Download Problem Set 4 Solutions

Problem Set 4 Solutions
Section 3.4
2. Prove that if u is an odd integer, then the equation x2 + x − u = 0 has no solution that is an
integer.
We will prove this by contradiction. Assume that is this quadratic equation as an integer
solution then there exists n ∈ Z such that n2 + n − u = 0. Then, u = n(n + 1). If n is odd,
n + 1 is even, making u even. If n is even, then n + 1 is odd and hence u is even. In either
case, this contradicts the assumption that u is odd and therefore we have no integer solution
to x2 + x − u = 0 is u is odd.
3. Prove that if n is an odd integer, then n = 4k + 1 for some integer k or n = 4k + 3 for some
integer k.
Let n be an odd integer. If we divide n by 4, then the Division Algorithm tells us that the
possible remainders are 0, 1, 2 and 3. If the remainder is 0 or 2, then n is even, so it must be
that the remainder is 1 or 3.
Case 1: The remainder is 1. Then, there exists k ∈ Z such that n = 4k + 1.
Case 2: The remainder is 3. Here, there exists k ∈ Z such that n = 4k + 3. Then,
n = 4k + 3 = 4k + (4 − 1)
= 4(k + 1) − 1
Hence, there exists an integer q such that n = 4q − 1. The proof is complete.
6. Are the following propositions true or false? Justify all your conclusions. If a biconditional
statement is found to be false, you should clearly determine if one of the conditional statements
within it is true. In that case, you should state an appropriate theorem for this conditional
statement and prove it.
(a) For all integers m and n, m and n are consecutive integers if and only if 4 divides
(m2 + n2 − 1).
For the implication ‘If m and n are consecutive integers, then 4 | (m2 + n2 − 1)’, notice
that exactly one of m and n must be even. We have two cases to consider; the case
where m is even and the case where m is odd.
Case 1: m is even. Here, n therefore is odd. So, for some k ∈ Z, let m = 2k and
n = 2k + 1. Then,
m2 + n2 − 1 = (2k)2 + (2k + 1)2 − 1
= 4k 2 + 4k 2 + 4k + 1 − 1
= 8k 2 + 4k
= 4(2k 2 + k)
Since 2k 2 + k ∈ Z, we have that 4 | (m2 + n2 − 1) when m is even and n is odd.
Case 2: m is odd. Here, n is even.Then there exits k ∈ Z such that m = 2k + 1 and
n = 2k. Here,
m2 + n2 − 1 = (2k + 1)2 + (2k)2 − 1
= 8k 2 + 4k
= 4(2k 2 + 4k)
Since 2k 2 + k ∈ Z, we have that 4 | (m2 + n2 − 1) when m is odd and n is even.
Now suppose 4 | (m2 + n2 − 1). We want to show that m and n are consecutive integers.
This statement is false. In m = 4 and n = 1. Then we have
m2 + n2 − 1 = 42 + 12 − 1 = 16
and 4 | 16. So, we have found a counterexample.
Therefore, the biconditional statement is false.
(b) For all integers m and n, 4 divides (m2 − n2 ) if and only if m and n are both even or if
m and n are both odd.
This biconditional statement is true.
First, we will prove that is m and n are both even or are both odd, then 4 divides
(m2 − n2 ). We do this in two cases.
Case 1: m and n are both even. Then, there exists integers p and q such that m = 2p
and n = 2q. Then,
m2 − n2 = (2p)2 − (2q)2
= 4p2 − 4q 2
= 4(p2 − q 2 )
Since p2 − q 2 ∈ Z, 4 | (m2 − n2 ).
Case 2: m and n are both odd. Then, there exists integers p and q such that m = 2p + 1
and n = 2q 1 . Then,
m2 − n2 = (2p + 1)2 − (2q + 1)2
= 4p2 + 4p + 1 − (4q 2 + 4q + 1)
= 4p2 + 4p − 4q 2 − 4q
= 4(p2 + p − q 2 − q)
Since p2 + p − q 2 − q ∈ Z, 4 | (m2 − n2 ).
In either case, we have that 4 | (m2 − n2 ), so the implication is proved.
Now we prove that if 4 | (m2 − n2 ) then either m and n are both even or m and n are
both odd. We will prove this by contrapositive. That is, we will show that is m and n
are both not even or both odd, then 4 6 |(m2 − n2 ).
If m and n are both not the same parity, then we have two cases to consider: when m
is even and when m is odd.
Case 1: m is even. Then, n is odd and there exists integers p and q such that m = 2p
and n = 2q + 1. Then,
m2 − n2 = (2p)2 − (2q + 1)2
= 4p2 − (4q 2 + 4q + 1)
= 4(p2 − q 2 − q) − 1
And since p2 − q 2 − q ∈ Z, 4 does not divide m2 − n2 .
Case 2: m is odd. Then, n is even and there exists integers p and q such that m = 2p + 1
and n = 2q. Then,
m2 − n2 = (2p + 1)2 − (2q)2
= 4p2 + 4p + 1 − 4q 2
= 4(p2 + p − q 2 ) + 1
Since p2 + p − q 2 ∈ Z, 4 does not divide m2 − n2 .
In either case, we have that if m and n have different parity, then 4 does not divide
m2 − n2 and the proof is complete.
7. Is the following proposition true of false? Justify your conclusion with a counterexample or a
proof.
For each integer n, if n is odd, then 8 | (n2 − 1).
The statement is true.
If n is an odd integer, then there exists k ∈ Z such that n = 2k + 1. Then,
n2 − 1 = (2k + 1)2 − 1 = 4k 2 + 4k = 4k(k + 1)
Notice that k and k + 1 are consecutive integers. We have two cases to consider.
Case 1: k is even. Then, k + 1 is odd and there exists an integer p such that k = 2p and
k + 1 = 2p + 1. Now,
4k(k + 1) = 4(2p)(2p + 1) = 8(2p2 + p)
Since 2p2 + p ∈ Z, 8 | (n2 + 1).
Case 2: k is odd. Then, k + 1 is even and so there exists p ∈ Z such that k = 2p + 1 and
k + 1 = 2p + 2 = 2(p + 1). Now,
4k(k + 1) = 4(2p + 1)(2(p + 1)) = 8(2p2 + 3p + 1)
and since 2p2 + 3p + 1 ∈ Z, 8 | (n2 − 1).
In either case we have that 8 | (n2 − 1) when n is an odd integer.
10.
(a) Prove Part (2) of Proposition 3.23.
For each x ∈ R, | − x| = |x|.
There are three cases to consider.
Case 1: x = 0
In this case, −x = 0 = x, so trivially, we have that | − x| = |x|.
Case 2: x > 0
Here, −x < 0 and so we have | − x| = −(−x) = x = |x|.
Case 3: x < 0
In this case, −x > 0. So, | − x| = −x = |x|.
In each case, we have that | − x| = |x|, proving the statement.
(b) Prove Part (2) of Theorem 3.25.
For all real numbers x and y, |xy| = |x||y|.
We will prove this using cases.
Case 1: a ≥ 0 and b ≥ 0.
Here, ab ≥ 0. So we have
|ab| = ab = |a||b|
as desired.
Case 2: a ≥ 0 and b < 0.
In this case, ab ≤ 0. So, we have
|ab| = −ab = a(−b) = |a||b|
Case 3: a < 0 and b ≥ 0.
We have that ab ≤ 0 and so
|ab| = −ab = (−a)b = |a||b|
Case 4: a < 0 and b < 0.
We have that ab > 0 and so
|ab| = ab = (−a)(−b) = |a||b|
In all cases, ab| = |a||b| as desired.
11. Let a be a positive real number. In Part (1) of Theorem 3.25, we proved that for each real
number x, |x| < a if and only if −a < x < a. It is important to realize that the sentence
−a < x < a is actually the conjunction of two inequalities. That is, −a < x < a means that
−a < x and x < a.
(a) Complete the following statement: For each real number x, |x| ≥ a if and only if ...
For each real number x, |x| ≥ a if and only if x ≥ a or x ≤ −a.
(b) Prove that for each real number x, |x| ≤ a if and only if −a ≤ x ≤ a.
There are two cases to consider: x ≥ 0 and x < 0.
Case 1: x ≥ 0
Notice that if x ≥ 0, then |x| = x. So, we have that if |x| ≤ a then x = |x| ≤ a, or x ≤ a.
On the other hand, if x ≤ a then we have |x| = x ≤ a.
Case 2: x < 0
Here, |x| = −x. So, if |x| ≤ a, then
−x = |x| ≤ a
x ≥ −a
And, if we have x ≥ −a, then
|x| = −x ≤ −(−a) = a
Taking these cases together we have that |x| ≤ a if and only if x ≤ a and x ≥ −a, as
desired.
(c) Complete the following statement: For each real number x, |x| > a if and only if ...
For each real number x, |x| > a if and only if x > a or x < −a.
12. Prove each of the following:
(a) For each nonzero real number x, x−1 =
Here, we have two cases to consider.
Case 1: x > 0
If x > 0 then x = |x|. Now,
1
|x| .
−1 1 x = x
1
=
x
1
=
|x|
Case 2: x < 0
If x < 0 then |x| = −x. It follows that
−1 1 x = x
1
=−
x
1
=
−x
1
=
|x|
Since we have that x−1 =
nonzero real numbers x.
1
|x|
in each case, we have shown that this is true for all
(b) For all real numbers x and y, |x − y| ≥ |x| − |y|.
Hint: An idea that is often used by mathematician is to add 0 to an expression “intelligently”. In this case, we know that −y + y = 0. Start by adding this “version” of 0
inside the absolute value sign of |x|.
Let x and y be real numbers. Consider
|x| − |y| = |x − y + y| − |y|
≤ |x − y| + |y| − |y|
= |x − y|
where |x − y + y| ≤ |x − y| + |y| by the Triangle Inequality. This completes the proof.
(c) For all real numbers x and y, ||x| − |y|| ≤ |x − y|.
Let x and y be real numbers. There are two cases to consider.
Case 1: |x| − |y| ≥ 0
If |x| − |y| ≥ 0, then ||x| − |y|| = |x| − |y|. Then,
||x| − |y|| = |x| − |y|
≤ |x − y|
by part (b).
Case 2: |x| − |y| < 0
Here, |x| − |y| < 0 implies that |y| − |x| > 0. Hence,
||x| − |y|| = |y| − |x|
≤ |y − x|
= | − (x − y)|
= |x − y|
again, by part (b). Since both cases show that ||x| − |y|| ≤ |x| − |y|, we have shown that
this statement is true for all real numbers x and y.
Section 3.5
3. Prove the symmetric property of congruence stated in Theorem 3.30.
Let n ∈ N. We will show for a, b ∈ Z, if a ≡ b (mod n), then b ≡ a (mod n).
If a ≡ b (mod n), then there exists an integer k such that a − b = nk. Now,
a − b = nk
−(b − a) = nk
b − a = −nk
b − a = n(−k)
where −k ∈ Z. That is, b ≡ a (mod n).
4.
(a) Let n ∈ N and let a ∈ Z. Explain why n divides a if and only if a ≡ 0 (mod n).
If n|a then there exists k ∈ Z such that a = nk. Then, since a = a − 0 = nk, we have
that a ≡ 0 (mod n).
(b) Let a ∈ Z. Explain why if a 6≡ 0 (mod 3), then a ≡ 1 (mod 3) or a ≡ 2 (mod 3).
If a 6≡ 0 (mod 3), then by the Division Algorithm, there exists q ∈ Z such that a = 3q + 1
or a = 3q + 2. If a = 3q + 1, then a − 1 = 3q and so a ≡ 1 (mod 3). If a = 3q + 2, then
a − 2 = 3q and so a ≡ 2 (mod 3).
(c) Is the following proposition true or false? Justify your conclusion.
For each a ∈ Z, a 6≡ 0 (mod 3) if and only if a2 ≡ 1 (mod 3).
First, suppose a 6≡ 0 (mod 3). There are two cases to consider.
Case 1: a ≡ 1 (mod 3)
Then, there exists k ∈ Z such that a = 3k + 1. Now,
a2 − 1 = (3k + 1)2 − 1
= 9k 2 + 6k + 1 − 1
= 3(3k 2 + 2k)
Since 3k 2 + 2k ∈ Z, 3 | (a2 − 1). That is, if a ≡ 1 (mod 3), a2 ≡ 1 (mod 3).
Case 2: a ≡ 2 (mod 3)
If a ≡ 2 (mod 3), then there exists k ∈ Z such that a = 3k + 2. Then,
a2 − 1 = (3k + 2)2 − 1
= 9k 2 + 12k + 4 − 1
= 3(3k 2 + 4k + 1)
Since 3k 2 + 4k + 1 ∈ Z, 3 | (a2 − 1). So, if a ≡ 2 (mod 3), then a2 ≡ 1 (mod 3).
When taking these cases together, it has been shown that a2 ≡ 1 (mod 3) whenever
a 6≡ 0 (mod 3).
Now, suppose a2 ≡ 1 (mod 3). We will show that a 6≡ 0 (mod 3) by contrapositive. That
is, suppose a ≡ 0 (mod 3). We will show a2 6≡ 1 (mod 3).
If a ≡ 0 (mod 3), then there exists k ∈ Z such that a = 3k. now,
a2 = (3k)2 = 3(3k 2 )
Since 3k 2 ∈ z, a2 ≡ 0 (mod 3), so a2 6≡ 1 (mod 3) as desired.
7. Prove the following proposition by proving its contrapositive. (Hint: Use case analysis. There
are several cases.)
For all integers a and b, if ab ≡ 0 (mod 3), then a ≡ 0 (mod 3) or b ≡ 0 (mod 3).
The contrapositive statement to be proven is ‘If a 6≡ 0 (mod 3) and b 6≡ 0 (mod 3), then
ab 6≡ 0 (mod 3). There are 4 cases to consider.
Case 1: a ≡ 1 (mod 3) and b ≡ 1 (mod 3)
Here, there exists integers m and n such that a = 3m + 1 and b = 3n + 1. Then, we have
ab = (3m + 1)(3n + 1)
= 9mn + 3m + 3n + 1
= 3(3mn + m + n) + 1
Since 3mn + m + n ∈ Z, then ab 6≡ 0 (mod 3) in this case.
Case 2: a ≡ 1 (mod 3) and b ≡ 2 (mod 3)
If a and b are as described above, there are integers m and n such that a = 3m + 1 and
b = 3n + 2. Then,
ab = (3m + 1)(3n + 2)
= 9mn + 6m + 3n + 2
= 3(3mn + 2m + n) + 2
Since 3mn + 2m + n ∈ Z, ab 6≡ 0 (mod 3).
Case 3: a ≡ 2 (mod 3) and b ≡ 1 (mod 3)
If a and b are as described above, then there exits integers n and m such that a = 3m + 2
and b = 3n + 1. Then,
ab = (3m + 2)(3n + 1)
= 9mn + 3m + 6n + 2
= 3(3mn + m + 2n) + 2
Since 3mn + m + 2n ∈ Z, ab 6≡ 0 (mod 3).
Case 4: a ≡ 2 (mod 3) and b ≡ 2 (mod 3)
For a and b both equivalent to 2 (mod 3), there exists integers m and n such that a = 3m + 2
and b = 3n + 2. Then
ab = (3m + 2)(3n + 2)
= 9mn + 6m + 6n + 4
= 3(3mn + 2m + 2n + 1) + 1
Since 3mn + 2m + 2n + 1 ∈ Z, ab 6≡ 0 (mod 3).
In each case, a 6≡ 0 (mod 3) and b 6≡ 0 (mod 3) implies that ab 6≡ 0 (mod 3), as desired.
9.
(a) Prove that the real number
√
3 is an irrational number. That is, prove that
If r is a positive real number such that r2 = 3. Then r is irrational.
We will prove this by contradiction. So, assume to the contrary that
√
3=
p
q
for integers
2
p and q with q 6= 0 and where p and q share no common factors. Then, 3 = pq2 , or
3q 2 = p2 . Then, 3 | p2 , which implies that 3 | p. So, there exists k ∈ Z such that p = 3k.
Then, 3q 2 = (3k)2 = 9k 2 , so we have that q 2 = 3k 2 , implying that 3 | q 2 . This gives that
3 | q and so there exists an integer m such that q = 3m. But, this implies that p and
q are both divisible by 3, contradicting our assumption that
√ p and q have no common
factors. Therefore,
there
are
no
integers
p
and
q
such
that
3 is rational and so it must
√
be the case that 3 is irrational.
√
(b) Prove that the real number 12 is an irrational number.
√
√
√
√
Notice that 12 = 2 3. If 12
We will prove that 12
√ is irrational by contradiction.
√
is rational, then so is 212 , which is to say that 3 is rational. But, in the last part of
√
this
√ exercise, we showed 3 to be irrational, contradicting our assumption. Therefore,
12 is irrational.
12.
(a) Prove that for each integer a, if a 6≡ 0 (mod 7), then a2 6≡ 0 (mod 7).
If a 6≡ 0 (mod 7), then by the Division Algorithm, a ≡ b (mod 7), where b = 1, 2, 3, 4, 5, 6.
We will consider these as separate cases.
Case 1: a ≡ 1 (mod 7).
Then, there exists k ∈ Z such that a = 7k + 1. Now,
a2 = (7k + 1)2
= 49k 2 + 14k + 1
= 7(7k 2 + 2k) + 1
Since 7k 2 + 2k ∈ Z, 7 6 |a2 when a ≡ 1 (mod 7).
Case 2: a ≡ 2 (mod 7).
Then, there exists k ∈ Z such that a = 7k + 2. Now,
a2 = (7k + 2)2
= 49k 2 + 28k + 4
= 7(7k 2 + 4k) + 4
Since 7k 2 + 4k ∈ Z, 7 6 |a2 when a ≡ 2 (mod 7).
Case 3: a ≡ 3 (mod 7).
Then, there exists k ∈ Z such that a = 7k + 3. Now,
a2 = (7k + 3)2
= 49k 2 + 42k + 9
= 7(7k 2 + 6k + 1) + 2
Since 7k 2 + 6k + 2 ∈ Z, 7 6 |a2 when a ≡ 3 (mod 7).
Case 4: a ≡ 4 (mod 7).
Then, there exists k ∈ Z such that a = 7k + 4. Now,
a2 = (7k + 4)2
= 49k 2 + 56k + 16
= 7(7k 2 + 8k + 2) + 2
Since 7k 2 + 8k + 2 ∈ Z, 7 6 |a2 when a ≡ 4 (mod 7).
Case 5: a ≡ 5 (mod 7).
Then, there exists k ∈ Z such that a = 7k + 5. Now,
a2 = (7k + 5)2
= 49k 2 + 70k + 25
= 7(7k 2 + 10k + 3) + 4
Since 7k 2 + 10k + 3 ∈ Z, 7 6 |a2 when a ≡ 5 (mod 7).
Case 6: a ≡ 6 (mod 7).
Then, there exists k ∈ Z such that a = 7k + 6. Now,
a2 = (7k + 6)2
= 49k 2 + 84k + 36
= 7(7k 2 + 12k + 5) + 1
Since 7k 2 + 12k + 5 ∈ Z, 7 6 |a2 when a ≡ 6 (mod 7).
When these cases are all considered, it has been shown that if a 6≡ 0 (mod 7) then
a2 6≡ (mod 7).
(b) Prove that for each integer a, if 7 divides a2 , then 7 divides a.
Consider the contrapositive of this statement, which says that if 7 does not divide a,
then 7 does not divide a2 . This contrapositive statement was proven in part (a).
√
(c) Prove that the real number 7 is irrational.
√
√
So
assume
7 is rational. Then
We will prove that 7 is irrational by contradiction.
√
there are non-zero integers p and q such that 7 = pq where p and q share no common
2
factors. Then, 7 = pq2 , or 7q 2 = p2 . Then, 7 | p2 , and as has been shown above, 7 | p.
So, there exists k ∈ Z such that p = 7k. The, 7q 2 = (7k)2 = 49k 2 , or q 2 = 7k 2 . So,
7 |q and so there is an integer m such that q = 7m. But, this implies that p and q are
√
both divisible by 7, contradicting that p and q share no common factors. Therefore, 7
is irrational.
Section 4.1
3. Use mathematical induction to prove each of the following:
(a) For each natural number n, 2 + 5 + 8 + . . . + (3n − 1) =
Let P (n) be the proposition that
n
X
n(3n + 1)
(3j − 1) =
2
j=1
n(3n+1)
.
2
for all n ∈ N. We will prove this by induction on n.
Base Case: P (1) is true because 3(1) − 1 = 2 = 3(1)+1
.
2
Inductive step: Assume for some k ∈ Z that
k
X
(3j − 1) =
j=1
We will show that
k(3k + 1)
2
k+1
X
(k + 1)(3k + 4)
(3j − 1) =
2
j=1
is also true. Consider
k+1
k
X
X
(3j − 1) =
(3j − 1) + 3(k + 1) − 1
j=1
j=1
k(3k + 1)
+ 3k + 2
2
3k 2 + k 6k + 4
=
+
2
2
3k 2 + 7k + 4
=
2
(k + 1)(3k + 4)
=
2
So, P (k + 1) is true whenever P (k) is true. Therefore, by mathematical induction P (n)
is true for all natural numbers n.
(b) For each natural number n, 1 + 5 + 9 + . . . + (4n − 3) = n(2n − 1).
Let P (n) be the statement that
=
n
X
(4j − 3) = n(2n − 1)
j=1
We will show that holds by mathematical induction.
Base case: P (1) is true because 4(1) − 3 = 1 = 1(2(1) − 1).
Inductive step: Suppose P (k) is true for some natural number k. We will show P (k + 1)
is true, which is to say that
k+1
X
(4j − 3) = (k + 1)(2k + 1)
j=1
Consider
k+1
X
(4j − 3) =
X
j = 1k (4j − 3) + [4(k + 1) − 3]
j=1
= k(2k − 1) + 4k + 1
= 2k 2 + 3k + 1
= (k + 1)(2k + 1)
This shows that P (k + 1) is true whenever P (k) is true. So, by mathematical induction,
n
X
(4j − 3) = n(2n − 1)
j=1
for all natural numbers n.
(c) For each natural number n, 13 + 23 + 33 + . . . + n3 =
h
n(n+1)
2
i2
.
Let P (n) be the statement
n
X
j=1
n(n + 1)
j =
2
3
2
for any n ∈ N. We will show this is true by mathematical induction.
.
Base case: P (1) is true because 13 = 1 = 1(1+1)
2
Inductive step: Assume P (k) is true for some natural number k. We will show that
P (k + 1) is true, which states that
k+1
X
j=1
(k + 1)(k + 2)
j =
2
3
2
Consider
k+1
X
3
j =
j=1
k
X
j 3 + (k + 1)3
j=1
2
4(k + 1)3
k(k + 1)
+
2
4
2
2
4(k + 1)3
k (k + 1)
+
=
4
4
k 2 (k + 1)2 + 4(k + 1)3
=
4
(k + 1)2 (k 2 + 4k + 4)
=
4
(k + 1)2 )(k + 2)2
=
4
(k + 1)(k + 2) 2
=
2
=
as desired. So, since P (k + 1) is true whenever P (k) is true, then by mathematical
induction,
n
X
n(n + 1) 2
3
j =
2
j=1
for any n ∈ N.
6.
(a) Calculate 1 + 3 + 5 + . . . + (2n − 1) for several natural numbers n.
For n = 1, we have a sum of 1.
For n = 2, we have a sum of 4.
For n = 3, we have a sum of 9.
(b) Based on your work in Exercise (6a), if n ∈ N, make a conjecture about the value of the
sum
n
X
1 + 3 + 5 + . . . + (2n − 1) =
(2j − 1)
j=1
n
X
(2j − 1) = n2
j=1
(c) Use mathematical induction to prove your conjecture in Exercise (6b).
Let P (n) be the proposition that
n
X
(2j − 1) = n2
j=1
for all natural numbers n. We will show this is true by induction on n.
Base case: P (1) is true because 2(1) − 1 = 1 = 12 .
Inductive step: Assume P (k) is true. We will show that in this case, P (k + 1) is also
true, which states that
k+1
X
(2j − 1) = (k + 1)2
j=1
Consider
k+1
X
(2j − 1) =
j=1
k
X
(2j − 1) + 2(k + 1) − 1
j=1
= k 2 + 2k + 2 − 1
= k 2 + 2k + 1
= (k + 1)2
This shows that P (k = 1) is true whenever P (k) is true. So, by mathematical induction,
n
X
(2j − 1) = n2
j=1
for all natural numbers n.
8. Use mathematical induction to prove each of the following:
(a) For each natural number n, 3 divides (4n − 1).
Let P (n) be the proposition that 3 divides (4n − 1) for any natural number n. We will
show this is true by mathematical induction.
Base case: P (1) is true since 3 divides 41 − 1 = 3.
Inductive step: Assume P (k) is true, which says that 3 divides (4k − 1). We will show
P (k + 1) is true, which says that 3 divides (4k+1 − 1). Now, if 3 divides (4k − 1), then
there exists an integer m such that 4k − 1 = 3m. Hence,
4k = 1 + 3m
4k · 4 = 4(1 + 3m)
4k+1 = 4 + 12m
4k+1 − 1 = 3 + 12m
4k+1 − 1 = 3(1 + 4m)
So, if 3 divides (4k − 1), then 3 divides (4k+1 − 1). Therefore, 3 divides (4n − 1) for any
natural number n.
(b) For each natural number n, 6 divides (n3 − n).
Let P (n) be the proposition that 6 divides (n3 − n) for each natural number n. We will
show this is true by mathematical induction.
Base case: P (1) is true because 6 divides 13 − 1 = 0.
Inductive step: Assume 6 divides k 3 − k for some natural number k. We will show that
this implies that 6 divides (k + 1)3 − (k + 1). Now, if 6 divides (k 3 − k), then there exists
an integer m such that k 3 − k = 6m. Then,
(k + 1)3 − (k + 1) = (k 3 + 3k 2 + 3k + 1) − k − 1
= (k 3 − k) + (3k 2 + 3k)
= 6m + 3k(k + 1)
It remains to show that k(k + 1) is even. Now, if k is even, then k + 1 is odd and their
product is even. If k + 1 is even, then k + 1 − 1 = k is odd and so their product is even
as well. So, there exists q ∈ Z such that k(k + 1) = 2q. Then, we have
(k + 1)3 − (k + 1) = 6m + 6q
= 6(m + q)
That is, 6 divides (k + 1)3 − (k + 1) whenever 6 divides k 3 − k. So, by mathematical
induction, 6 divides n3 − n for each natural number n.
11.
(a) Calculate the value of 5n − 2n for n = 1, n = 2, n = 3, n = 4, n = 5 and n = 6.
n = 1: 51 − 21 = 3
n = 2: 52 − 22 = 21
n = 3: 53 − 23 = 117
n = 4: 54 − 24 = 609
n = 5: 55 − 25 = 3093
n = 6: 56 − 26 = 15561
(b) Based on your work in Part (a), make a conjecture about the value of 5n − 2n for each
natural number n.
The conjecture is that 3 divides 5n − 2n for each natural number n.
(c) Use mathematical induction to prove your conjecture in Part (b).
Let P (n) be the proposition that 3 divides 5n − 2n for each natural number n.
Base case: P (1) is true because 3 divides 51 − 21 = 3.
Inductive step: Suppose 3 divides 5k − 2k for some natural number k. We will show
that 3 divides 5k+1 − 2k+1 . Now, if 3 divides 5k − 2k , then there exists m ∈ Z such that
5k − 2k = 3m. Then,
5k = 3m + 2k
5k+1 = 5 3m + 2k
= 15m + 5 · 2k
= 15m + (3 + 2) · 2k
k
k
= 3 · 5m
+ 3 · 2 + 2 · 2
= 3 5m + 2k + 2k+1
That is 5k+1 − 2k+1 = 3(5m + 2k ), which shows that 3 divides 5k+1 − 2k+1 when 3 divides
5k − 2k . So, by mathematical induction, 3 divides 5n − 2n for each natural number n.
15. Let a be a real number. We will explore the derivatives of the function f (x) = eax . By using
the chain rule, we see that
d ax
(e ) = aeax
dx
Recall that the second derivative of a function is the derivative of the derivative function.
Similarly, the third derivative is the derivative of the second derivative.
(a) What is
d2
dx2
(eax ), the second derivative of eax ?
d2 ax
(e ) = a2 eax
dx2
(b) What is
d3
dx3
(eax ), the third derivative of eax ?
d3 ax
(e ) = a3 eax
dx3
(c) Let n be a natural number. Make a conjecture about the nth derivative of the function
dn
ax )? this conjecture should be written as a selff (x) = eax . That is, what is dx
n (e
contained proposition including an appropriate quantifier.
For each natural number n,
dn ax
(e ) = an eax
dxn
(d) Use mathematical induction to prove your conjecture.
Let P (n) be the proposition that
dn ax
(e ) = an eax
dxn
for each natural number n. We will prove this is true by mathematical induction.
d
Base case: P (1) is true because dx
(eax ) = aeax .
Inductive step: Assume
dk
dxk
(eax ) = ak eax . We will show
dk+1
dxk+1
(eax ) = ak+1 eax . Now,
dk ax
(e )
dxk
d k ax =
a e
dx
d ax
(e )
= ak
dx
= ak · aeax
d
dk+1 ax
(e ) =
k+1
dx
dx
= ak+1 eax
This shows that P (k + 1) is true whenever P (k) is true. So, by mathematical induction,
dn
ax
n ax for all natural numbers n.
dxn (e ) = a e