Problem Set 1 Solutions

Problem Set 1 Solutions
1.1.4. Define P (n) to be the statement that for all k ∈ N with 1 ≤ k ≤ n,
Q(k) is true. We will apply the principle of (weak) induction to P (n). Note
that the assertion ‘P (n) holds for all n ∈ N’ is equivalent to the assertion ‘Q(n)
holds for all n ∈ N.’
Base step: We have assumed that Q(1) is true. Q(1) is equivalent to P (1),
so we conclude that P (1) is true.
Induction step: Assume P (n) is true, so by definition Q(k) is true for all k ∈ N
with 1 ≤ k ≤ n. Thus our assumption guarantees Q(n + 1) is true. We see
that both P (n) and Q(n + 1) are both true, and since the statement ‘P (n) and
Q(n + 1) are both true’ is equivalent to P (n + 1), we conclude P (n + 1) is true.
Thus the principle of weak induction applied to P (n) guarantees P (n) holds
for all n ∈ N. Thus Q(n) holds for all n ∈ N and the proof is complete.
1.1.5. There are at least two good ways of finding a formula for
Pn
k
k=1 2
.
1. Combinatorial approach: Let A(n) = {S ⊆ {1, 2, . . . , n + 1} : |S| = 3} be
the family of all subsets of the first n + 1 natural numbers that have size
3. For example,
A(3) = {{1, 2, 3} , {1, 2, 4} , {1, 3, 4} , {2, 3, 4}} .
Then from the definition of the binomial function, we have A(n) = n+1
3
(n)
(the set A(n) has n+1
members). Let Bk = S ∈ A(n) : max S = k + 1
3
for k ∈ {1, . . . , n}. For example,
(4)
B4
= {{1, 2, 5} , {1, 3, 5} , {1, 4, 5} , {2, 3, 5} , {2, 4, 5} , {3, 4, 5}} .
Then since the largest element in any member of A(n) is at most n + 1 and
Sn
(n)
at least 2 (in fact, at least 3), we have A(n) = k=1 Bk . Furthermore,
(n)
(n)
(n)
it is clear that if j 6= k, then Bj ∩ Bk = ∅. How large is each Bk ?
Well, if the largest element must be k + 1, then we must still choose 2
more elements
the numbers {1, . . . , k}, and any choice will suffice.
from
(n) (n) Hence Bk = k2 (in particular, note B1 = 12 = 0 since no subsets
(n)
have largest element 2). Since the sets Bk are disjoint, this immediately
yields
n n n
n + 1 X
X
k
(n) [ (n) (n) =
Bk = A =
.
Bk = 2
3
k=1
k=1
k=1
2. More direct approach: Using the fact that
1
k
2
=
k(k−1)
2
and using the
identities for sums discussed in class, we compute
n X
k
2
k=1
!
n
n
1 X 2 X
=
=
k −k =
k −
k
2
2
2
k=1
k=1
k=1
k=1
1 2n3 + 3n2 + n 3n2 + 3n
1 n(2n + 1)(n + 1) n(n + 1)
−
=
−
=
2
6
2
2
6
6
3
n −n
n(n + 1)(n − 1)
n+1
=
=
=
.
6
6
3
n
X
k(k − 1)
n
X
1
2
After finding a formula,P
our task is now to
verify it using induction. So let P (n)
n
denote the statement “ k=1 k2 = n+1
3 ”. Since
1 X
k
k=1
2
2(2 − 1)(2 − 2)
2
1+1
1
1(1 − 1)
=0=
=
=
,
=
=
2
6
3
3
2
P (1) is true. Now let us assume that P (n) is true. Then
n+1
X
k=1
k
2
n X
k
n+1
n+1
n+1
n(n + 1)(n − 1) n(n + 1)
+
=
+
=
+
=
6
2
2
2
3
2
k=1
=
n(n + 1)
n(n + 1)(n + 2)
((n − 1) + 3) =
=
6
6
n+2
.
3
Hence
P(n + 1). So by induction, we have P (n) for all n ∈ N, and so
Pn Pk(n) ⇒
n+1
=
for all n ∈ N.
k=1 2
3
1.1.6. Generalizing the result from 1.1.5 we obtain for all n, k ∈ Z≥0
n+1
= # of ways to pick k + 1 integers from {1, . . . , n + 1}
k+1
n
X
=
# of ways to pick k + 1 integers from {1, . . . , n + 1} the biggest one being j + 1
=
=
j=1
n
X
# of ways to pick k integers from {1, . . . , j}
j=1
n X
j=1
j
k
Fix a k ∈ Z≥0 . For X ∈ Z, X ≥ k we have
X
X · (X − 1) · · · · · (X − k + 1)
=
.
k
k!
2
By expanding the numerator we see that X
k considered as a function in X is
1
.
a polynomial of degree k with leading coefficient k!
X
k
X
=
ci X i
k
i=0
(1)
So, for n ∈ Z≥0 we obtain
!
X
n
k
k
n
k
n X
X
X
X
X
n+1
j
i
=
=
ci j =
ci
ji =
ci Si (n)
k+1
k
j=1
j=1
i=0
i=0
j=1
i=0
for suitable ci ∈ Q and ck =
1
k! .
We recall from lecture that the Si (n) can be rewritten as polynomials in n of
degree i + 1. So, when denoting the leading coefficient of Sk (n) by a the right
hand side of the above equation can be rewritten as a polynomial
in n of degree
k + 1 with leading coefficient ck a = a/k!. As before n+1
is
also
a
polynominal
k+1
in n of degree k + 1 with leading coefficient 1/(k + 1)!. Subsequently,
1
1
1
= a ⇒ a= .
(k + 1)!
k!
k
1.2.3. Let x = am am−1 · · · a1 .b1 b2 · · · be a positive real number with a repeating
decimal expansion i.e. there is a natural number N and some natural number
j such that for all k ≥ N , bj+k = bk . Consider the following two real numbers:
10N −1+j x = am · · · a1 b1 · · · bN +j−1 .bN +j · · ·
10N −1 x = am · · · a1 b1 · · · bN −1 .bN bN +1 · · ·
Note that since bj+k = bk for all k ≥ N , we have
10N −1+j x − 10N −1 x = c1 · · · cm .000 · · ·
In particular, 10N −1+j x−10N −1 x ∈ Z is an integer. Moreover, 10N −1+j −10N −1
is an integer. Therefore,
x=
10N −1+j x − 10N −1 x
10N −1+j − 10N −1
is a quotient of two integers. Hence, x ∈ Q is a rational number.
1.2.4. Let x, y be such that tn (x) = z, tn (y) = w. Since truncation rounds
towards 0, let us consider the non-negative and the negative cases separately.
Case 1: z, w ≥ 0.
We have tn (x) ≤ x < tn (x) + 1/10n and tn (y) ≤ y < tn (y) + 1/10n , so adding
these 2 inequalities, we get
tn (x) + tn (y) ≤ x + y < tn (x) + tn (y) + 2/10n .
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Notice the inequality on the right is strict and that all terms above are nonnegative. Taking the truncation gives tn (x + y) = z + w or z + w + 1/10n .
Case 2: z, w < 0.
We have tn (x) − 1/10n < x ≤ tn (x) and tn (y) − 1/10n < y ≤ tn (y), so adding
these 2 inequalities, we get
tn (x) + tn (y) − 2/10n < x + y ≤ tn (x) + tn (y).
This time, the inequality on the left is strict and that all terms above are negative. Taking the truncation gives tn (x + y) = z + w − 1/10n or z + w.
Case 3: z ≥ 0, w < 0.
We have tn (x) ≤ x < tn (x) + 1/10n and tn (y) − 1/10n < y ≤ tn (y). Adding
these 2 inequalities gives
tn (x) + tn (y) − 1/10n < x + y < tn (x) + tn (y) + 1/10n
. Taking the truncation gives
1. tn (x + y) = z + w − 1/10n or z + w, if z + w ≥ 1/10n ,
2. tn (x + y) = z + w or z + w + 1/10n , if z + w ≤ −1/10n ,
3. tn (x + y) = z + w, if −1/10n < z + w < 1/10n (or equivalently, z + w = 0).
Of course, we still have the case z < 0, w ≥ 0, but that is essentially the same
as case 3 due to commutativity of addition.
We have shown above that the only possibilities for tn (x + y) are z + w − 1/10n ,
z+w and z+w+1/10n . Let us exhibit examples to show that all the possibilities
we found above can indeed occur. It should be obvious that tn (x + y) = z + w
can always occur regardless of what z, w are (simply take x = z and y = w), so
we will omit this in the examples below.
Case 1: z, w ≥ 0.
Let x = z + (1/10n )/2, y = w + (1/10n )/2, then
tn (x + y) = tn (z + w + 1/10n ) = z + w + 1/10n .
Case 2: z, w < 0.
Let x = z − (1/10n )/2, y = w − (1/10n )/2, then
tn (x + y) = tn (z + w − 1/10n ) = z + w − 1/10n .
Case 3: z ≥ 0, w < 0.
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1. Subcase: z + w ≥ 1/10n .
Let x = z, y = w − (1/10n )/2, then
tn (x + y) = tn (z + w − (1/10n )/2) = z + w − 1/10n .
2. Subcase: z + w ≤ −1/10n .
Let x = z + (1/10n )/2, y = w, then
tn (x + y) = tn (z + w + (1/10n )/2) = z + w + 1/10n .
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