0.4 Equivalences
1.
a. It is an equivalence by Example 4.
c. Not an equivalence. x ≡ x only if x = 1, so the reflexive property fails.
e. Not an equivalence. 1 ≡ 2 but 26 ≡ 1, so the symmetric property fails.
g. Not an equivalence. x ≡ x is never true. Note that the transitive property also fails.
i. It is an equivalence by Example 4. [(a, b)] = {(x, y) 
y − 3x = b − 3a} is the line with slope 3 through (a, b).
y − 3x = b − 3a} is the line with slope 3 through (a, b).
2. In every case (a, b) ≡ (a1, b1) if α(a, b) = α(a1, b1) for an appropriate function 
. Hence ≡ is the kernel equivalence of α.
. Hence ≡ is the kernel equivalence of α.
a. The classes are indexed by the possible sums of elements of U.
c. The classes are indexed by the first components.
3.
a. It is the kernel equivalence of 
where α(n) = n2. Here [n] = { − n, n} for each n. Define 
by σ[n] = |n|, where |n| is the absolute value. Then [m] = [n] 
m ≡ n |m| = |n|. Thus σ is well-defined and one-to-one. It is clearly onto.
where α(n) = n2. Here [n] = { − n, n} for each n. Define by σ[n] = |n|, where |n| is the absolute value. Then [m] = [n] m ≡ n |m| = |n|. Thus σ is well-defined and one-to-one. It is clearly onto.
c. It is the kernel equivalence of 
where α(x, y) = y. Define 
by σ[(x, y)] = y. Then
where α(x, y) = y. Define by σ[(x, y)] = y. Then
so σ is well-defined and one-to-one. It is clearly onto.
e. Reflexive: 
; Symmetric: 
; Transitive: x ≡ y and 
and 
. Hence
; Symmetric: ; Transitive: x ≡ y and and . Hence
Now define 
by σ[x] = x− 
x 
where x denotes the greatest integer ≤x. Then [x] = [y] ⇒ x ≡ y ⇒ x − y = n, 
. Thus x = y + n, so x = y + n. Hence,
by σ[x] = x− x where x denotes the greatest integer ≤x. Then [x] = [y] ⇒ x ≡ y ⇒ x − y = n, . Thus x = y + n, so x = y + n. Hence,
and σ is well-defined. To see that σ is one-to-one, let σ[x] = σ[y], that is x− x = y − y . Then 
, so x ≡ y, that is x = y . Finally, σ is onto because, if 0 ≤ x < 1, x = 0, so x = σ[x].
x = y − y . Then , so x ≡ y, that is x = y . Finally, σ is onto because, if 0 ≤ x < 1, x = 0, so x = σ[x].
5.
a. If a 
A, then a Ci and a Dj for some i and j, so a Ci ∩ Dj. If 
, then either i ≠ i′ or j ≠ j′. Thus
A, then a Ci and a Dj for some i and j, so a Ci ∩ Dj. If , then either i ≠ i′ or j ≠ j′. Thus
in either case.
7.
a. Not well defined: 
and 
.
and .
c. Not well defined: 
and 
.
and .
9.
a.[a] = [a1] a ≡ a1 α(a) = α(a1). The implication ⇒ proves σ is well defined; the implication ⇐ shows it is one-to-one. If α is onto, so is σ.
a ≡ a1 α(a) = α(a1). The implication ⇒ proves σ is well defined; the implication ⇐ shows it is one-to-one. If α is onto, so is σ.
c. If we regard σ : A≡ → a(A), then σ is a bijection.
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