HOMEWORK ASSIGNMENT 8
ACCELERATED PROOFS AND PROBLEM SOLVING [MATH08071]
Each problem will be marked out of 4 points.
Exercise 1 ([1, Exercise 18.2]). Describe equivalence classes for the following equivalence
relations on the given set S.
(i) S = R, and a ∼ b ⇐⇒ a = b or −b.
(iii) S = R, and a ∼ b ⇐⇒ a2 + a = b2 + b.
(v) S is the set of all points in the plane, and a ∼ b means a and b are the same
distance from the origin.
(vi) S = N, and a ∼ b ⇐⇒ ab is a square.
(viii) S = R × R, and (x, y) ∼ (a, b) ⇐⇒ x2 + y 2 = a2 + b2 .
Solution.
(i) The set of equivalence classes consists of subsets {a, −a}, where a ∈ R. Since
either a or −a is non-negative, we see also describe the set of equivalence classes
as all subsets {a, −a}, where a ∈ R such that a > 0.
(iii) For every a ∈ R, we have
n
o
cl(a) = b ∈ R a2 + a = b2 + b
by definition. Once we fix a ∈ R, if a2 + a = b2 + b, then
p
√
−1 ± (2a + 1)2
−1 + ±|2a + 1|
−1 ± 4a2 + 4a + 1
=
=
∈ {a, −1 − a}.
b=
2
2
2
Vice versa, if b = −1 − a or a, than a2 + a = b2 + b. Thus, cl(a) = {a, −1 − a}.
Thus, the set of equivalence classes consists of subsets {a, −1 − a}, where a ∈ R.
Since either a > −1/2 or −1 − a > −1/2, we see that the set of equivalence
classes consists of subsets {a, −1 − a}, where a ∈ R such that a > −1/2.
(v) The set of equivalence classes consists of circles centered at the origin, where we
consider the origin as a circle of radius 0. Each circle correspond to exactly one
equivalence class.
(vi) Take any a ∈ N. Then
n
o
cl(a) = b ∈ N ab is a square
by definition. How to describe cl(a) more neatly?
It is clear that cl(1) consists of all squared in N. This is good description.
Suppose that a > 2. Then it follows from [1, Theorem 11.1] (the Fundamental
Theorem of Arithmetics) that
a = pa11 pa22 · · · pamm
for some prime numbers 2 6 p1 < p2 < · · · < pm and some positive integers
a1 , . . . , am , where m is also a positive integers. Put
0 if ai is even,
bi =
1 if ai is odd.
This assignment is due on Thursday 19th November 2015.
1
bm , because
Then a ∼ pb11 p2b2 · · · pm
a
bm
pb11 p2b2 · · · pm
a1 −b1
2
= pa11 −b1 pa22 −b2 · · · pamm −bm =
p1
a2 −b2
2
p2
am −bm
2
!2
· · · pm
Thus, we see that p1b1 pb22 · · · pbmm ∈ cl(a).
We proved that for every n ∈ N such that n 6= 1, the equivalent class cl(n)
contains an element, say n̂, such that n̂ is not divisible by a square of a prime
number, i.e. n̂ is a product of finitely many different primes. This implies that n̂
is the smallest element in cl(n). Moreover, this implies that every element in cl(n)
is a product of n̂ and a square of √
a natural number. Indeed, if ñ is any element
in cl(n), then n̂ñ is a square (i.e. n̂ñ ∈ N), which implies that ñ is divisible by
every prime that divides n̂ (see [1, Proposition 11.4]), which implies that n̂ divides
ñ and
√ !2
n̂ñ
ñ
n̂ñ
,
= 2 =
n
n
n
because n̂ is a product of finitely many different primes.
Therefore, the set of equivalence classes can be descried as sets of positive integers of the form
p1 p2 . . . pm n2 ,
where 2 6 p1 < p2 < . . . < pm are prime numbers (fixed for the given equivalence
class), and n runs through all natural numbers. Fir example, we have
n
o
2
2
2
2
2
cl(2 × 3) = 2 × 3, 2 × 3 × 2 , 2 × 3 × 3 , 2 × 3 × 4 , 2 × 3 × 5 , 2 × 3 × 6 , . . . .
In this description we allow m to be 0. In this case we get cl(1), i.e. the set of all
squares.
Thus, we have a 1–1 correspondence between the equivalence classes of ∼ and
natural numbers that are not divisible by a square of any prime number.
(viii) If we treat R × R as a plane R2 , then (x, y) ∼ (a, b) ⇐⇒ (x, y) and (a, b) are the
same distance from the origin (0, 0) (this follows from the Pythagoras theorem).
Thus, the set of equivalence classes consists of circles centered at the origin, where
we consider the origin as a circle of radius 0.
Exercise 2 ([1, Exercise 18.6]). Let S = {1, 2, 3, 4}, and suppose that ∼ is an equivalence
relation on S. You are given information that 1 ∼ 2 and 2 ∼ 3. Show that there are
exactly two possibilities for the relation ∼, and describe both (i.e., for all a, b ∈ S, say
whether or not a ∼ b).
Solution. We must fill in the following “equivalence” table:
∼
1
2
3
4
1
? ∼
?
?
2
?
?
∼ ?
3
?
?
?
?
4
?
?
?
?
where we put ∼ (6∼, respectively) at the intersection of the raw containing i and the column
containing j if i ∼ j (i 6∼ j, respectively).
Let us fill in this table slowly. Step by step.
Since ∼ is an equivalence relation, it is reflexive. Thus, we must have 1 ∼ 1, 2 ∼ 2,
3 ∼ 3, and 4 ∼ 4. This helps a bit:
2
∼
1
2
3
4
1
∼ ∼
?
?
2
?
∼ ∼
?
3
?
?
∼
?
4
?
?
?
∼
Since ∼ is an equivalence relation, it is symmetric. Thus, we must have 2 ∼ 1 and
3 ∼ 2, because we are given information that 1 ∼ 2 and 2 ∼ 3. So, our table now looks
like this:
∼ 1 2 3 4
1
∼ ∼
?
?
2
∼ ∼ ∼
?
3
?
∼ ∼
?
4
?
?
∼
?
Since ∼ is an equivalence relation, it is transitive. Namely, we must have 1 ∼ 3, since
we are given information that 1 ∼ 2 and 2 ∼ 3. Then 3 ∼ 1, since ∼ is symmetric. Hence,
we have a almost completed table:
∼
1
3
4
1
∼ ∼ ∼
?
2
∼ ∼ ∼
?
3
∼ ∼ ∼
?
4
?
∼
2
?
?
Now we have a freedom: either 1 ∼ 4 or 1 6∼ 4. We have to chose which way to go. If
1 ∼ 4, then we have this simple looking table:
∼
1
1
∼ ∼ ∼ ∼
2
∼ ∼ ∼ ∼
3
∼ ∼ ∼ ∼
4
∼ ∼ ∼ ∼
2
3
4
using transitivity and symmetricity of ∼ as we already did before. Similarly, if 1 6∼ 4, then
we have a slightly less boring table:
∼
1
1
∼ ∼ ∼ 6∼
2
∼ ∼ ∼ 6∼
3
∼ ∼ ∼ 6∼
4
6∼ 6∼ 6∼ ∼
2
3
4
using transitivity and symmetricity of ∼
Exercise 3 ([1, Exercise 19.2]). The functions f : R → R and g : R → R are defined as
follows:
(
2x if 0 6 x 6 1,
x2 if 0 6 x 6 1,
g(x) =
f (x) =
1 otherwise;
0 otherwise.
3
Give formulae describing the functions g ◦ f and f ◦ g. Draw the graph of these functions.
Solution. The (sketchy) graph of the function f (x) is given here
.
The graph of the function g(x) is given here
.
Easy case by case analysis implies that
(
f ◦ g(x) =
2x2 if 0 6 x 6 1,
0 otherwise.
4
The graph of the function f ◦ g(x) is given here
.
Less easy case by case analysis implies that
g ◦ f (x) =

1 if x < 0,





0 if x = 0,




2


 4x if 0 < x < 1/2,
1



0





0




1
if x = 1/2,
if 1/2 < x < 1,
if x = 1,
if x > 1,
which can be simplified a bit to
 1 if x < 0,



 4x2 if 0 6 x 6 1/2,
g ◦ f (x) =
 0 if 1/2 < x 6 1,



1 if x > 1.
Its graph is given here
5
.
Exercise 4 ([1, Exercise 19.4]). Let X, Y , Z be sets, and let f : X → Y and g : Y → Z
be functions.
(a) Given that g ◦ f is onto, can you deduce that f is onto? Give a proof or a counterexample.
(b) Given that g ◦ f is onto, can you deduce that g is onto?
(c) Given that g ◦ f is 1–1, can you deduce that f is 1–1?
(d) Given that g ◦ f is 1–1, can you deduce that g is 1–1?
Solution.
(a) Put X = {1}, Y = {1, 2}, and Z = {1}. Take f such that f (1) = 1, take g such that
g(1) = g(2) = 1. Then g ◦ f is an identity function, which implies that it is onto. But
f is not onto.
(b) Suppose that g ◦ f is onto. Then g is also onto. Indeed, for every z ∈ Z, there exists
x ∈ X such that g ◦ f (x) = z, because g ◦ f is onto. Then g(f (x)) = z, which implies
that g is onto (we found y ∈ Y such that g(y) = z).
(c) If g ◦ f is 1–1, then f is 1–1. Indeed, if f is not 1–1, then there exists x1 and x2 in X
such that x1 6= x2 and f (x1 ) = f (x2 ), which implies that g(f (x1 )) = g(f (x2 )), which
implies that g ◦ f is not 1–1 as well.
(d) Put X = {1}, Y = {1, 2}, and Z = {1}. Take f such that f (1) = 1, take g such that
g(1) = g(2) = 1. Then g ◦ f is an identity function, which implies that it is 1–1. But
g is not 1–1 by construction.
Exercise 5 ([1, Exercise 19.6]). (a) Find an onto function from N to Z.
(b) Find a 1–1 function from Z to N.
Solution. Let us construct a bijective functions f : N → Z and g : Z → N such that f ◦ g
and g ◦ f are identity functions. Put
(
2m if m > 0,
n/2 if n is even,
f (n) =
g(m) =
2(−m) + 1 if m 6 0,
(1 − n)/2 if n is odd;
6
where n ∈ N and m ∈ Z. Then f ◦ g and g ◦ f are identity functions. Hence, both f and
g are bijections (see Exercise 4).
References
[1] M. Liebeck, A concise introduction to pure mathematics
Third edition (2010), CRC Press
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