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4

$\triangle ABC \sim \triangle DEF$.


4

Consider the multiply-by-$2$ map on the integers.


4

It's not. Take $j=k=1$ and $l=m=2$. Then $\delta_{jk}\delta_{lm}=1$, but $$\sum_{i=1}^N \delta_{ij}\delta_{ik}\delta_{il}\delta_{im}=\delta_{11}\delta_{11}\delta_{12}\delta_{12}+\delta_{21}\delta_{21}\delta_{22}\delta_{22}=0+0=0.$$ What is true is that $$\sum_{i=1}^N \delta_{ij}\delta_{ik}\delta_{il}\delta_{im}=\delta_{jklm},$$ i.e. the function that is ...


3

Strictly speaking, the universal quantification should be there in all three cases. It's often just left out as a shorthand. Edit: strictly speaking, all statements are implicitly universally quantified. This is true even if the statement is written as an implication e.g. $$\text{ if } a \in A \text{ then } (a,a) \in R$$ is really just $$\forall a \in ...


2

If the set has finite cardinality, yes. If not, then see the counterexamples in the other answers/comments.


2

Regarding your first point. To evaluate transitivity you have to look at two couples $(a,b),(b,c)$ in your relation set and verify that $(a,c)$ also belongs to it. Hence, you don't need to evaluate $(2,4),(3,4)$. Regarding your second point. Name $R=\{(0, 0), (0, 2), (2, 0), (2, 2), (2, 3), (3, 2), (3, 3)\}$. You have $(0,2) \in R$ and $(2,3) \in R$. If ...


2

Thanks for uploading the screen shot! Now it's clear that the author of the problem made a mistake somewhere. You could try to guess what they meant, but if possible, it would be better to notify them of the mistake and ask them to fix it.


1

In fewer symbols, the relation you are trying to show is an equivalence relation is that two functions $f$ and $g$ are equivalent if their derivatives are the same. The definition of an equivalence relation is a relation which is symmetric, reflexive, and transitive, so all you need to do is to prove the relation is each of these three things. Again in ...


1

You are not completely missing the point, but you're a bit off the mark. Firstly, let go of the fact that you know nothing about the elements of the set $A$. It really is not important. (Incidentally, the claim remains true even if $A$ is empty.) What you have to do is construct the function $f$. To construct a function you must specify its domain and ...


1

We find $M^{[n]}$ in the same way we would find $M^n$, only we replace addition with $\vee$ and multiplication with $\wedge$. So, for example: we have $$ M_R = \pmatrix{ 1&0&\color{blue}1\\ \color{red} 0& \color{red} 1& \color{purple}0\\ 1&1&\color{blue}0 } $$ So, the entry of $M_R^{[2]}$ in the $\color{red}2$nd row and ...


1

The symbol for denoting similar triangles is ($\color{blue}{\sim}$) Notice, suppose $\triangle ABC$ & $\triangle PQR$ are similar then in LaTex it is written as $\text{"\triangle ABC \sim \triangle PQR"}$ surrounded in-between by 2 or 4 dollar signs which appears as follows $$\color{blue}{\triangle ABC \sim \triangle PQR}$$


1

A pair $\langle a,b\rangle$ belongs to $R^2=R\circ R$ if there is an $x\in A$ such that $\langle a,x\rangle\in R$ and $\langle x,b\rangle\in R$. If $G$ is the digraph for $R$, that means that there’s an edge from $a$ to $x$ and another from $x$ to $b$. In other words, $\langle a,b\rangle\in R^2$ if there is a path of length $2$ in $G$ from $a$ to $b$. For ...


1

Suppose $R$ is reflexive and transitive. Then the relation defined by $$ x\mathrel{R_s}y \qquad\text{if and only if}\qquad x\mathrel{R}y\text{ and }y\mathrel{R} x $$ is an equivalence relation. The proof is easy. Consider $B=A/R_s$ and define, for $[x],[y]\in B$ ($[a]$ denotes the equivalence class with respect to $R_s$ of $a\in A$), $$ [x]\mathrel{S}[y] ...


1

Reflexive: Does $xD_nx \forall x\in S$? Yes, because $x|x \implies x|x$. Anti-symmetric: Does $xD_ny \land yD_nx \implies x=y$? Your proof here is correct, you might want to add $jk=1 \implies j=1,k=1$. Transitive: Does $xD_ny \land yD_nz \implies xD_nz$? We have $jx=y$ and $ky=z$ so $z=ky=kjx \implies jkx=z \implies xD_nz$. For the Hasse diagrams, I ...


1

You are done though! $$xm=y, yn=z \implies xmn= z \implies x|z \implies x D_n z$$ Substitution, you were doing so well.


1

Your intuition is correct. To describe the equivalence classes explicitly, it often helps to find the equivalence classes of numbers that are easy to work with. Let's find, for example, the equivalence classes $[0]$ and $[1]$. $0+a$ is even for which integers $a$? All these will form $[0]$. $1+b$ is even for which integers $b$? All these will form $[1]$. ...


1

Reflexive means EACH $a\in A$ is in relation with itself. Symmetric means AS LONG AS $aRb$, then $bRa$. Transitive means IF $aRb$, $bRc$, then $aRc$. A relation that is reflexive, symmetric and transitive is called an equivalence relation. I would like to think "friendship" as an example of equivalent relation (in a perfect world in which 1) you are ...


1

Probably the best-known example is the collinearity relation among three points $C(x,y,z)$, which is an equivalence relation in that it satisfies the axioms of... Symmetry: $C(x,y,z)$ is invariant under permutations of $x, y, z$ Reflexivity: $C(x,y,y)$ Transitivity: $C(a,x,y)\land C(a,y,z)\to C(a,x,z)$ And there doesn't seem to be a reasonable way to ...


1

I'm not sure if "$\rho \circ \phi$" means we do $\rho$ first or not, but it won't matter. Let $(x,y) \in \rho \circ \phi$. Case 1: We do $\rho$ first. Then there exists $z \in A$ such that $(x,z) \in \rho$ and $(z,y) \in \phi$. But then $(x,z) \in \sigma$, and so $(x,y) \in \sigma \circ \phi$. Case 2: We do $\phi$ first. Then there exists $z \in A$ such ...



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