(b) Determine if the following sets are bounnded above or below.
In each case, if the set is bounded above, find the supremum; if
the set is bounded below, find the inflmum.
i
The set is formed by two sequences:
A={3+~,3+~,... ~
and
A is a decreasiiig sequences with limit 3. So sup A = 3~
(first term) and inf A = 3. Similarly, supB = —i~ (first
term) and inf B = —2 (the limit). Hence the set has supre
mum 3~ and infimum —2.
ii {zeRIx>Oandz2—4x+3>Q}
x2~4x+3=(x—1)(x—3)=O=*~x=i,3
It is easy to show that x2 4x + 3 > 0 when x < 1 or
z > 3. Combine with the condition x > 0, we know that
the set is (0,1) U (3, oo). So it is not bounded above but
bounded below with inf = 0.
—
MATH3283W EXAM1 SOLUTIONS
3
iii{xERIx3—z<O}
x3—x=x(z—1)(x+1Y=O~r=—1,O,1
It is easy tq show that x3 x < 0 when x < —1 or
o < x < 1. So the set is (—oo,—1)U(0,1). It is not
bounded below but bounded above with sup = 1.
—
iv {1—.3,2—.33,3—.333,4—.3333,5--.33333,~ }U{~fri E
N}
The first set is an increasing sequence. It goes to oo when
n
co. The second set is a decreasing sequence with limit
o by Archirnedean property (or its corollary). So the set is
hot bounded above butbounded below with inf = 0.
—~
```
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