Download Problem Set 3 Solutions

Problem Set 3 Solutions
Section 3.1
2. For each of the following, use a counterexample to prove the statement is false.
(a) For each odd natural number n, if n > 3 then 3 divides (n2 − 1).
Consider n = 9. Then, n2 − 1 = 80 and 3 6 |80.
(b) For each natural number n, (3 · 2n + 2 · 3n + 1) is a prime number.
For n = 6, 3 · 26 + 2 · 36 + 1 = 1651 = 13 · 127.
p
(c) For all real numbers x and √
y, x2 + y 2 > 2xy.
If x = y = 1, then we have 2 > 2, which is false.
(d) For each integer a, if 4 divides (a2 − 1), then 4 divides (a − 1).
The statement can be shown to be false if we take a = 3, which gives that a2 − 1 = 8
and 4|8, but a − 1 = 2 and 4 6 |2.
3. Determine if each of the following statements is true or false. If a statement is true, then write
a formal proof of that statement, and if it is false, then provide a counterexample that shows
it is false.
(a) For all integers a, b and c with a 6= 0, if a|b then a|(bc).
True.
Let a, b, c ∈ Z with a 6= 0. We will show that if a|b then a|(bc). If a|b, then ∃k ∈ Z such
that b = ka. Then, bc = (ak)c = a(kc) and hence, a|bc.
(b) For all integers a and b with a 6= 0, if 6|(ab) then 6|a or 6|b.
False. Consider a = 2 and b = 3. then, 6|6, but 6 6 |2 and 6 6 |3.
(c) For all integers a, b, and c with a 6= 0, if a divides (b − 1) and a divides (c − 1), then a
divides (bc − 1).
True
Let a, b, c ∈ Z. If a|(b − 1) and a|(c − 1) then there exists integers m, n such that
am = b − 1 and an = c − 1. Then
bc − 1 = (am + 1)(an + 1) − 1
= a2 mn + am + an
= a(amn + m + n)
and since amn + m + n ∈ Z, a|(bc − 1).
(d) For each integer n, if 7 divides (n2 − 4), then 7 divides (n − 2).
False. Let n = 5. Then, 7|21 but 7 6 |3. The trick is to note that n2 − 4 = (n + 2)(n − 2)
and to look for an n such that 7|(n + 2) as it will therefore not be able to divide n − 2.
(e) For every integer n, 4n2 + 7n + 6 is an odd integer.
False. If n is even, then for some k ∈ Z, we have that n = 2k. Then,
4n2 + 7n + 6 = 4(2k)2 + 7(2k) + 6 = 16k 2 + 14k + 6 = 2(8k 2 + 7k + 3)
and since 8k 2 + 7k + 3 ∈ Z, 4n2 + 7n + 6 is even whenever n is even.
(f) For every odd integer n, 4n2 + 7n + 6 is an odd integer.
True
Let n be an odd integer. Then, there exists k ∈ Z such that n = 2k + 1. Now,
4n2 + 7n + 6 = 4(2k + 1)2 + 7(2k + 1) + 6
= 4(4k 2 + 4k + 1) + 14k + 7 + 6 = 16k 2 + 30k + 17
= 2(8k 2 + 15k + 8) + 1
Since 8k 2 + 15k + 3 ∈ Z, 4n2 + 7n + 6 is odd whenever n is odd.
(g) For all integers a, b, and d with d 6= 0, if d divides both a − b and a + b, then d divides a.
False. Let a = 3, b = 1 and d = 2. Then, d|4 and d|2 but d 6 |3.
(h) For all integers a, b, and c with a 6= 0, if a|(bc) then a|b or a|c.
False. Consider a = 6, b = 4 and c = 3. Then we have 6|12 but 6 6 |4 and 6 6 |3.
6. Determine if each of the following statements is true or false. If a statement is true, then write
a formal proof of that statement, and if it is false, then provide a counterexample that shows
it is false.
(a) For each integer a, if there exists an integer n such that a divides (8n + 7) and a divides
(4n + 1), then a divides 5.
True.
If a|(8n + 7) and a|(4n + 1), then there exists integers k, m such that ak = 8n + 7 and
am = 4n + 1. Multiplying the second equation by 2 gives 2am = 8n + 2 and subtracting
the second equation from the first yields ak − 2am = 5, or 5 = a(k − 2m). Since
k − 2m ∈ Z, we have that a|5.
(b) For each integer a, if there exists an integer n such that a divides (9n + 5) and a divides
(6n + 1), then a divides 7.
True
If a|(9n + 5) and a|(6n + 1) then there exists integers k, m such that ak = 9n + 5 and
am = 6n + 1. Multiplying the first by 2 and the second by 3 yields 2ak = 18n + 10
and 3am = 18n + 3. Subtracting the second from the first gives 2ak − 3am = 7, or
7 = a(2k − 3m). Since 2k − 3m ∈ Z, we have that a|7.
(c) For each integer n, if n is odd, then 8 divides (n4 + 4n2 + 11).
True. Suppose n is an odd integer. we will show 8|(n4 + 4n2 + 11). If n is odd, then
∃k ∈ Z such that n = 2k + 1. Consider
n4 + 4n2 + 11 = (2k + 1)4 + 4(2k + 1)2 + 11
16k 4 + 32k 3 + 24k 2 + 8k + 1 + 16k 2 + 16k + 4 + 11
= 16k 4 + 32k 3 + 40k 2 + 24k + 16
= 8(2k 4 + 4k 3 + 5k 2 + 3k + 2)
and since 2k 4 + 4k 3 + 5k 2 + 3k + 2 is an integer, 8|(n4 + 4n2 + 11).
(d) For each integer n, if n is odd, then 8 divides (n4 + n2 + 2n).
False. Let n = 1. Then, we have 8|4, which is not true.
7. Let a be an integer and let n ∈ N.
(a) Prove that if a ≡ 0 (mod n), then n|a.
If a ≡ 0 (mod n), then n|(a − 0), or n|a.
(b) Prove that if n|a, then a ≡ 0 (mod n).
If n|a, then n|(a − 0) and hence a ≡ 0 (mod n).
9. Let a and b be integers. Prove that if a ≡ 7 (mod 8) and b ≡ 3 (mod 8), then
(a) a + b ≡ 2 (mod 8)
If a ≡ 7 (mod 8) and b ≡ 3 (mod 8), then there exists integers k, m such that a − 7 = 8k
and b − 3 = 8m. Then,
a + b − 2 = 8k + 7 + 8m + 3 − 2 = 8k + 8m + 8 = 8(k + m + 1)
That is, a + b − 2 ≡ 0 (mod 8), or a + b ≡ 2 (mod 8).
(b) a · b ≡ 5 (mod 8)
If a ≡ 7 (mod 8) and b ≡ 3 (mod 8), then there exists integers k, m such that a − 7 = 8k
and b − 3 = 8m. Then,
ab − 5 = (8k + 7)(8m + 3) − 5 = 64km + 56m + 24k + 21 − 5 = 8(8km + 7m + 3k + 2)
That is, ab − 5 ≡ 0 (mod 8), or ab ≡ 5 (mod 8).
16. Determine if each of the following statements is true or false. Provide a counterexample for
statements that are false and provide a complete proof for those that are true.
√
(a) For all real numbers x and y, xy ≤ x+y
2 .
False. For any x, y ∈ R such that x < 0 and y < 0, we would have that xy > 0, but
x+y
2 < 0, making the inequality false.
2
(b) For all real numbers x and y, xy ≤ x+y
.
2
x+y 2
xy ≤
2
2
x + 2xy + y 2
xy ≤
4
4xy ≤ x2 + 2xy + y 2
0 ≤ x2 − 2xy + y 2
0 ≤ (x − y)2
This is true for all real numbers x and y, so the original statement is true.
√
(c) For all nonnegative real numbers x and y, xy ≤ x+y
2 . True
√
√
Let x, y ∈ R. If xy ≤ x+y
,
then
2
xy
≤
x
+
y.
Now,
2
√
2 xy ≤ x + y
4xy ≤ x2 + 2xy + y 2
0 ≤ x2 − 2xy + y 2
0 ≤ (x − y)2
which is always true for nonnegative real numbers x and y.
Note: we covered the failure of this if x and y are negative in (a). If only one of x and
√
y are negative, then xy is imaginary, so the hypothesis would be invalid.
Section 3.2
3.
(a) Write the contrapositive of the following statement:
√
For all positive real numbers a and b, if ab 6= a+b
2 , then a 6= b.
√
For all positive real numbers a and b, if a = b, then ab = a+b
2 .
(b) Is this statement true or false? Prove the statement if it is true and provide a counterexample if it is false.
True
a+a
2a
Suppose for positive real numbers a and b, a = b. Then, a+b
2 = 2 = 2 = a and
√
√
ab = a2 = a.
4. Are the following statements true or false? Justify your conclusions.
(a) For each a ∈ Z, if a ≡ 2 (mod 5), then a2 ≡ 4 (mod 5).
True
If a ≡ 2 (mod 5), then ∃k ∈ Z such that a − 2 = 5k, or a = 2 + 5k. Consider
a2 = (2 + 5k)2 = 4 + 20k + 25k 2
That is, a2 − 4 = 5(4k + 5k 2 ), or that a2 ≡ 4 (mod 5).
(b) For each a ∈ Z, if a2 ≡ 2 (mod 5), then a ≡ 4 (mod 5).
False. There is no integer a such that a2 ≡ 2 (mod 5), so the hypotheses cannot be
satisfied. All squares (modulo 5) have remainders 1, 4 or 5.
(c) For each a ∈ Z, a ≡ 2 (mod 5), if and only if a2 ≡ 4 (mod 5).
False. Part (b) shows this is false.
7. Consider the following proposition: For each integer a, a ≡ 2 (mod 8) if and only if (a2 + 4a) ≡
4 (mod 8).
(a) Write the proposition as the conjunction of two conditional statements.
(∀a ∈ Z)([(a ≡ 2 (mod 8)) → ((a2 + 4a) ≡ 4 (mod 8))]
∧[((a2 + 4a) ≡ 4 (mod 8)) → ((a ≡ 2 (mod 8))])
(b) Determine Determine if the two conditional statements in Part (a) are true or false. If a
conditional statement is true, write a proof, and if it is false, provide a counterexample.
(a ≡ 2 (mod 8)) → ((a2 + 4a) ≡ 4 (mod 8))
The statement is true.
If a ≡ 2 (mod 8), then there exists k ∈ Z such that a − 2 = 8k, or a = 8k + 2. Then
a2 + 4a − 4 = (8k + 2)2 + 4(8k + 2) − 4
= 64k 2 + 32k + 4 + 32k + 8 − 4
= 64k 2 + 8
= 8(8k 2 + 1)
Since 8k 2 + 1 ∈ Z, a2 + 4a − 4 ≡ 0 (mod 8), or (a2 + 4a) ≡ 4 (mod 8).
((a2 + 4a) ≡ 4 (mod 8)) → ((a ≡ 2 (mod 8))
The statement is false.
Let a = 6. Then we have 36 + 24 = 60 ≡ 4 (mod 8), but 6 6≡ 2 (mod 8).
(c) Is the given proposition true or false? Explain.
The given proposition is false since the statement Q → P is false.
9. A real number x is defined to be a rational number provided
there exists integers m and n with n 6= 0 such that x =
m
n.
A real number that is not rational is called an irrational number. It is known that if x is
a positive rational number, then there exists positive integers m and n with n 6= 0 such that
x= m
n.
Is the following proposition true or false? Explain.
For each positive real number x, if x is irrational, then
√
x is irrational.
The proposition is true. We will prove by contrapositive.
√
√
Suppose x is rational. We will show x is rational as well. If x is rational then there exists
2
2
some rational number pq with p, q ∈ Z and q 6= 0. Then, pq = pq2 = x, and since p2 and q 2
are integers with q 2 6= 0, x is rational.
14. Are the following propositions true or false? Justify your conclusion.
(a) There exists integers x and y such that 4x + 6y = 2.
The statement is true.
4x + 6y = 2(2x + 3y) and for x = 2, y = −1, we have that 2x + 3y = 1. Therefore,
4x + 6y = 2.
(b) There exists integers x and y such that 6x + 15y = 2.
The statement is false.
6x + 15y = 3(2x + 5y). Now, 2x + 5y ∈ Z, so we have that 3(2x + 5y) = 3n for some
n ∈ Z. But, if 3n = 2, then n = 23 , and 23 6∈ Z, so we contradict that n ∈ Z.
(c) There exists integers x and y such that 6x + 15y = 9.
The statement is true. Note that 6x + 15y = 9 is the same as 2x + 5y = 3. If we let
x = 4 and y = −1, we see that this equation is satisfied.
17. Let a and b be rational numbers such that a2 = b3 . Prove each of the propositions in Parts
(6a) through (6d).
(a) If a is even, then 4 divides a.
Assume that a is even. Then, there exists k ∈ Z such that a = 2k. So, 4k 2 = b3 . Since
b ∈ Z and b3 is even, b must be even as well. So, there exists m ∈ Z such that b = 2m.
The, we have that 4k 2 = 8m3 , or that k 2 = 2m3 . Thus, k 2 is even. Since k ∈ Z and k 2
is even, k is even as well. So, there exists q ∈ Z such that k = 2q, and since a = 2k, we
have that a = 4q, or that 4|a.
(b) If 4 divides a, then 4 divides b.
Since 4|a, there is an integer n ∈ Z such that a = 4n. Since a2 = b3 , we have that
b3 = 16n2 . So, b3 is even and so b is even. Then, there exists m ∈ Z such that b = 2m.
This implies that 8m3 = 16n2 , or m3 = 2n2 . So, since m3 is even and m ∈ Z, m is even
as well. So, there exists k ∈ Z such that m = 2k. But since b = 2m, it follows that
b = 4k, or that 4|b.
(c) If 4 divides b, then 8 divides a.
If 4|b, then since a2 = b3 , we can conclude that a2 is even. But this implies that a is
even. So, by (a), 4|a. So there exists integers s, t such that b = 4s and a = 4t. Then,
16t2 = 64s3 , or t2 = 4s3 . So, t2 is even and hence t must be even. Since a = 4t and t is
even, it follows that 8|a.
(d) If a is even, then 8 divides a.
If a is even, then by (a), 4|a. If 4|a, then by (b), 4|b. And if 4|b, then by (c), 8|a.
(e) Give an example of natural numbers a and b such that a is even and a2 = b3 , but b is
not divisible by 8.
This is the case when a = 8 and b = 4.
Section 3.3
2. Are the following statements true or false? Justify each conclusion.
(a) For all integers a and b, if a is even and b is odd, then 4 does not divide (a2 + b2 ).
The statement is true.
If a is even and b is odd, then there exists s, t ∈ Z such that a = 2s and b = 2t + 1.
Then,
a2 + b2 = (2s)2 + (2t + 1)2 = 4s2 + 4t2 + 4t + 1 = 4(s2 + t2 + t) + 1
where s2 + t2 + t ∈ Z. Therefore, a2 + b2 ≡ 1 (mod 4), or 4 6 |(a2 + b2 ).
(b) For all integers a and b, if a is even and b is odd, then 6 does not divide (a2 + b2 ).
The statement is true.
From (a), we see that we can write a2 + b2 as 4(s2 + t2 + t) + 1. Notice that this is odd,
so 2 6 |(a2 + b2 ). Since this is the case, no multiple of can divide a2 + b2 either.
(c) For all integers a and b, if a is even and b is odd, then 4 does not divide (a2 + 2b2 ).
The statement is true.
If a is even and b is odd, then there exists s, t ∈ Z such that a = 2s and b = 2t + 1.
Then,
a2 + 2b2 = (2s)2 + 2(2t + 1)2 = 4s2 + 8t2 + 8t + 2 = 2(2s2 + 4t2 + 4t + 1)
Since 2s2 + 4t2 + 4t + 1 ∈ Z, 2|(a2 + 2b2 ). But, notice that 2s2 + 4t2 + 4t + 1 =
2(s2 +2t2 +2t)+1, so this expression is odd and is therefore not divisible by 2. Therefore,
4 6= a2 + 2b2 .
(d) For all integers a and b, if a is odd and b is odd, then 4 divides (a2 + 3b2 ).
The statement is true.
If a and b are odd, then there exists integers k and m such that a = 2k+1 and b = 2m+1.
Consider
a2 +3b2 = (2k+1)2 +3(2m+1)2 = 4k 2 +4k+1+12m2 +12m+3 = 4(k 2 +k+3m2 +3m+1)
Since k 2 + k3 m2 + 3m + 1 ∈ Z, 4|(a2 + 3b2 ).
4. Prove that the cube root of 2 is an irrational number. That is, prove that if r is a real number
such that r3 = 2, then r is an √
irrational number.
3
Suppose
to
the
contrary
that
2 is rational. Then, there exists p, q ∈ Z with q 6= 0 such that
√
3
p
2 = q and p and q have no common divisors. Then,
2=
3
p
⇒ 2q 3 = p3
q
So, p3 must be even and since p ∈ Z, p is even as well. Therefore there exists k ∈ Z such that
p = 2k. Then, we have
2q 3 = (2k)3 ⇒ 2q 3 = 8k 3 ⇒ q 3 = 4k 3
That is, q 3 is even, and since q ∈ Z, q is even as well. But, if p and q are both even, then they
share the
common factor of 2, contradicting our assumption that they were relatively prime.
√
3
Thus, 2 is irrational.
7.
(a) Give an example that shows that the sum of two irrational numbers can be a rational
number. √
√
√
√
Consider 2 and − 2. The, 2 + (− 2) = 0, and 0 ∈ Q.
√
√ (b) Now explain why the following proof that
2 + 5 is an irrational number is not a
√
√
valid proof: Since 2√and √ 5 are both irrational numbers, their sum is an irrational
number. Therefore,
2 + 5 is an irrational number.
The proof is invalid because we are simply stating the conclusion as fact with no justification. They assume that irrational numbers are closed under addition, but (a) points
out that this is not the case.
√
√
(c) Is the real number 2 + 5 a rational number or an irrational number” Justify your
conclusion.
√
√
√
√
2 + 5 is irrational. We will prove this √
by contradiction.
So
suppose
2
+
5 is
√
rational.
Then
there
exists
a
∈
Q
such
that
2
+
5
=
a.
Then,
we
can
rewrite
this
as
√
√
5 = 2 − a, and squaring both sides gives
√
5 = a2 − 2a 2 + 2
We can rewrite this to give
√
2=
a2 − 3
2a
Since a2 − 3 and 2a are both integers, this shows that
√
2 is rational, a contradiction.
17. Is the following statement true or false? Justify your conclusion.
For each integer n that is greater than 1, if a is the smallest positive factor of n
that is greater than 1, then a is prime.
The statement is true.
If n is prime, then it has only one prime factor greater than 1, namely n, so the smallest
positive factor is indeed prime. Suppose n is composite. We will prove the assertion by
contradiction. So suppose to the contrary that a is the smallest positive factor of a, but
that a is composite. Then, there exists positive integers k, m such that 1 < k, m < a where
a = mk. Hence, a has positive factors smaller than a, and therefore k and m are factors of n
as well. But, k and m are both positive integers smaller than a, contradicting the assumption
that a is the smallest positive factor of n. Therefore, if a is the smallest positive factor of an
integer n, a is prime.