Solutions of Practice Test 1 MA225 ав бд гегзжй гзжй ¥ ¨ ¦! © ¦! " ¨
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Solutions of Practice Test 1
MA225
Directions: On the test no aid (books, notes, calculators or any other kind) will be allowed. You will have 75 minutes. Make sure that you read the questions carefully, you
write understandably in full sentences, and that you show all work to get partial credit.
1. Determine whether the following is a tautology, contradiction or neither. Prove your
answer.
Solution: The truth table is
thus it is neither a tautology nor a contradiction.
2. Give a denial of the following sentences. Translate your denials from English into symbolic sentences. Make sure that your answers are in a form with no negations in front of
quantifiers.
(a) All people are honest or no one is honest.
(b) There is a smallest integer.
(c)
"!$#%'&()*#% +-,./01,-
2#45 3 ,-76
(d) Between any real number and any larger real number, there is a rational number.
Solution:
(a) Some people are not honest and some people are honest, or symbolically
, where the universe is all people, and
" is honest".
+8#%/94:#%;
+8#%'194:#%
94*#% 5 #
1
exist an integer = which is smaller than < , or symbolically
>! <@?BA '+ =C?BA ' =E< there
D< .
+8#%'&F)*#%;
>!%,-'):,.G2# 5 ,.H6 , or equivalently +IJ#%K)*#% .
(c)
+8# ?L /+-, ?L '& # D ,
M>!%N ?BO /1NQPR#ST,UPNV76 .
(d)
(b) For all integer
3. Find the converse and the contrapositive of the following propositions and determine
the truth value of the original statement and its converse.
<@]R^ or @< ]_a` . (<@?L )
(b) The differentiability of b is sufficient for f to be continuous. ( b
(a)
WYX<QZ\[
only if
is a real function)
Solution:
(a) Converse: If
or
then
Contrapositive: If
and
then
original statement, and both are False.
<@]R^ P <c]_a` P WYX<QZ\[ . False.
< ^ < _a`
WDd<@Z[ . The contrapositive is equivalent to
b
(b) Converse: The differentiability of is necessary for f to be continuous, or equivalently,
if is continuous then is differentiable. This is False.
Contrapositive: If is not continuous then is not differentiable. The contrapositive is
equivalent to original statement, and both are True.
b
b
b
b
4. Give a direct proof for the following conditionals:
< egfihj<klfm<nfm[ is odd.
# #
s ?L
(b) If oqpr]` and ptsuD` then o klfip fms 5 ` has two real solutions. ( o p u
(a) If
<
is a natural number then
.)
Solution:
(a) Proof: Case 1. Assume that is odd. Then
is odd, since it is the product of odd
is odd since it is the sum of an odd and an even number. Then
numbers. Then
is even, since it is the sum of two odd numbers. Then
is odd
since it is the sum of an even and an odd number.
Case 2. Assume is even. Then
is even since all the three terms are even,
thus their sum is even. Then
is odd since it is the sum of an even and an
odd number.
<ef4hj<kf<
< e fhj< k
<
<e
<
<exfyhv<kzf<
<elfhj<%kgf\<0fi[
2
<ef4hv<kwf<f[
o # k{f|p # f|s 5 `
oqpr]` ptsuD`
Z^(o~s]`
p k X`
p?L
(b) Proof: One has to work backward, and first note that
has two real
solutions if and only if
. Now assume that
and
. This implies that
and has different signs, thus
, so
. Since
for all
, we get that
. Therefore
has two real solutions.
p}kZG^(o~su]`
#o k fmp # o~fisus D5 ` `
o
s
p k Z4^oqsr]`
# kzZ_
#
5. Prove the following statement by using contrapositive.
Claim: If does not divide
then is even.
#
# 5 hj=f
_ for some integer = . Then # kZ_ 5
Then
hv=f_ kZi_ 5 V^ =Qkwf^is= odd.
. Since =Qkf= is even (sum of two odd numbers),
5
^
Q
=
k
f
=
#
therefore ^ = k f= is divisible by , i.e. k Zy_ is divisible by . Therefore, if does not
#
#
divide kxZR_ then must be odd.
Solution:
Proof: assume that
6. Assign a grade A (correct), C (partially correct), or F (failure) to the following. Justify
your grades other that A.
#
Z #
)f #
# 5
# Z\ 5 `
# fi 5 h hv
o
p
hv 5
s } # 5 # k
hv
Z f4
N
N
N
(b) Claim: If the natural number has the property that divides o or divides p whenN
N
ever divides o~p , then is prime.
N
N
"Proof." Suppose that the natural number has the property that for all o and p if divides
N
N
N
o~p then either divides o or divides
p . Assume
that is composite. Then there exist }
N
N
N
5
natural numbers not equal _ or such that
. This implies that divides . Since
D N and D N we have 5that q doesn’t
divide neither nor . This is a contradiction to
5
and p . Thus with our assumptions N must be prime.
our assumption, taking o
(a) Claim: If is a real number then either
is irrational or
is irrational.
"Proof." It is known that is an irrational number, that is cannot be written in the form
of for any integers
. Let
. Then
rational, but
. If
were
rational then
for some integers
, but then
, so is also rational. This is
impossible, so
is irrational. Therefore, either
or
is irrational.
Solution:
(a) F. Cannot prove a universal statement by an example. This "proof" only examines the
case
.
# 5
(b) A.
7. Be able recite a proof for all of the following classical theorems:
h
is irrational;
3
The set of primes is infinite;
Rolle’s Theorem.
One of them will be on the test.
4
