#reflexive symmetric transitive relations

[edusquaremaths]

Q10 Exercise1.1 I Class 12 Maths NCERT Chapter 1 Relations and Functions | NCERT solutions

NCERT Class 12Chapter: Relations and FunctionsExercise 1.1Question 10: Give an example of a relation. Which is(i) Symmetric but neither reflexive nor transitive.(ii) Transitive but neither reflexive nor symmetric.(iii) Reflexive and symmetric but not transitive.(iv) Reflexive and transitive but not symmetric.(v) Symmetric and transitive but not reflexive.

[maybeasunflower]

When I see "transitive", I can't help think of equivalence relations (a maths thing), which I know are defined by three properties: they are transitive (A~B and B~C => A~C), reflexive (A~B => B~A), and the other one (A~A is true), but I can't remember the name of the other one.

Then I look up equivalence relations and find the "other one" is actually "reflexive", what I think is called "reflexive" is actually called "symmetric", and it's fine.

if shes your girl then why have i slowly been replacing her parts until there’s nothing left of her original body? is she then still your girl?

[catgirl-lucy]

Cardinal Numbers

Nyaa!~

I have too many drafts :c

Hi everyone! Today, we are going to learn about infinite cardinal numbers! I hope this'll be fun! ^^

I was also planning to explain something about the axiom of choice, but I decided that it's better if that becomes its own blog-post.

So, first: what is a cardinal number?

✦ Motivation ✦

Well, cardinal numbers are numbers that represent the size of a set of things. For example, {owo, uwu} is a set of cardinality 2: it has two elements, owo and uwu.

However, we can't simply count the elements of a set to figure out its cardinality. What if we come across a set like {0, 1, 2, 3, ...}? I mean... you can try counting the elements, but it's gonna take you a veryy long time.

So! What do we do?

Well,,, first, observe that cardinals can't exist in a vacuum. If we say that {owo, uwu} has cardinality 2, what does that actually mean? ‘2’ is just a symbol we use to describe the cardinality of {owo, uwu}, but has no meaning on its own. Where it gets interesting is when we compare the cardinality of {owo, uwu} to {10, 3}: they're the same! Now, that is interesting.

So, we use ‘2’ to describe the cardinality of sets like {owo, uwu} and {10, 3}, and we use this symbol so we can more easily recognize when two sets have the same cardinality, without having to search for statements like “this set has the same size as this other set” exhaustively having to apply transitivity of (cardinal) equality.

But what makes {owo, uwu} and {10, 3} so similar that they can be considered ‘equinumerous’?

Well,,, observe that the cardinality of a set doesn't change when we replace some of the elements with other elements: the cardinality of a set is completely independent from what the members of that set actually are. If we have a set like {catgirl lucy, my dearest friend, Semi}, and I replace ‘my dearest friend’ with ‘my worst enemy’ (so we get {catgirl lucy, my worst enemy, Semi}), the size of the set doesn't change!

I hope these two observations are enough to motivate the definition of a cardinal number:

～ Cardinal Equality ～

[Definition] Two sets, A and B, have the same cardinality, denoted |A| = |B|, iff there exists a bijection f: A → B.

There are three kinds of functions that will be important to us in this blog-post:

An injection is a function f: A → B from some set A to some set B so that, for all x,y ∈ A, if f(x) = f(y), then x = y.

A surjection is a function f: A → B from some set A to some set B so that, for all y ∈ B, there is some x ∈ A so that f(x) = y.

A bijection is a function f: A → B from some set A to some set B that is both an injection and a surjection.

A bijection thus pairs every element of A with a unique element of B, and vice versa.

So now, we need to show that this definition of cardinal numbers actually makes sense. I.e., that it is an equivalence relation:

An equivalence relation is a relation ~ such that:

~ is reflexive: x ~ x for all x

~ is symmetric: if x ~ y then y ~ x

~ is transitive: if x ~ y and y ~ z, then x ~ z

...

What?

OK, fine... I'll give you the answer: reflexivity of cardinal equality is witnessed by the identity function on a set, symmetry is from the functions inverse and transitivity comes from composition. I do want you to start thinking about how to solve these problems in the future!

Given an equivalence relation ~ on some class C, we can define a function F with domain C so that, for all x,y ∈ C, F(x) = F(y) iff x ~ y. If ~ fails one of the conditions of an equivalence relation, such a function does not exist.

This time, I will leave you to ponder why~

So! We can choose some function |·| that maps sets X to objects |X| such that |X| = |Y| iff X and Y have the same cardinality, the choice of mapping doesn't really matter.

The cardinality of ℕ, the set of natural numbers, is denoted ℵ₀ (‘aleph-nought’), this is also the cardinality of ℤ, the set of integers, and ℚ, the set of rational numbers. ℵ₀ is an infinite cardinal (not 0, 1, 2, 3, etc), but it isn't the only infinite cardinal!

|P(ℕ)| = 𝔠, the continuum, is another infinite cardinal. It is the cardinality of the set of subsets of ℕ, i.e. the set of sets of natural numbers.

Why are these cardinals different? Well, given some function f: ℕ → P(ℕ), try to find some subset of ℕ that is not in the range of f~ Thus, show that f cannot be surjective!~

Here is a hint for when you get stuck: if there is some x for which A and B disagree on whether they contain it (x is in A but not in B, or x is in B but not in A), then A and B are different.

Now, that we have gone over the basics of cardinal equality, let's look at something more exciting~ cardinal comparison:

⊰ Cardinal Comparison ⊱

[Definition] For two sets A and B, |A| ≤ |B| iff there exists an injection f: A → B.

≤ is a partial order on cardinal numbers:

≤ is reflexive: x ≤ x for every cardinal x

≤ is transitive: if x ≤ y and y ≤ z, then x ≤ z

≤ is antisymmetric: if x ≤ y and y ≤ x, then x = y

Reflexivity and transitivity are easy exercises for the reader~ (the proof is literally just the same as refl and trans for cardinal equality :p). Antisymmetry, on the other hand, is a lot more difficult!

That cardinals are antisymmetric is known as the Shröder-Bernstein theorem. The proof may be hard to follow, so good luck! :D

Let A and B be sets and suppose there are injections f: A → B and g: Β → A. We want to define a bijection h: A → B.

We can see that f and g form chains a₁ ↦{f} b₁ ↦{g} a₂ ↦{f} b₂ ↦{g} ... alternating between elements of A and elements of B. Since g is an injection, for every a ∈ A, we either have that it comes from some b (i.e. g(b) = a), or from nothing. The same holds for f. So every element in A and B belongs to a unique chain ... ↦ * ↦ * ↦ ... defined by f and g. Some of these chains have a sudden start (i.e. some a ∈ A or some b ∈ B that doesn't have an inverse under g or f), and some of these chains are infinite.

For a ∈ A, we can define h(a) based on what kind of chain a belongs to. If the chain a belongs to starts in A, we can set h(a) = f(a). If the chain a belongs to starts with some element in B, we can set h(a) = g⁻¹(a) (i.e. we set h(a) to an element b in B such that g(b) = a). If a belongs to a chain that goes infinitely far left, we do whatever. I'm just going to choose h(a) = f(a).

I'll leave it to you to verify that h is indeed a bijection.

We can also define strict comparison:

[Definition] For cardinals x and y, x < y iff x ≤ y and x ≠ y.

Since ℵ₀ ≠ 𝔠, as shown in the previous chapter, and n ↦ {n} forms an injection from ℕ to P(ℕ), we have ℵ₀ < 𝔠. In fact, for every cardinal x, there is some cardinal y so that x < y.

Here is some basic terminology for partial orders:

[Definition] Let ≤ be a partial order on some class P and let A ⊂ P be a subclass of P. A maximum element of A is some x ∈ A so that, for all y ∈ A, x ≤ y → x = y. A minimal element of A is some x ∈ A so that, for all y ∈ A, y ≤ x → x = y. The greatest element of A, if it exists, is some x ∈ A so that, for all y ∈ A, y ≤ x. The least element of A, if it exists, is some x ∈ A so that, for all y ∈ A, x ≤ y. An upper bound of A is some x ∈ P so that, for all y ∈ A, y ≤ x. A lower bound of A is some x ∈ P so that, for all y ∈ A, x ≤ y. The infimum of A, denoted inf(A), if it exists, is the greatest lower bound of A. The supremum of A, denoted sup(A), if it exists, is the least upper bound of A.

Given x,y ∈ P, the meet of x and y, often denoted x ∧ y, is the infimum of {x,y}. The join of x and y, often denoted x ∨ y, is the supremum of {x,y}. If every pair of elements in a poset (partial ordered set) has a meet and a join, then that poset is called a lattice. x and y are comparable iff x ≤ y or y ≤ x, x ⊥ y is used to denote that x and y are incomparable. Examples of lattices are: the powerset lattice (P(X),⊂) of any set X, where x ∧ y = x ∩ y and x ∨ y = x ∪ y, linear orders like ℚ, where x ∧ y = min(x,y) and x ∨ y = max(x,y), etc.

Without assuming the axiom of choice, two cardinals needn't have a meet or join. Now you know random useless stuff about posets!!

Oh, and: ℵ₀ is a minimum infinite cardinal. Proof is left as an exercise~

❈ Cardinal Arithmetic ❈

Cardinal numbers are numbers, so it'd make sense if we could do arithmetic on them. And we can!

[Definition] For sets A and B, |A| + |B| is the cardinality of the set A ⊔ B, i.e. the disjoint union of A and B. Members of A ⊔ B are (0,a) and (1,b) for a ∈ A and b ∈ B.

[Definition] For sets A and B, |A| · |B| is the cardinality of the Cartesian product A × B of A and B. Members of A × B are ordered pairs (a,b) for a ∈ A and b ∈ B.

Arithmetic on finite cardinals works as you'd expect: 6 + 3 = 9 and 12 · 2 = 24. Addition and multiplication on infinite cardinals, however, is actually quite boring, as x+y and xy are simply max(x,y). Well, that is, if you assume AC. Without AC, cardinal arithmetic can actually get very interesting (and weird af).

I am not that familiar with cardinal arithmetic without the axiom of choice, so you'll have to do more research on that on your own if you're interested!

Cardinal addition and multiplication are commutative, x+y = y+x, xy = yx, associative, (x+y)+z = x+(y+z), (xy)z = x(yz), and multiplication distributes over addition, x(y+z) = xy+xz.

Personally, I think cardinal exponentiation is more interesting than addition or multiplication:

[Definition] For sets A and B, |A|^|B| is the cardinality of the set of functions from B to A.

For cardinals x and y, if x ≥ 2, then we can prove that x^y > y. 𝔠, the cardinality of the continuum, is equal to 2^ℵ₀ and to ℵ₀^ℵ₀. Also, it's the cardinality of the set of real numbers ℝ. Here's a vid I found that explains why.

I think I might be getting too eepyy to write. I'll write more tomorrow.

[zzz]

Good morning!

Cardinal exponentiation has all the properties you'd think it has: x^y · x^z = x^(y+z) and (x^y)^z = x^(yz). Also, 0⁰ = 1.

There are also infinite sums and infinite products:

[Definition] Σ_(i ∈ I) A_i is the set of tuples (i,a) for i ∈ I and a ∈ A_i. Σ_(i ∈ I) |A_i| = |Σ_(i ∈ I) A_i| is the cardinality of this set.

[Definition] Π_(i ∈ I) A_i is the set of functions f with I as domain and f(i) ∈ A_i for all i. Π_(i ∈ I) |A_i| = |Π_(i ∈ I) A_i| is the cardinality of this set.

Σ_(i ∈ {1,...,k}) x_i = x₁ + ... + xₖ and Π_(i ∈ {1,...,k}) x_i = x₁ · ... · xₖ, so this is a valid extension of sums and products. The sum of x many y's, i.e. Σ_(i ∈ I) x for |I| = y, is the same as the product of x and y. The product of x many y's, Π_(i ∈ y) x, is equal to x^y. If X is a family of sets, I also sometimes write ΣX and ΠX for Σ_(A ∈ X) A and Π_(A ∈ X) A.

Here is a fun fact that idk where to place in this blog post: a Dedekind infinite cardinal is a cardinal x = |A| for which there exists an injection f: A → A that is not surjective. In other words, x+1 > x. Without the axiom of choice, infinite Dedekind finite cardinal numbers (aka mediate/Dedekind cardinals) can exist, and these cardinals are incomparable to ℵ₀.

Here is a fun question: is the product of any number of non-zero cardinals always non-zero?

Well?~

◈ Axiom of Choice ◈

The axiom of choice (AC) states that the product of any number of non-zero cardinals is always non-zero. I.e. for any family X of non-empty sets, ΠX is non-empty (a member of ΠX is called a choice function for X).

Eh... I didn't really plan what to write in this chapter besides what the axiom of choice is. Oh, well!

...

Wait-

Wait, WHAT THE F---

So.. apparently, you cannot take infinite products of cardinals without assuming the axiom of choice. ℵ₀ is the cardinality of ℕ, but ℕ is not the only set with cardinality ℵ₀. If A has cardinality ℵ₀, then in how many ways does it have cardinality ℵ₀? Well, if a bijection f: A → ℕ exists, then 𝔠 many of those bijections exist. If we take ℵ₀ different sets A₀, A₁, A₂, ... all with cardinality ℵ₀, we have, for each k, multiple bijections f: Aₖ → ℕ. We would expect |Π_(k ∈ ℕ) Aₖ| = ℵ₀^ℵ₀ = 𝔠, but how do we choose a bijection f: Aₖ → ℕ for each natural number k? If we only had a finite amount of Aₖ, this wouldn't be a problem. However, we have an infinite amount of Aₖ... so we need to make an infinite amount of choices. So, we can define the set X = {{f | f: Aₖ → ℕ is a bijection} | k ∈ ℕ}, and a choice function for X gives us a bijection for each Aₖ! But... we need AC to prove such a choice function exists.

...

HOW AM I SUPPOSED TO LEARN AD WHEN THIS F----D UP SH-- HAPPENS?

..sorry for screaming at you.

Luckily, infinite cardinal addition still works as normal without AC... probably.

Maybe.

Nope, it doesn't.

【 Aleph Cardinals 】

As you can see, we have run out of symbols to put on the sides of the chapter titles.

Anyways! So, a well order is an order (A,≤) where:

≤ is a partial order on A.

≤ is total/linear: for all x and y in A, x and y are comparable by ≤.

≤ is well-founded: all non-empty S ⊂ A have a minimum element.

Well, ig those random facts about posets weren't completely useless.

I might make a blog post about ordinals later, which'll go more in-depth on well-orders and linear orders.

[Definition] An aleph cardinal is an infinite cardinal number |A| for which there exists a well-order (A,≤).

An example of an aleph cardinal is ℵ₀: ℵ₀ = |ℕ| and the usual order on ℕ is a well-order. The n-th aleph cardinal is written as ℵₙ, starting at 0th of course :3

The statement ‘ℵ₁ = 𝔠’ is known as the continuum hypothesis (CH), which is both unprovable and undisprovable. I might make a blog post about forcing in the future! ^^

Although cardinals (without AC) don't need to be linearly ordered, aleph cardinals are linearly ordered and even well-ordered. A proof of this is left as an exercise~ Because, well, of course it is :P

The well-ordering principle states that every set A has a well-order (A,≤). Equivalently, every infinite cardinal is an aleph cardinal.

It turns out: AC and the well-ordering principle are equivalent! Although AC, when written in its original form, seems more obviously true (to me at least), I think the well-ordering principle is a lot more useful.

I might go more in depth on why this is in my post about ordinal numbers. (why AC and the well ordering principle are equivalent)

❖ Cofinality ❖

There are two definitions of cofinality: one for cardinals, and one for ordinals. Since this blog-post is about cardinals, I'll give the one for cardinals:

[Definition] The cofinality of a cardinal x = |A| is the minimum cardinality of a partition X of A into sets of cardinality <x.

I often use cf(x) to denote the cofinality of x and cof(x) to denote the class of cardinals with cofinality x, though people also often use cof(x) to denote the cofinality of x. A partition of a set A is a set X such that all members of X are subsets of A, none of the members of X intersect and the union of all members of X is A.

We have cf(0) = 0, cf(1) = 1, cf(2) = 2 and for every finite n > 2, cf(n) = 2. 3 = |{a,b,c}| can be partitioned into {{a},{a,b}}, the partition {{a,b,c}} doesn't work as this has a set {a,b,c} of cardinality 3, which is not < 3.

[Definition] A cardinal κ is regular iff cf(κ) = κ.

Equivalently, every partition of κ either has cardinality κ or an element of cardinality κ.

We have cf(cf(x)) = cf(x) for every cardinal number x, so cf(x) is always regular.

0, 1 and 2 are examples of finite regular cardinals, though often, 2 is not regarded as regular. ℵ₀ also is a regular cardinal.

A cardinal that is not regular is called a singular cardinal. An example of a singular cardinal is ℵ_ω, which is the sum of ℵₙ for finite n. cf(ℵ_ω) = ℵ₀ as {ℵ₀,ℵ₁,ℵ₂,...} is a partition of ℵ_ω = Σ_(n ∈ ω) ℵₙ into ℵ₀ many cardinals each of which is <ℵ_ω.

Here is the ordinal definition of cofinality, which is (a lot) more often used if you have the axiom of choice:

[Definition] The cofinality of an order (A,≤) is the minimum cardinality of a cofinal subset S ⊂ A. S is cofinal if ∀x ∈ A ∃y ∈ S, x ≤ y.

Assuming the axiom of choice, the cofinality of a cardinal κ is often defined as the cofinality of the minimum ordinal α such that |α| = κ. An ordinal is the order-type of a well-order, and one ordinal is less than another if there exists an order preserving injection from one to the other. cf(κ) with the above definition agrees with the one I gave earlier for infinite κ, but if κ > 0 is finite, then cf(κ) is 1 instead of 2. I might talk more about this in my blog-post about ordinal numbers. For now, we'll just use the partition definition of cofinality.

If x is infinite Dedekind finite, then cf(x) is 2, which I think is kinda funni. We can also have cf(𝔠) = 2 if we omit choice :3 Though with choice, cf(𝔠) is uncountable (i.e. >ℵ₀).

I don't know how to end this blog, so here is some random terminology:

A cardinal x is finite iff κ is 0, 1, 2, 3, etc.

A cardinal x is countable iff it's ≤ℵ₀. It is uncountable if it's not ≤ℵ₀.

An alpeh cardinal is an infinite well-orderable cardinal.

A strong limit cardinal is a cardinal x for which, for all y < x, we have 2^y < x.

ℶ_0 = ℵ_0, ℶ_(n+1) = 2^(ℶ_n) and ℶ_λ = sup{ℶ_α | α < λ} for limit ordinal λ. A cardinal of the form ℶ_n is called a beth cardinal and ℶ₁ = 𝔠.

Assuming AC, x⁺ is the next cardinal after x, called the successor cardinal. AC is needed to ensure there exists a cardinal directly after κ.

The continuum hypothesis (CH) states that ℵ₁ = 𝔠. The generalized continuum hypothesis (GCH) states that 2^x = x⁺ for all infinite x. Both are unprovable and undisprovable from the usual axioms of ZFC.

Assuming AC, a weak limit cardinal (or simply limit cardinal) is a cardinal x for which, for all y < x, y⁺ < x. Assuming GCH, weak and strong limit cardinals are the same.

A strongly inaccessible cardinal (or simply, inaccessible cardinal) is a regular strong limit cardinal. Assuming AC, a weakly inaccessible cardinal is a regular weak limit cardinal. The existence of an inaccessible cardinal implies the consistency of ZFC, and thus, by Gödel's incompleteness theorems, ZFC cannot prove the existence of an inaccessible cardinal.

That's all I had to say about cardinals (for now), bye!~

[bubbloquacious]

A partial equivalence relation is a reflexive, symmetric, and transitive binary relation ≡, which allows you to compare some, but not all pairs of elements x, y of the domain. A total equivalence relation is similar, but we also require that for all x, y we have that x ≡ y or y ≡ x.

[codeshive]

Homework 3 CSCI 301 solved

Use the method of proof by contradiction. 1. [4 points] Prove that √6 is irrational. 2. [4 points] If �, � �ℤ , then �) − 4� − 2 ≠ 0. 3. [4 points] Suppose Suppose A≠ Ø. Since Ø ⊆A×A, the set R= Ø is a relation on A. Is R reflexive? Symmetric? Transitive? If a property does not hold, say why. 4. [4 points] Define a relation R on Z as xRy if and only if 4 | (x + 3y) Prove R is an equivalence…

[codingprolab]

Homework 3 CSCI 301

Use the method of proof by contradiction. 1. [4 points] Prove that √6 is irrational. 2. [4 points] If �, � �ℤ , then �) − 4� − 2 ≠ 0. 3. [4 points] Suppose Suppose A≠ Ø. Since Ø ⊆A×A, the set R= Ø is a relation on A. Is R reflexive? Symmetric? Transitive? If a property does not hold, say why. 4. [4 points] Define a relation R on Z as xRy if and only if 4 | (x + 3y) Prove R is an equivalence…

[class12maths]

Relations and Functions: Class 12 Maths Video Tutorials — Mathyug

Are you a Class 12 student gearing up for your Maths exams? Do the concepts of relations and functions seem daunting? Fret not! Ashish Sir from MathYug has designed an exceptional self-study course that covers everything you need to master these topics. This comprehensive course includes video tutorials, detailed notes, and assignments that ensure thorough understanding and practice.

What You’ll Learn

Ashish Sir’s course on MathYug is meticulously structured to provide a deep understanding of relations and functions, a crucial component of Class 12 Mathematics. The course is divided into two main parts:

1. Relations

In this segment, you will explore various types of relations. Here’s a brief overview of what you’ll learn:

Empty Relations: Understanding the concept of relations where no element of a set is related to any element of another set.

Universal Relations: Learning about relations where every element of a set is related to every element of another set.

Trivial Relations: Delving into the simplest form of relations that only relate to the identical pairs.

Reflexive Relations: Understanding relations where every element is related to itself.

Symmetric Relations: Exploring relations where if one element is related to another, the second element is also related to the first.

Transitive Relations: Learning about relations where if one element is related to a second, and the second to a third, then the first is related to the third.

Equivalence Relations: Discovering relations that are reflexive, symmetric, and transitive.

Equivalence Classes: Understanding the partition of a set into disjoint subsets where each subset is an equivalence class.

To ensure comprehensive understanding, the course provides practice questions from diverse sources such as NCERT Textbook exercises, NCERT Examples, Board’s Question Bank, RD Sharma, and NCERT Exemplar. You can download the PDF of assignments within the course for additional practice.

2. Functions

The second part of the course focuses on functions, including:

One to One and Onto Functions: Understanding functions where each element of one set is paired with a unique element of another set, and functions that map elements from one set onto every element of another set.

Composite Functions: Learning how to combine two functions to form a new function.

Inverse of a Function: Exploring how to find the function that reverses the effect of the original function.

Similar to the relations segment, this part also includes practice questions from various reputable sources. The assignments are designed to reinforce your understanding and are available for download within the course.

Sample Videos

To give you a glimpse of the quality of instruction, Ashish Sir has shared sample videos covering some of these essential topics. Watching these will help you understand his teaching methodology and the depth of content provided.

If you’re aiming to excel in your Class 12 Maths exams, Ashish Sir’s course on MathYug is your go-to resource. With detailed video tutorials, comprehensive notes, and extensive assignments, you will gain a solid understanding of relations and functions. Start your journey towards mastering these crucial topics today!

For more information and to access the course, visit MathYug. Happy learning!

[mathyug]

Unlock a deeper understanding of Reflexive, Symmetric, Transitive, and Equivalence Relations, and master Exercise 1.1 in Relations and Functions Class 12 Maths with expert instructor Ashish Sir on MathYug.Explore additional video lectures on our website: https://mathyug.com/class-12-maths

[dorothytheexplorothy]

alternatively always use the same X in the same font, but make sure it's clear from context which one you're referring to.

Definition. two subsets X,X ⊆ X of the natural numbers are called equivalent mod finite, denoted X=*X, if their symmetric difference XΔX is finite, i.e. if there exists a natural number X such that |XΔX| = X.

Proposition. equivalence mod finite is an equivalence relation.

proof. reflexivity holds by ∀X.|XΔX|=0; symmetry by that of the symmetric difference XΔX=XΔX; and transitivity, by the following "triangle inequality" for sets:

∀X,X,X⊆X.∀X,X∊X.(|XΔX|=X∧|XΔX|=X)⇒|XΔXΔX|≤X+X. ▢

Lemma. the finite sets of natural numbers form an equivalence class mod finite.

proof. for any subset X of the naturals, we have XΔX=X, where X is the empty set; therefore, XΔX is finite (i.e. X is equivalent to the empty set) exactly when X is finite. by the properties of equivalence relations any two subsets X,X are equivalent iff there's a third subset equivalent to both, and so when X is finite, X=*X iff X is also finite, since just then they're both equivalent to the empty set. ▢

Proposition. the class of finite sets of natural numbers is the only class with a least element under the usual ordering of sets by inclusion.

proof. the least element in the class of finite sets is the empty set, since it's a subset of every set. for any other equivalence class, assume by way of contradiction there's a least element X. by assumption, X is not equivalent to the empty set, and in particular is infinite. taking X := X\{minX} (the minimum of every set exists since the natural numbers are well-ordered), we have X⊆X and XΔX={minX}, or in other words X<X and X=*X, contradicting the assumption that X was minimal. ▢

Only represent variables by x in various fonts.

[quietimtryingtostudy]

9. Let W denote the set of words in the English dictionary. Define the relation R by R= {(x, y) ∈ W x W such that x and y have at least one letter in common}. Show that this relation R is reflexive and symmetric, but not transitive.

(i vaguely remember my teacher saying that we don't have word problems from this chapter or smtg. but just in case, ya know?)

Took help from one of my friends for this one- but I understood the entire procedure by the end!

(Yeah, I understand. Thanks for sending it though!!)

[edusquaremaths]

Q9 Exercise1.1 I Class 12 Maths NCERT Chapter 1 Relations and Functions | NCERT solutions

NCERT Class 12Chapter: Relations and FunctionsExercise 1.1Question 9: Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by(i) R = {(a, b) : |a – b| is a multiple of 4}(ii) R = {(a, b) : a = b}is an equivalence relation. Find the set of all elements related to 1 in each case.

[bhargaw1]

Equivalence relation

#equivalencerelations Definition 🙂 An equivalence relation on a set A, is a relation “R” in A which is #reflexive, #symmetric, and #transitive.

Definition 🙂 An equivalence relation on a set A, is a relation “R” in A which is reflexive, symmetric, and transitive. Reflexive relation ARB is a relation in set A, then R is called Reflexive relation R is reflexive, i.e., for all a belongs to R (a, a) also belongs to R a=1 & b=a then (1,1) also belongs to R ⭐⭐⭐⭐ Rating: 4 out of 5. Symmetric relation (mirror) Let R be a relation in set…

[mathhombre]

Pat Ballew adds "Equal The mathematical use of equal means that two things are related in a transitive, symmetric, and reflexive way in relation to some specified properties.  The meaning is rooted in the Latin word aeques for level. Chaucer's 1400 Treatise on the Astrolabe uses the term.  Cajori's History of Mathematical Notation gives many early symbols used in the West, both before and after Recorde wrote the now ubiquitous "=" for the symbol in 1557, but even for another century, many mathematicians used no symbol at all, or used a collection of other symbols.In fact, after 1557, Recorde's symbol  did not appear again in print until 1618.  Viete, writing 1n 1571 used the same symbol for difference (the absolute  value of a-b) and this symbol was widely repeated in this use across Europe. Another early symbol for equality was the script conjunction of ae."

When Was The Equals Sign (=) Invented?

The equals sign (=) was introduced by Robert Recorde in 1557 to denote equality between mathematical expressions.

More: https://riverainventions.com/who-invented-mathematical-symbols/

[noelements-setempty]

Word rhyming is an equivalence relation

Take the definition that two words rhyme if and only if they end with the same sound.

Reflexive: Every word rhymes with itself.

Well, if two words are the same, all their sounds have to match, including the final one, so this point holds.

Symmetric: If A rhymes with B, B rhymes with A.

This one’s really hard to prove, because it’s so obvious. If A rhymes with B, then the final sounds of A and B are the same. They will still be the same if we swap the words around. Please don’t make me explain it more, I’ll cry.

Transitive: If A rhymes with B and B rhymes with C, then A rhymes with C.

Call the sound at the end of word A ‘&’. If A rhymes with B, then B also has to end with ‘&’. If B rhymes with C, and B ends with ‘&’, then C also has to end with ‘&’. This means that both A and C end with ‘&’, and so A rhymes with C.

There we go. The argument no one cares about but me has been made. Rhyme is an equivalence relation. You can all go home.

[lipshits-continuous]

If your relation(ship) is

Reflexive

Symmetric

Transitive

It's not romantic, it's an equivalence

[wuggen]

An innocuous little math thing that I find really neat for some reason: a partial equality relation (symmetric and transitive) almost implies a full equality relation (symmetric, transitive, and reflexive). If a=b, then by symmetry b=a, and by transitivity a=a, and then you've got reflexivity. The only way to have a partial equality relation that is not a full equality relation is if there exists some element that just doesn't appear in the relation at all