Core Focus on Rational Numbers & Equations Adding Integers Lesson 2.2.
MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...
Transcript of MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...
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MATH 301INTRODUCTION TO PROOFS
Sina HazratpourJohns Hopkins University
Fall 2021
- integers- rational numbers
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Relevant sections of the textbook
• Section B.2. (incomplete!)
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Quotients by relations
Recall from problem 5 of homework #4 that for each a binary relation R on aset X we can construct a set X/R whose elements are R-classes
[x ]R = {y ∈ X | R(x , y )}
for all x ∈ X .
Now collect all such R-classes into one set:
X/R =def { [x ]R | x ∈ X}
We call the set X/R the quotient of X by the relation R.
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Quotients by relations
Recall from problem 5 of homework #4 that for each a binary relation R on aset X we can construct a set X/R whose elements are R-classes
[x ]R = {y ∈ X | R(x , y )}
for all x ∈ X . Now collect all such R-classes into one set:
X/R =def { [x ]R | x ∈ X}
We call the set X/R the quotient of X by the relation R.
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Quotients by relations
Recall from problem 5 of homework #4 that for each a binary relation R on aset X we can construct a set X/R whose elements are R-classes
[x ]R = {y ∈ X | R(x , y )}
for all x ∈ X . Now collect all such R-classes into one set:
X/R =def { [x ]R | x ∈ X}
We call the set X/R the quotient of X by the relation R.
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ExampleConsider the set of natural numbers with the usual ordering ⩽ : N → N → 2defined for m ∈ N recursively by
m ⩽ 0 if and only if m = 0, and
m ⩽ succ(n) if and only if m = succ(n) or m ⩽ n.
We have classes [n] forming a chain in the subset relation ordering:[0] ⊃ [1] ⊃ [2] ⊃ .... Note that 0 ⩽ 1 but [0] ̸= [1].
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ExampleConsider the set of natural numbers with the usual ordering ⩽ : N → N → 2defined for m ∈ N recursively by
m ⩽ 0 if and only if m = 0, and
m ⩽ succ(n) if and only if m = succ(n) or m ⩽ n.
We have classes [n] forming a chain in the subset relation ordering:[0] ⊃ [1] ⊃ [2] ⊃ ....
Note that 0 ⩽ 1 but [0] ̸= [1].
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ExampleConsider the set of natural numbers with the usual ordering ⩽ : N → N → 2defined for m ∈ N recursively by
m ⩽ 0 if and only if m = 0, and
m ⩽ succ(n) if and only if m = succ(n) or m ⩽ n.
We have classes [n] forming a chain in the subset relation ordering:[0] ⊃ [1] ⊃ [2] ⊃ .... Note that 0 ⩽ 1 but [0] ̸= [1].
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So far R is just a general binary relation. In general we do not have that
PropositionProve that if R is reflexive then ∀x ∈ X , x ∈ [x ].
PropositionProve that if R is transitive then ∀x , y ∈ X , R(x , y ) ∧ R(y , x) ⇒ [x ] = [y ].
PropositionProve that if R is symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇒ [x ] = [y ].
PropositionProve that if R is reflexive, symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇔ [x ] = [y ].
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So far R is just a general binary relation. In general we do not have that
PropositionProve that if R is reflexive then ∀x ∈ X , x ∈ [x ].
PropositionProve that if R is transitive then ∀x , y ∈ X , R(x , y ) ∧ R(y , x) ⇒ [x ] = [y ].
PropositionProve that if R is symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇒ [x ] = [y ].
PropositionProve that if R is reflexive, symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇔ [x ] = [y ].
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So far R is just a general binary relation. In general we do not have that
PropositionProve that if R is reflexive then ∀x ∈ X , x ∈ [x ].
PropositionProve that if R is transitive then ∀x , y ∈ X , R(x , y ) ∧ R(y , x) ⇒ [x ] = [y ].
PropositionProve that if R is symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇒ [x ] = [y ].
PropositionProve that if R is reflexive, symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇔ [x ] = [y ].
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So far R is just a general binary relation. In general we do not have that
PropositionProve that if R is reflexive then ∀x ∈ X , x ∈ [x ].
PropositionProve that if R is transitive then ∀x , y ∈ X , R(x , y ) ∧ R(y , x) ⇒ [x ] = [y ].
PropositionProve that if R is symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇒ [x ] = [y ].
PropositionProve that if R is reflexive, symmetric and transitive then∀x , y ∈ X , R(x , y ) ⇔ [x ] = [y ].
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Suppose we have a graph G. Define a relation R on vertices of G byimposing that
R(a, b) ⇔ there is an edge from a to b.
• What is a class [a] for a vertex a?
• Show that if R(a, b) ∧ R(b, a) then it is not necessarily true that [a] = [b].
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Suppose we have a graph G. Define a relation R on vertices of G byimposing that
R(a, b) ⇔ there is an edge from a to b.
• What is a class [a] for a vertex a?
• Show that if R(a, b) ∧ R(b, a) then it is not necessarily true that [a] = [b].
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Suppose we have a graph G. Define a relation R on vertices of G byimposing that
R(a, b) ⇔ there is an edge from a to b.
• What is a class [a] for a vertex a?
• Show that if R(a, b) ∧ R(b, a) then it is not necessarily true that [a] = [b].
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Suppose we have a graph G. Define a relation R on vertices of G byimposing that
R(a, b) ⇔ there is a path from a to b.
• What is a class [a] for a vertex a?
• Show that if R(a, b) ∧ R(b, a) then [a] = [b].
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Suppose we have a graph G. Define a relation R on vertices of G byimposing that
R(a, b) ⇔ there is a path from a to b.
• What is a class [a] for a vertex a?
• Show that if R(a, b) ∧ R(b, a) then [a] = [b].
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Suppose we have a graph G. Define a relation R on vertices of G byimposing that
R(a, b) ⇔ there is a path from a to b.
• What is a class [a] for a vertex a?
• Show that if R(a, b) ∧ R(b, a) then [a] = [b].
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PropositionProve that the function q : X → X/R assigning to each x in X the class [x ] inX/R is a surjection.
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The universal mapping property of quotient construction
Proposition
Let R be a symmetric and transitive relation on a set X . For any set Y ,precomposing with q yields a bijection
(X/R → Y ) ∼= {f : X → Y | ∀x , y ∈ X , R(x , y ) ⇒ f (x) = f (y )}
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Recall that a relation R on a set A is called an equivalence if it satisfies thefollowing conditions:
• reflexivity: ∀a ∈ A, R(a, a),
• symmetry: ∀a, b ∈ A, R(a, b) → R(b, a), and
• transitivity: ∀a, b, c ∈ A, R(a, b) ⇒ R(b, c) ⇒ R(a, c).
We usually denote an equivalence relation by the symbol ∼ (instead of R).
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Recall that a relation R on a set A is called an equivalence if it satisfies thefollowing conditions:
• reflexivity: ∀a ∈ A, R(a, a),
• symmetry: ∀a, b ∈ A, R(a, b) → R(b, a), and
• transitivity: ∀a, b, c ∈ A, R(a, b) ⇒ R(b, c) ⇒ R(a, c).
We usually denote an equivalence relation by the symbol ∼ (instead of R).
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Quotients by equivalence relations
For each equivalence ∼ on a set X we can construct a set X/ ∼ whoseelements are equivalence classes
[x ]∼ = {y ∈ X | x ∼ y}
for all x ∈ X .
Now collect all such equivalence classes into one set:
X/ ∼=def {[x ]∼ | x ∈ X}
We call the set X/ ∼ the quotient of X by equivalence relation ∼.
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Quotients by equivalence relations
For each equivalence ∼ on a set X we can construct a set X/ ∼ whoseelements are equivalence classes
[x ]∼ = {y ∈ X | x ∼ y}
for all x ∈ X . Now collect all such equivalence classes into one set:
X/ ∼=def {[x ]∼ | x ∈ X}
We call the set X/ ∼ the quotient of X by equivalence relation ∼.
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Quotients by equivalence relations
For each equivalence ∼ on a set X we can construct a set X/ ∼ whoseelements are equivalence classes
[x ]∼ = {y ∈ X | x ∼ y}
for all x ∈ X . Now collect all such equivalence classes into one set:
X/ ∼=def {[x ]∼ | x ∈ X}
We call the set X/ ∼ the quotient of X by equivalence relation ∼.
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For an equivalence relation ∼, the surjection q : X → X/ ∼ has an extra niceproperty:
q(x) = q(y ) ⇔ R(x , y )
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Every function f : A → B gives rise to an equivalence relation ∼f on Adefined by
a ∼f b ⇔ f (a) = f (b) .
That is to say that a ∼f b if a, b are in the same fibre of f .
ExerciseShow that the relation above is indeed an equivalence relation.
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Every function f : A → B gives rise to an equivalence relation ∼f on Adefined by
a ∼f b ⇔ f (a) = f (b) .
That is to say that a ∼f b if a, b are in the same fibre of f .
ExerciseShow that the relation above is indeed an equivalence relation.
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Every function f : A → B gives rise to an equivalence relation ∼f on Adefined by
a ∼f b ⇔ f (a) = f (b) .
That is to say that a ∼f b if a, b are in the same fibre of f .
ExerciseShow that the relation above is indeed an equivalence relation.
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Image factorization
PropositionSuppose f : A → B is a function. We can factor f into a surjection followed bya bijection followed by an injection.
A B
C D
f
q
g
i
that is there are functions q, g, i such that f = i ◦ g ◦ q, where q is asurjection, g is a bijection, and i is an injection.
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i .
Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f .
We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.
We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B.
Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ).
We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection.
g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ].
Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Proof.We have to construct the sets C, D and a surjection q, a bijection g and aninjection i . Now we define C to be A/ ∼f , and we define D to be the imagef∗(A) of A under f . We also define q to be the obvious quotient map and i tobe the obvious inclusion. As shown before, q is surjective and i is injective.We define g to be the assignment which takes an equivalence class [x ] tothe element f (x) ∈ B. Note that g is well-defined, since if [x ] = [y ] thenx ∼f y and therefore, by the definition of ∼f , we have f (x) = f (y ). We nowshow that g is a bijection. g is injective since for every x , y ∈ A, ifg([x ]) = g([y ]) then f (x) = f (y ) and therefore, [x ] = [y ]. Also, g is surjective:given b in f∗(A) there is some a ∈ A such that b = f (a) = g([a]).
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Our factorization diagram becomes
A B
A/ ∼ f∗(A)
f
q
g
i
In fact, g ◦ q = p : X → Im(f ).
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Our factorization diagram becomes
A B
A/ ∼ f∗(A)
f
q
g
i
In fact, g ◦ q = p : X → Im(f ).
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DefinitionFor a function f : X → X, define
Fix(f ) =def {x ∈ X | f (x) = x} .
We call Fix(f ) the set of fix-points of f .
DefinitionA function f is called an idempotent if f ◦ f = f .
ExerciseShow that if f : X → X is idempotent, then Fix(f ) ∼= Im(f ).
ExerciseFor an idempotent function f : X → X, show that
X/ ∼f∼= Fix(f ) ∼= Im(f )
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DefinitionFor a function f : X → X, define
Fix(f ) =def {x ∈ X | f (x) = x} .
We call Fix(f ) the set of fix-points of f .
DefinitionA function f is called an idempotent if f ◦ f = f .
ExerciseShow that if f : X → X is idempotent, then Fix(f ) ∼= Im(f ).
ExerciseFor an idempotent function f : X → X, show that
X/ ∼f∼= Fix(f ) ∼= Im(f )
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DefinitionFor a function f : X → X, define
Fix(f ) =def {x ∈ X | f (x) = x} .
We call Fix(f ) the set of fix-points of f .
DefinitionA function f is called an idempotent if f ◦ f = f .
ExerciseShow that if f : X → X is idempotent, then Fix(f ) ∼= Im(f ).
ExerciseFor an idempotent function f : X → X, show that
X/ ∼f∼= Fix(f ) ∼= Im(f )
![Page 45: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/45.jpg)
DefinitionFor a function f : X → X, define
Fix(f ) =def {x ∈ X | f (x) = x} .
We call Fix(f ) the set of fix-points of f .
DefinitionA function f is called an idempotent if f ◦ f = f .
ExerciseShow that if f : X → X is idempotent, then Fix(f ) ∼= Im(f ).
ExerciseFor an idempotent function f : X → X, show that
X/ ∼f∼= Fix(f ) ∼= Im(f )
![Page 46: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/46.jpg)
ExerciseSuppose r : A → B is a retraction. Show that
B ∼= A/ ∼r
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ExerciseSuppose r : A → B is a retraction. Show that
B ∼= A/ ∼r
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Integers as quotient by an equivalence relation
Consider the relation ∼ on N× N. where
(m, n) ∼ (m′, n′) ⇔ m + n′ = n + m′ .
Prove that this relation is an equivalence.
The equivalence class [(0, 0)] is the set {(0, 0), (1, 1), (2, 2), ...}. What is theequivalence class [(0, 1)]?
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Integers as quotient by an equivalence relation
Consider the relation ∼ on N× N. where
(m, n) ∼ (m′, n′) ⇔ m + n′ = n + m′ .
Prove that this relation is an equivalence.The equivalence class [(0, 0)] is the set {(0, 0), (1, 1), (2, 2), ...}.
What is theequivalence class [(0, 1)]?
![Page 50: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/50.jpg)
Integers as quotient by an equivalence relation
Consider the relation ∼ on N× N. where
(m, n) ∼ (m′, n′) ⇔ m + n′ = n + m′ .
Prove that this relation is an equivalence.The equivalence class [(0, 0)] is the set {(0, 0), (1, 1), (2, 2), ...}. What is theequivalence class [(0, 1)]?
![Page 51: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/51.jpg)
1 2 3 4
1
2
3
4
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Representing integers
We define the set Z of integers as the quotient N× N/ ∼.
• In other words, a pair (m, n) represents the would-be integer m − n.
• In this case, there are canonical representatives of the equivalenceclasses: those of the form (n, 0) or (0, n).
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Representing integers
We define the set Z of integers as the quotient N× N/ ∼.
• In other words, a pair (m, n) represents the would-be integer m − n.
• In this case, there are canonical representatives of the equivalenceclasses: those of the form (n, 0) or (0, n).
![Page 54: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/54.jpg)
Representing integers
We define the set Z of integers as the quotient N× N/ ∼.
• In other words, a pair (m, n) represents the would-be integer m − n.
• In this case, there are canonical representatives of the equivalenceclasses: those of the form (n, 0) or (0, n).
![Page 55: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/55.jpg)
Addition on integersWe can define the operation of addition on Z by an assignment+∼ : Z× Z → Z which assigns to the pair ([(m, n)], [(m′, n′)]) the class[(m + m′, n + n′)].
ExerciseShow that the assignment +∼ is well-defined, i.e. it defines a function.
Exercise• Show that for all integers a we have 0 + a = a = a + 0.
• Show that for all integers a, b, c we have (a + b) + c = a + (b + c).
• Show that for all integers a, b we have a + b = b + a.
ExerciseConstruct an idempotent f : N× N → N× N such that Fix(f ) is in bijectionwith the set of integers.
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Addition on integersWe can define the operation of addition on Z by an assignment+∼ : Z× Z → Z which assigns to the pair ([(m, n)], [(m′, n′)]) the class[(m + m′, n + n′)].
ExerciseShow that the assignment +∼ is well-defined, i.e. it defines a function.
Exercise• Show that for all integers a we have 0 + a = a = a + 0.
• Show that for all integers a, b, c we have (a + b) + c = a + (b + c).
• Show that for all integers a, b we have a + b = b + a.
ExerciseConstruct an idempotent f : N× N → N× N such that Fix(f ) is in bijectionwith the set of integers.
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Addition on integersWe can define the operation of addition on Z by an assignment+∼ : Z× Z → Z which assigns to the pair ([(m, n)], [(m′, n′)]) the class[(m + m′, n + n′)].
ExerciseShow that the assignment +∼ is well-defined, i.e. it defines a function.
Exercise• Show that for all integers a we have 0 + a = a = a + 0.
• Show that for all integers a, b, c we have (a + b) + c = a + (b + c).
• Show that for all integers a, b we have a + b = b + a.
ExerciseConstruct an idempotent f : N× N → N× N such that Fix(f ) is in bijectionwith the set of integers.
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We can define the operation of multiplication on Z by an assignment·∼ : Z× Z → Z which assigns to the pair ([(m, n)], [(m′, n′)]) the class[(m · m′, n · n′)].
ExerciseShow that the assignment ·∼ is well-defined, i.e. it defines a function.
Exercise• Show that for all integers a we have 0 + a = a = a + 0.
• Show that for all integers a, b, c we have (a + b) + c = a + (b + c).
• Show that for all integers a, b we have a + b = b + a.
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We can define the operation of multiplication on Z by an assignment·∼ : Z× Z → Z which assigns to the pair ([(m, n)], [(m′, n′)]) the class[(m · m′, n · n′)].
ExerciseShow that the assignment ·∼ is well-defined, i.e. it defines a function.
Exercise• Show that for all integers a we have 0 + a = a = a + 0.
• Show that for all integers a, b, c we have (a + b) + c = a + (b + c).
• Show that for all integers a, b we have a + b = b + a.
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Induction for integers
We identify a natural number n with the corresponding non-negative integer,i.e. with the image of (n, 0) ∈ N× N in Z.
Lemma
Suppose P : Z → Prop is a predicate over integers, and
• P(0) holds,
• ∀n : N, P(n) ⇒ P(succ(n)), and
• ∀n : N, P(−n) → P(−succ(n)).
Then we have ∀z : Z, P(z).
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Induction for integers
We identify a natural number n with the corresponding non-negative integer,i.e. with the image of (n, 0) ∈ N× N in Z.
Lemma
Suppose P : Z → Prop is a predicate over integers, and
• P(0) holds,
• ∀n : N, P(n) ⇒ P(succ(n)), and
• ∀n : N, P(−n) → P(−succ(n)).
Then we have ∀z : Z, P(z).
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We constructed the integers Z as a quotient of N×N, and observed that thisquotient is generated by an idempotent.
We construct the field of rationals Q along the same lines as well, namely asthe quotient
Q =def (Z× N)/≈
where(u, a) ≈ (v , b) =def (u(b + 1) = v (a + 1)).
In other words, a pair (u, a) represents the rational number u/(1 + a).Here too we have a canonical choice of representatives, namely fractions inlowest terms.
![Page 63: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/63.jpg)
We constructed the integers Z as a quotient of N×N, and observed that thisquotient is generated by an idempotent.We construct the field of rationals Q along the same lines as well, namely asthe quotient
Q =def (Z× N)/≈
where(u, a) ≈ (v , b) =def (u(b + 1) = v (a + 1)).
In other words, a pair (u, a) represents the rational number u/(1 + a).Here too we have a canonical choice of representatives, namely fractions inlowest terms.
![Page 64: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/64.jpg)
We constructed the integers Z as a quotient of N×N, and observed that thisquotient is generated by an idempotent.We construct the field of rationals Q along the same lines as well, namely asthe quotient
Q =def (Z× N)/≈
where(u, a) ≈ (v , b) =def (u(b + 1) = v (a + 1)).
In other words, a pair (u, a) represents the rational number u/(1 + a).
Here too we have a canonical choice of representatives, namely fractions inlowest terms.
![Page 65: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/65.jpg)
We constructed the integers Z as a quotient of N×N, and observed that thisquotient is generated by an idempotent.We construct the field of rationals Q along the same lines as well, namely asthe quotient
Q =def (Z× N)/≈
where(u, a) ≈ (v , b) =def (u(b + 1) = v (a + 1)).
In other words, a pair (u, a) represents the rational number u/(1 + a).Here too we have a canonical choice of representatives, namely fractions inlowest terms.
![Page 66: MATH 301 INTRODUCTION TO PROOFS - - integers - rational ...](https://reader030.fdocuments.us/reader030/viewer/2022021406/6209767b4d498950b860d3b4/html5/thumbnails/66.jpg)
The arithmetic of rational numbers
We write down the arithmetical operations on Q so that we can compute withfractions.
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The order on rational numbers
We equip Q with a total order.
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Questions
Thanks for your attention!