18.014 Problem Set 2 Solutions
Sam Elder
September 30, 2015
Problem 1. Let n be an integer and let x be a real number satisfying n < x < n + 1. Prove that x is not
an integer. (You may assume without proof that the sum or difference of integers is an integer.)
Solution. First, we need some information about the integers.
Lemma. 1 is the smallest positive integer, i.e. every positive integer n ≥ 1.1
Proof. To show this, we must use the property that the positive integers are contained in every inductive
set, and then prove that one such inductive set is {x : x ≥ 1}. Indeed, if x ≥ 1, x + 1 ≥ 1 + 1 ≥ 1 and 1 ≥ 1,
so this satisfies the two properties of an inductive set. This shows that every positive integer is at least 1,
as desired.
Now we proceed with the main proof by contradiction. Suppose that x is an integer satisfying n < x <
n + 1. Then x − n ∈ Z as a difference of two integers. By definition, n < x implies x − n > 0, so x − n ∈ Z+ .
By the lemma, this shows that x − n ≥ 1. Adding n to both sides, x ≥ n + 1, which contradicts x < n + 1.
Therefore, x must not be an integer.
Problem 2 (I 3.12-4). If x is an arbitrary real number, prove that there is exactly one integer n which
satisfies the inequalities n ≤ x < n + 1. This n is called the greatest integer in x and is denoted by [x]. For
example, [5] = 5, [5/2] = 2, and [−8/3] = −3.
Solution. First we show that there is such an integer n, and then we show that this integer is unique. We
offer two proofs of the existence claim:
Lemma. There exists some integer n such that n ≤ x < n + 1.
Proof 1. Suppose for sake of contradiction that no such integer n exists. By Exercise I 3.12-2, there is
an integer m such that m < x. We prove by induction that every integer n ≤ x. This is easily true by
transitivity for n ≤ m, so we take m as our base case.
Base Case. If n = m, m < x so the claim is satisfied.
Inductive Step. Suppose that n ≤ x, and we will show that n+1 ≤ x. Indeed, otherwise we have n ≤ x < n+1,
but we have assumed that no such integer n exists. Therefore, n + 1 ≤ x.
So we have shown that every integer n ≤ x. But this directly contradicts Theorem I.29. So we conclude
that some such integer n ≤ x < n + 1 exists.
Proof 2. Define the set S = {m ∈ Z : m ≤ x}. Since x is an upper bound for S and S is nonempty by
Exercise I 3.12-2, we can define sup S = n. As x is an upper bound for S, we must have n ≤ x.
We now wish to show that n is an integer. By Theorem I.32(a) with h = 1 > 0, there is some m ∈ S
with m > n − 1. Since m ∈ S, we must have m ≤ sup S = n. If there is some m < k ∈ S, then k − m ∈ Z+
so k − m ≥ 1 by the lemma from Problem 1. Then k ≥ m + 1 > n but this contradicts the definition of
1 Since
positive integers aren’t particularly well-covered in the textbook and weren’t the focus of the lectures, students may
assume the result of this lemma without proof.
1
n = sup S. So there is no such k and m is an upper bound for S. As n is the least upper bound, n ≤ m, so
we conclude that n = m and n is therefore an integer.
Since n is an upper bound for S, the integer n + 1 6∈ S, which implies that x < n + 1, and we have
n ≤ x < n + 1 as desired.
Finally, we show uniqueness. Suppose m and n are two distinct integers satisfying m ≤ x < m + 1 and
n ≤ x < n + 1. If m > n, we also have m ≤ x < n + 1 so n < m < n + 1. But this contradicts problem 1
above. Similarly, if n < m, we have m < n < m + 1, another contradiction. So there cannot be two distinct
integers satisfying this pair of inequalities, as desired.
Problem 3 (I 4.10-4). Use induction to prove the binomial theorem
(a + b)n =
n X
n
k=0
k
ak bn−k .
Then use the theorem to derive the formulas
n n
X
X
n
n
k n
= 2 and
(−1)
= 0.
k
k
k=0
k=0
(Binomial coefficients are defined at the beginning of section 4.10. You can use Exercise I 4.10-3.)
Solution. As the problem suggests, we induct on n.
Base Case. We could take n = 0 as a base case, where the formula collapses to (a + b)0 = 1 = 00 a0 b0 , but
1
1!
1
1!
= 1·1
= 1 and 11 = 0!1!
= 1·1
= 1 so
we will also show n = 1 for clarity. When n = 1, 10 = 1!0!
1 X
1 k 1−k
1 0 1
1 1 0
a b
=
a b +
a b = b + a = (a + b)1 ,
k
0
1
k=0
as desired.
Inductive Step. Assume that
(a + b)n =
n X
n
k=0
k
ak bn−k ,
2
and we will show this formula for n + 1.
We start with the right hand side, and apply Exercise I 4.10-3:
n+1
X
k=0
n+1 n + 1 k n+1−k X
n
n
a b
=
+
ak bn+1−k
k
k−1
k
k=0
n+1
n+1
X n X n
k−1 n−(k−1)
=a
a
b
+b
ak bn−k .
k−1
k
k=0
n
−1
k=0
n
n+1
Since
=
= 0, we can ignore the first term of the first sum and the last term of the second sum.
Then we can reparameterize the first sum by replacing k − 1 with k. The sum will now go from 1 − 1 = 0
to (n + 1) − 1 = n. We then use the induction hypothesis:
n+1
X
k=0
n n X
X
n + 1 k n+1−k
n k n−k
n k n−k
a b
=a
a b
+b
a b
= a(a + b)n + b(a + b)n = (a + b)n+1 .
k
k
k
k=0
k=0
2 For clarity and to match the notation of Exercise I 4.10-3, I won’t be writing the formula for n = m and showing it for
n = m + 1, but this is also valid and perhaps more preferable.
2
Having proved the statement for n + 1 from the statement for n, induction is complete.
Finally, we verify those formulas. By taking x = a = 1, we get
n n X
X
n
n k n−k
=
1 1
= (1 + 1)n = 2n ,
k
k
k=0
k=0
as desired. By taking x = −1 and a = 1, we get
X
n
n X
n
k n
(−1)
=
(−1)k 1n−k = (−1 + 1)n = 0n = 0,
k
k
k=0
k=0
as desired.
Problem 4 (I 4.10-16). The numbers 1, 2, 3, 5, 8, 13, 21, . . . , in which each term after the second is the sum
of its two predecessors, are called Fibonacci numbers. They may be defined by induction as follows:
a1 = 1,
a2 = 2,
an+1 = an + an−1
Prove that
an <
if n ≥ 2.
√ !n
1+ 5
2
for every n ≥ 1.
Solution. As with most Fibonacci number related proofs, we proceed by induction.
Base √
Case. We need two base cases here: n = 1 and n = 2, as will become clear in a moment. Indeed, we
have 5 > 1 as 5 > 1, so
√
√ !1
1+1
1+ 5
1+ 5
<
=
.
a1 = 1 =
2
2
2
√
√ !2
6+2
6+2 5
1+ 5
a2 = 2 =
<
=
,
4
4
2
as desired.
√ n−1
√ n
and an−1 < 1+2 5
. By taking two base
Inductive Step. For n ≥ 2, we assume that an < 1+2 5
cases, we guarantee that this holds for all n ≥ 2. Then
√ !n
√ !n−1
1+ 5
1+ 5
an+1 = an + an−1 <
+
2
2
!
!
√
√ !n−1
√
√ n−1
1+ 5
1+ 5
3+ 5
1+ 5
+1 =
=
2
2
2
2
!
!
!
√ n−1
√
√ n−1
√ 2
√ !n+1
1+ 5
6+2 5
1+ 5
1+ 5
1+ 5
=
=
=
,
2
4
2
2
2
as desired, completing the induction step.
Induction is complete.3
3 This
solution is correct, but not quite written in the inductive framework that we have established. To match that framework
√ n
√ n−1
properly, we should instead prove by induction that for n ≥ 2, both an < 1+2 5
and an−1 < 1+2 5
. This redefinition
of the claim to be proved by induction also makes it clear that the base case should include bounds on both a1 and a2 , but is
not required of students’ solutions on this problem, as long as they include both base cases.
3
Problem 5. This exercise develops some fundamental properties of polynomials. Let f (x) =
a polynomial of degree n. Prove each of the following statements.
Pn
k=0 ck x
k
be
Part 5.1. If n ≥ 1 and f (0) = 0, then f (x) = xg(x) where g is a polynomial of degree n − 1.
Solution. Plugging in x = 0, we have 00 = 1 and 0k = 0 if k ≥ 1, so 0 = f (0) = c0 . Therefore, we can leave
out this term of the sum and
f (x) =
n
X
ck xk =
k=1
n−1
X
ck+1 xk+1 = x
k=0
n−1
X
ck+1 xk .
k=0
Pn−1
Define g(x) = k=0 ck+1 xk . This is clearly a polynomial, so it remains to show it is degree n − 1. Since f
is degree n ≥ 1, cn 6= 0. But cn is also the coefficient on the highest-degree term, xn−1 , in g(x), so g(x) is
degree n − 1, as desired.
Part 5.2. For each real a, the function p given by p(x) = f (x + a) is a polynomial of degree n.
Solution. We induct on n to prove this statement.4
Base Case. We could use n = 0 as the base case, but for clarity will show n = 1. When n = 1, f (x) = c0 +c1 x,
so p(x) = f (x + a) = c0 + c1 (x + a) = (c0 + c1 a) + c1 x. This is in the form of a polynomial of degree, and
since f (x) is degree 1, c1 6= 0 and so p(x) is also degree 1.
Inductive Step. Suppose the statement is true for all polynomials of smaller degree than n. Define g(x) =
Pn−1
k
n
k=0 ck x , so f (x) = g(x) + cn x . Now g(x) is a polynomial of lower degree so we can apply the induction
hypothesis to it: If q(x) = g(x + a), q is also polynomial of the same degree as g, and we can write
Pn−1
q(x) = k=0 dk xk .
Therefore
p(x) = f (x + a) = g(x + a) + cn (x + a)n = q(x) + cn (x + a)n .
n X
n k n−k
By Binomial theorem (Problem 3 on this problem set), (x + a)n =
x a
. Thus we have
k
k=0
p(x) =
n−1
X
k=0
dk xk + cn
n X
n
k=0
k
xk an−k =
n−1
X
k=0
dk + cn
n n−k
a
xk + cn xn .
k
This is also the form of a polynomial, and since cn is also the leading coefficient of f (x), it is nonzero, so
p(x) is also degree n, as desired.
Induction is complete.
Part 5.3. If n ≥ 1 and f (a) = 0 for some real a, then f (x) = (x − a)h(x), where h is a polynomial of degree
n − 1.
Solution. Define the function p by p(x) = f (x + a), which is a polynomial of degree n by the previous part.
We know p(0) = f (0+a) = f (a) = 0, so by the first part of this problem, p(x) = xq(x) for some polynomial q
of degree n − 1. Then f (x) = p(x − a) = (x − a)q(x − a) = (x − a)q(x + (−a)) and by the previous part again,
q(x + (−a)) is also a polynomial of degree n − 1. Letting h(x) = q(x + (−a)), we have f (x) = (x − a)h(x),
as desired.
Part 5.4. If f (x) = 0 for n + 1 distinct real values of x, then every coefficient ck is zero and f (x) = 0 for
all real x.
4 It is not necessary to induct on n here, but it might be a little simpler. One alternative is to simply apply binomial theorem
P
Pn
Pk
k l k−l
k
to get p(x) = f (x + a) = n
and then rearrange the terms of this sum.
k=0 ck (x + a) =
k=0 ck
l=0 l x a
4
Solution. We prove this by induction on n.
Base Case. If n = 0, then f (x) = c0 so if f (x) = 0 for 1 real value of x, c0 = 0 and f (x) = 0 for all real x,
as desired.
Inductive Step. Suppose n ≥ 1 and the statement is true for all smaller values of n. Let f (x) = 0 for the
n + 1 distinct real values a1 , a2 , . . . , an+1 . By the previous part, since f (an+1 ) = 0, f (x) = (x − an+1 )h(x)
Pn−1
for some degree n − 1 polynomial h(x) = k=0 dk xk .
We also know that 0 = f (ai ) = (ai − an+1 )h(ai ) for any i < n + 1. Since the ai are distinct, ai − an+1 6= 0,
so we conclude that h(ai ) = 0 for i = 1, 2, . . . , n. This is n = (n − 1) + 1 distinct real values, so by the
induction hypothesis, all the dk are zero and h(x) = 0 for all x. But then
f (x) = (x − an+1 )h(x) =
n−1
X
dk (x − an+1 )xk =
k=0
n−1
X
dk xk+1 −
k=0
n−1
X
dk an+1 xk =
k=0
n−1
X
0xk+1 −
k=0
n−1
X
0xk = 0,
k=0
so all of the coefficients of f (x) are 0 and f (x) = 0 for all x, as desired.5
Induction is complete.
Pm
Part 5.5. Let g(x) = k=0 bk xk be a polynomial of degree m, where m ≥ n. If g(x) = f (x) for m + 1
distinct real values of x then m = n, bk = ck for each k, and g(x) = f (x) for all real x.
Solution. We claim that h(x) = g(x) − f (x) is a polynomial of degree at most m. Indeed, we have
h(x) = g(x) − f (x) =
m
X
bk x k −
n
X
k=0
k=0
ck x k =
m
X
(bk − ck )xk ,
k=0
if we define ck = 0 for n < k ≤ m. Since there are no terms with xk for k > m, this is a polynomial of degree
d ≤ m, as desired.
Now, if g(x) = f (x) for m + 1 distinct real values of x, then h(x) = g(x) − f (x) = 0 on these m + 1 ≥ d + 1
distinct real values. By the previous part, this implies that the coefficients of h(x) are all 0 and h(x) = 0 for
all real x. Therefore bk = ck for all k as desired. Taking k = m, bm = cm , and since bm 6= 0, cm 6= 0 as well,
making f (x) of degree n = m as desired. Finally, h(x) = 0 for all real x implies g(x) = f (x) for all real x,
as desired.
Problem 6. In each case, find all polynomials p of degree ≤ 2 which satisfy the given conditions.
Part 6.1. p(0) = p(1) = p(2) = 1.
Answer. The only such polynomial is p(x) = 1.
Solution. Clearly p(x) = 1 satisfies these properties, so we must show that this is the only one. If q(x) = p(x)
is another potential polynomial of degree d ≤ 2, q(x) = p(x) for 3 ≥ d + 1 distinct values of x (namely,
0, 1, and 2). But then by part (e) of the previous problem, p(x) and q(x) have the same coefficients, as
desired.
Part 6.2. p(0) = p(1) = 1.
Answer. All solutions are of the form p(x) = 1 − ax + ax2 = 1 + ax(x − 1), where a ∈ R.
5 These last couple statements might seem redundant, but in other contexts, particularly finite fields, there are polynomials
which evaluate to 0 everywhere but have nonzero coefficients.
5
Solution. Indeed, if p(x) = 1 + ax(x − 1), then
p(0) = 1 + a · 0(0 − 1) = 1 + 0 = 1
p(1) = 1 + a · 1(1 − 1) = 1 + 0 = 1,
as desired. To show all polynomials are of this form, consider q(x) = p(x) − 1, which is also a polynomial of
degree at most 2. Then q(0) = 0, so by the last problem, q(x) = xr(x) where r(x) is a polynomial of degree
at most 1. Since 0 = q(1) = r(1), again by the last problem, r(x) = (x − 1)s(x) where s(x) is a polynomial
of degree at most 0. But such a polynomial is just a constant, so let s(x) = a. Then
p(x) = q(x) + 1 = xr(x) + 1 = ax(x − 1) + 1 = 1 − ax + ax2 ,
as desired.
Problem 7. Prove that 0.999 . . . = 1.
Solution. The number 0.999 . . . is defined (c.f. section I 3.15) as the supremum of the truncated decimal
expansions:
9
9
9
0.999 . . . = sup
+
+ ··· + k : k ≥ 1 .
10 100
10
Define xk =
9
10
+
9
100
Lemma. xk = 1 −
+ ··· +
9
.
10k
First we find a closed-form formula for these numbers xk :
1
.
10k
Proof. We prove this claim by induction.
Base Case. If k = 1, x1 =
9
10
=1−
1
10
Inductive Step. Suppose that xk = 1 −
xk+1
=1−
1
.
10k
1
,
10k
as desired.
Then
9
9
9
9
9
9
9
+
+ · · · + k+1 =
+
+ · · · + k + k+1
=
10 100
10
10 100
10
10
1
9
10
9
1
9
= xk + k+1 = 1 − k + k+1 = 1 − k+1 + k+1 = 1 − k+1 ,
10
10
10
10
10
10
as desired.
Induction is complete.
Now we have 0.999 . . . = sup{1 − 101k : k ≥ 1}. Clearly 1 is an upper bound for this set, as 1 − 101k < 1.
Suppose for sake of contradiction that there is a smaller upper bound y < 1, and we will show that there is
some xk > y. For this purpose, we use the following lemma:
Lemma. If k ≥ 1, then 10k > k.
Proof. We prove this by induction.
Base Case. k = 1. Then 10k = 10 > 1 = k, as desired.
Inductive Step. If 10k > k for k ≥ 1, then 10k+1 = 10 · 10k > 2 · 10k > 2k ≥ k + 1, as desired.
Induction is complete.
Now if y < 1, 1 − y > 0, so there is some integer k such that k(1 − y) > 1 by Theorem I.30. Therefore, by
this lemma, 10k (1 − y) > 1. Solving for y, we get 1 − y > 10−k , or y < 1 − 10−k = xk , as desired. Therefore,
no such y < 1 is an upper bound, making 1 the least upper bound, as desired.
6