# Tag Info

## Hot answers tagged relations

10

You don't need infinite sets at all - a set with two elements is enough. Suppose $a\not=b$. Then $a\not=b$ and $b\not=a$. But $a=a$! Indeed, "$\not=$" isn't transitive on any set with at least two elements.

9

For example $1\neq 2$ and $2\neq 1$, but is $1\neq 1$?

8

Hint: $|x|+|y|=|x+y|$ is true if $x$ and $y$ have the same sign or one of them is $0$. Thus, all positive numbers are related to each other, and all negative numbers are related to each other, and $0$ is related to everything ...

6

It's not so much about being a different relation. Both $\lt$ and $\lneq$ mean the same thing. The second one is emphasizing that it is not equal. In the answer, he emphasizes that it's not equal. $\subset$ is sometimes taken to mean $\subseteq$, so if you want to be clear, you can write $\subsetneq$.

5

If $a \in [[x]]_R$, then $xRa$. We know that $xRy$ and $R$ is symmetric, hence $yRx$. By transitivity, we get $yRa$. Thus $a \in [[y]]_R$. Now just exchange the roles.

5

Definition of a symmetric relation $R$: a relation $R$ on a set $X$ is said symmetric if for all $x,y\in X$ such that $xRy$, then $yRx$. This statement says that IF $x$ and $y$ are related, then $y$ and $x$ are related. The false statement is $xRy$ and $yRx$ for all $x,y\in X$. This statement makes the assumption that $xRy$ has a meaning ...

4

Let $x R^T y$ and $y R^T z$. Then $zRy$ and $yRx$ by definition of the transpose relation. Since $R$ is transitive, that means that $zRx$. Again, by definition of the transpose relation, $xR^Tz$.

4

This is form latin reflexio "back-bending"; the relation "bends" $a$ back to itself (and by the way, "itself" is a reflexive pronoun).

4

Recall the definition: $R$ is a relation on a set $A$ if and only if $R\subseteq A\times A$. Now ask yourself, how many elements, and therefore how many subsets, does $A\times A$ have?

3

Hint: A relation on $A$ is a subset of $A\times A$. In general not an element as you seem to think.

3

The same number of decimal digits sounds good to me. To make it harder to understand, we could write $x\sim y$ if $\lfloor \log_{10} x\rfloor=\lfloor \log_{10} y\rfloor$.

3

The proof you gave is correct. However, you only prove uniqueness where you should also prove existence. In general, when you are asked to prove that "there exists a unique" something, you have to prove that there actually exists such an object, and that it is unique. The proof of existence is not correct. You cannot start by assuming you have such a ...

3

Yes $<$ and $\lneq$ are the same relation. However, be careful about set inclusion: older texts, and some current mathematicians, use $\subset$ to mean $\subseteq$, and to indicate strict inclusion they might use $\subsetneq$ or $\subsetneqq$. Other authors who use both $\subset$ and $\subseteq$ will use $\subset$ to mean strict inclusion. These days, ...

3

Just a speculation. In case of partial ordering (for example, with respect to a cone) there are normally more situations than just $\ge$ and $>$. It may happen that a vector $a\ge 0$ (in the cone), $a\ne 0$ (not a vertex), but not yet $a>0$ (in the sense of the cone interior). It is where the demand for more notation symbols is most likely to come ...

3

We have, by associativity, that $$(R;S;R)^2 = R;S;R;R;S;R = R;S;R^2;S;R$$ Now, as $R$ is transitive, $R^2\subseteq R$, hence $$(R;S;R)^2 \subseteq R;S;R;S;R$$ As $S$ is reflexive $\operatorname{id}\subseteq S$, giving $$(R;S;R)^2 \subseteq R;S;R;S;R = R;S;R;S;R;\operatorname{id} \subseteq R;S;R;S;R;S = (R;S)^3$$

3

I'm confident the term comes from grammar, where reflexive pronouns are things like "self" in English, and reflexive verbs are verbs whose object is the same as its subject, e.g. Él se lavó (Spanish - He washes himself). Romance languages in particular have verbs which are only reflexive, e.g. "to sit down" in French is s'asseoir. The reflexive pronoun is ...

3

It is not transitive. So no. $1$ is related to $0$ but also $0$ is related to $-1$. Is it true that $1$ and $-1$ are related?

3

Your proof is ok. It can be proved it in a simpler way Since $$U_1\times\cdots\times U_n=U_1\times A_2\times\cdots\times A_n\cap A_1\times U_2\times\cdots\times A_n\cap\cdots \cap A_1\times A_2\times\cdots\times U_n$$ \begin{align} f^{-1}(U_1\times\cdots\times U_n)&=f^{-1}(U_1\times A_2\times\cdots\times A_n)\cap f^{-1}(A_1\times U_2\times\cdots\times ...

3

Let me see. Never heard of those, but try looking at this Wiki page; Absolutely correct; Correct when $X,Y$ are finite, might extend to infinite sets with some transfinite arithmetic, I don't know; e.g., $|\mathbb{Q}|^{|\mathbb{Q}|}$ is not really clear; on the other hand, $2^{|\mathbb{Q}|}=|\mathbb{R}|$, if we want this relation in your 3) to continue ...

3

The textbook is right. For example: $A=\{1,2,3,4,5,6,7,8,9\}$ and $R_1=\{(1,1),(2,3),(3,2),(4,3),(3,4)\}$ and $R_2=\{(2,2),(3,3),(1,2),(4,1),(3,4)\}$ Here $R_1$ is symmetric since $\forall\ (x,y) \in R_1$ there exists $(y,x) \in R_1$ but $R_2$ is not since $\forall\ (x,y) \in R_2$ there does not exist $(y,x) \in R_2$. The definition of symmetry does not ...

3

$R:=\{(1,2),(2,1)\}$ is not transitive. $R\circ R=\{(1,1),(2,2)\}$ is transitive.

3

You specified a function as follows: $$f(2)=3, f(3)=3, f(4)=2, f(5)=1$$ This is not completely defined on the domain, as $1$ is not sent anywhere. If instead you misunderstood the usual way functions are defined, you might argue that you instead defined the function $$f(3)=2, f(3)=3, f(2)=4, f(1)=5$$ This would of course be wrong, but also still not be a ...

3

Is $R_1$ reflexive? That means: Does $xx=5$ hold for every real number $x$? Is $R_1$ anti-reflexive? Does $xx\ne5$ hold for every real number $x$? Is $R_1$ symmetric? Does $xy=5$ imply $yx=5$ for real numbers $x,y$? Is $R_1$ transitive? Does $xy=5$ & $yz=5$ imply $xz=5$ for real numbers $x,y,z$? First, $R_1$ is not reflexive, because $17$ is a real ...

2

Let $X$ be the set $\mathbb{N}$ of natural numbers, and let $R$ be $=$, the usual equality relation. Then $=$ is symmetric, because $x=y$ and $y=x$ mean the same thing. However it is not true that all natural numbers $x$ and $y$ fulfill $x=y$. For example $1=2$ is not satisfied. If $x$ equals $y$, then $y$ equals $x$. But not all pairs $x,y$ are like that. ...

2

I think you have to assume you are distributing distinguishable objects into distinct containers. You need to use Stirling numbers of the second kind or something similar. The answer in the book seems to be $S_2(6,3)\times 3! + S_2(6,4)\times 4! = 90 \times 6 + 65 \times 24$. I might do it another way and say that you have to distribute $7$ accounts onto ...

2

Hint Go back to the definitions: $S$ is reflexive iff for every $x\in A$, we have $(x,x)\in S$. Since there are no pairs $(a,b)\in S$, what can you conclude? $S$ is symmetric iff whenever $(x,y)\in S$ then $(y,x)\in S$. Since there are no pairs $(x,y)\in S$, what can you conclude? $S$ is transitive iff ... . What can you conclude?

2

You can use the fact that $$A\subset B\implies f(A)\subset f(B)$$ Since for any $\alpha$ $D_{α}\subset \bigcup_{α\in Δ} D_{α}$, there is $$f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})\quad\text{and so}\quad \bigcup_{α\in Δ}f(D_{α})\subset f(\bigcup_{α\in Δ} D_{α})$$ On the other hand, for any $y\in f(\bigcup_{α\in Δ}D_{α})$, there is a $x\in \bigcup_{α\in ... 2 You can kill (1) and (3) with one blow by giving two equivalence relations whose composition is not transitive. Consider these equivalence relations$R$and$S$on the set$\{0,1,2\}$: $$R=\{(0,0),(1,1),(2,2),(0,1),(1,0)\}$$ $$S=\{(0,0),(1,1),(2,2),(1,2),(2,1)\}$$ In fact$R\circ S$is neither transitive nor symmetric. This follows from the facts:$(2,1)\in ...

2

In general, I try to explain abstract algebra as the study of common operations. For example, we use the "+" operation in many contexts: we can add integers, real numbers, complex numbers, vectors, polynomials, matrices, continuous functions, hours in the day, and in many other contexts. All of these "+"'s have similar properties, they are commutative and ...

2

Checking element by element you see it's antisymmetric. If (2,1) and (1,2) exist, there would be a problem, but it's not the case.

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