MATH CSE20 Test 3 Review Sheet - SOLUTIONS
Test Tuesday November 12 in lecture: CENTER 115, 3:30pm
Textbook sections: Unit NT Sections 1 and 2
(1) All questions from Homeworks 5 and 6.
(2) (NT Review Q1) Prove the statement “If k ą 1 is an integer, then 2k ´ 1 is not a perfect
square.”
Proof: Let k ą 1 be an integer. Suppose, towards a contradiction, that there
is an integer n such that
2k ´ 1 “ n2 .
Then n2 is an odd integer. By theorems discussed in class, this implies that n
is an odd integer. Therefore, let c P Z be such that n “ 2C ` 1. Substituting,
we get
2k ´ 1 “ p2C ` 1q2 “ 4C 2 ` 4C ` 1
and solving for 2k :
2k “ 4C 2 ` 4C ` 2.
We can factor 2 from both sides and still get integers (since k ą 1):
2k´1 “ 2C 2 ` 2C ` 1 “ 2pC 2 ` Cq ` 1 “ 2` ` 1,
an odd number since ` P Z (by closure of integers under addition and multiplication). But, the only integer power of 2 that is odd is 1 “ 20 , so k ´ 1 “ 0 and
k “ 1, contradicting the assumption that k ą 1. Thus, it cannot be the case
that 2k ´ 1 is a perfect square.
(3) (NT Review Q9) Which of the following statements are true? Prove or give counterexamples.
These statements are implicit universal statements.
(a) If a and b ‰ 0 are rational and r ‰ 0 is real, then a ` br is rational only if r is
irrational.
False. Symbolically, this statement is
@a P Q@b P Q‰0 @r P R p a ` br P Q Ñ r R Qq
Consider the counterexample a “ 1, b “ 1, c “ 1. Then b ‰ 0 and
a ` br P Q but it’s not the case that r R Q.
(b) If a and b ‰ 0 are rational and r ‰ 0 is real, then a ` br is irrational only if r is
irrational.
True. Symbolically, this statement is
@a P Q@b P Q‰0 @r P R p a ` br R Q Ñ r R Qq
which is equivalent to replacing the inner implication by its contrapositive:
@a P Q@b P Q‰0 @r P R p r P Q Ñ a ` br P Qq
2
To prove this, let a, b, r be in their respective domains and assume (as
in a direct proof) that r P Q. So, so there integers m, n, c, d, x, y (with
n, c, d, y all ‰ 0 ) such that
m
c
x
a“
b“
r“ .
n
d
y
Substitution gives a ` br “ m
` dc xy “ mdy`ncx
. Since the integers are
n
ndy
closed under addition and multiplication,the numerator and denominators of this expression and each integers and the denominator is nonzero
since it is a product of nonzero integers. Thus, a ` br P Q as well.
(c) If a and b ‰ 0 are rational and r ‰ 0 is real, then r is rational only if a ` br is
rational.
True. Similar to (b).
(d) If a and b ‰ 0 are rational and r ‰ 0 is real, then a ` br is rational only if r is
rational.
True. Similar to (b).
(e) If a and b ‰ 0 are rational and r ‰ 0 is real, then a ` br is irrational only if r is
rational.
False. Similar to (a).
(4) Prove or disprove each of the following statements. You might find it useful to translate
them to formal statements first.
(a) There is a perfect square that can be written as a sum of two other perfect squares.
True. An example suffices: consider 52 “ 25 “ 16 ` 9 “ 42 ` 32 .
(b) There is an integer n such that 2n2 ´ 5n ` 2 is prime.
Symbolically, this statement is
Dn P Zp Pp2n2 ´ 5n ` 2q q
True. An example suffices: consider n “ 3, then 2n2 ´ 5n ` 2 “ 18 ´
15 ` 2 “ 5, a prime number.
(c) The average of any two even integers is even.
Symbolically, this statement is
a`b
qq
2
False. A counter-example suffices: consider the average of 4 and 6 which
is 5.
(d) There exists an integer m ě 3 such that m2 ´ 1 is prime.
False. Assume towards a contradiction that there is such an integer
m ě 3. Call N “ pm2 ´ 1q; so, N is prime Factoring gives m2 ´ 1 “
pm ` 1qpm ´ 1q. By assumption on m, each of m ` 1 and m ´ 1 is an
integer greater than 1. By construction of N , each of m ` 1, m ´ 1 is a
factor of N . Thus, we have found positive factors of N other than itself
and 1, contradicting it being prime.
(e) The difference of two odd integers is odd.
Symbolically, this statement is
@a P Z@b P Zp rEvenpaq ^ Evenpbqs Ñ Evenp
@a P Z@b P Zp rOddpaq ^ Oddpbqs Ñ Oddpa ´ bq q
3
False. A counter-example suffices: consider the difference of 1 and itself,
0, which is even.
(f) Every integer is a rational number.
Symbolically, this statement is
@a P Zpa P Qq
(g)
(h)
(i)
(j)
True. Let n be an arbitrary but fixed integer. Then we can write n “ n1
and since 1 is a nonzero integer, we have found witnesses for the fact
that n is a rational number.
If r is any rational number then 3r2 ´ 2r ` 4 is rational.
True. Similar to previous question.
The product of any two even numbers is divisible by 8.
False. A counter-example suffices: consider product of 2 and 6, which is
12 , and is not divisible by 8.
For positive integers a, b, d, if a ă b then pa%dq ă pb%dq.
False. A counter-example suffices: consider a “ 2, b “ 4, d “ 3. Then
a ă b but pa%dq “ 2 ą 1 “ pb%dq.
There is no least positive rational number.
Symbolically, this statement is
„ Da P Q@b P Qp a ď b q
or, equivalently,
@a P QDb P Qp b ă a q
True. Suppose, towards a contradiction, that there was. Call this number q0 . Consider half of this number: q20 . Note that this is a positive
number, it is rational, and it is less than q0 , contradicting the assumption that q0 is the least positive rational number.
?
(k) 3 2 is irrational.
?
True. Similar to proof that 2 is irrational after proving the fact that
for any integer n, if n3 is even then so is n.