Solutions to the Homework VII (2016)
I601 Logic and Discrete Mathematics
1. Let fn be the nth Fibonacci number, n ∈ N. Prove the identity fn−1 fn+1 − fn2 =
(−1)n for every n > 1.
S o l u t i o n.
The predicate to be proved for all n ∈ N+ is P (n) ≡ fn−1 fn+1 − fn2 = (−1)n .
Base Case: The statement P (1) is true because
f1−1 f1+1 − f12 = 0 · 1 − 12 = −1 = (−1)1
.
Inductive Step: Let’s assume that P (k) is true for some k ∈ N. Then
2
2
fk fk+2 − fk+1
= fk (fk+1 + fk ) − fk+1
=
2
= fk2 + fk fk+1 − fk+1
=
= fk2 + fk+1 (fk − fk+1 ) =
= fk2 − fk+1 fk−1 =
= −(−1)k = (−1)k+1
This yields that P (k + 1) is true, and hence ∀nP (n).
2.
What is the following sum?
1
1
1
1
+
+
+ ··· +
1·2 2·3 3·4
(n − 1) · n
S o l u t i o n. Let’s compute some values in order to guess closed formula for the sum:
1
1
=
1·2
2
1
1
3+1
2
+
=
=
1·2 2·3
2·3
3
1
1
1
2
1
8+1
3
+
+
= +
=
=
1·2 2·3 3·4
3 3·4
3·4
4
1
1
1
1
3
1
15 + 1
4
+
+
+
= +
=
=
1·2 2·3 3·4 4·5
4 4·5
4·5
5
From these examples we can formulate a hypothesis P (n) that
1
1
1
n−1
1
1
+
+
+ ··· +
=
=1− .
1·2 2·3 3·4
(n − 1) · n
n
n
We prove this equality by induction. Actually, the base cases for P (2), P (3), P (4), and P (5) are
already proven. Remains only induction step. assuming , that P (k) is true, prove that so is also
P (k + 1):
1
1
1
1
1
1
1−k−1
1
+
+···+
+
= 1− +
= 1+
= 1−
.
1·2 2·3
(k − 1) · k k · (k + 1)
k k · (k + 1)
k · (k + 1)
k+1
This yields that P (k + 1) is true, and hence (∀n > 1).P (n).
3.
Find the cardinal number of each set:
a) A = {a, b, c, ..., y, z}
d) D = {6, 7, 8, 9, ...}
b) B = T
{x|x ∈ N, x2 = 5}
e) E = n∈N En
c) C = T
{10, 20, 30, 40, ...}
f ) F = n∈N Fn
1
1
1
where En = (− n+1
, n+1
) and Fn = (0, n+1
).
S o l u t i o n. The answers are as follows:
a) |A| = 26
d) |D| = ℵ0
4.
b) |B| = 0
e) |E| = |{0}| = 1
c) |C| = ℵ0
f ) |F | = |∅| = 0
Let R be the relation on N defined by x + 3y = 12, i.e. R = {(x, y)|x + 3y = 12}.
(a) Write R as a set of ordered pairs.
(b) Find the domain and range of R.
(c) Find R−1 .
(d) Find the composition relation R ◦ R.
S o l u t i o n.
(a) R = {(x, 4 − x3 )|x ∈ N} = {(0, 4), (3, 3), (6, 2), (9, 1), (12, 0)}.
(b) Dom(R) = {0, 3, 6, 9, 12} and Ran(R) = {0, 1, 2, 3, 4} .
(c) R−1 = {(x, 12 − 3x)|x ∈ N} = {(0, 12), (1, 9), (2, 6), (3, 3), (4, 0)}.
(d) R2 = R ◦ R = {(3, 3), (12, 4)}.
5. Let R is a relation on the coordinate plane, such that xRy iff the point A(x, y)
belongs to the set of interior points of the region bounded by the lines y = 6 − x ,
x = 0 and y = 0. Determine whether this relation is reflexive, symmetric or transitive.
Justify the answer.
S o l u t i o n.
The shaded area on the figure below represents the points which coordinates are related.
The region is symmetric according to the diagonal y = x, Thus, if a point (x, y) belongs to
the region, then also (x, y) does, and we can conjecture that R is a symmetric relation.
The relation is not reflexive (for instance the point (5, 5) 6∈ R), as well as not transitive (a
counterexample would be (4, 1) ∈ R and (1, 3) ∈ R, whereas (4, 3) 6∈ R).