Consider a relation R: A → A which is both symmetric and transitive.  The following proof shows that the relation is also reflexive: Take a ∈ A. If a ∼ b then b ∼ a by symmetry, and hence a ∼ a by transitivity. Therefore, the relation is reflexive."  Is the proof correct or not?  I think it is incorrect. I think it should be "For every a ∈ A, there exists b ∈ A that a ~ b and b ~ a.", not just "Take a ∈ A". After that we can use symmetry and transitivity to prove it is reflexive. That is what I think.  Am I right? Or what is the correct answer and the reason? THANKS!
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An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.  Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.  For any $a$ and $b$, $a R b$ implies $b R a$ by symmetry. Using transitivity, we have $a R a$.  Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al
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1) How do I prove the following:   Let A = {6a + 4b ∈ Z : a, b ∈ Z} and B = {2a ∈ Z : a ∈ Z}. Show that A = B.  Thank you all for the help!
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Consider a non-empty set A containing n objects. How many relations on A are both symmetric and reflexive?  The answer to this is $2^p$ where $p=$ $n \choose 2$. However, I dont understand why this is so. Can anyone explain this?
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I am reading  of the Book of Proof, and I'm stuck at the Exercise 10 of section 11.2. It is as follows.  Suppose $R$ and $S$ are two equivalence relations on a set $A$. Prove that $R \cap S$ is also an equivalence relation.  Thanks for helps!
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As part of a HW assignment in the course elementary set theory, I was given the following question:  Let $A$ be a set and let $T$ and $S$ be two equivalence relations on $A$. Prove: $S\circ T=T\circ S \iff S\circ T$ is an equivalence relation The $\Longleftarrow $ direction was not so difficult but I had a problem with the $\Longrightarrow $ direction.  I managed to prove reflexiveness and symmetry of $S\circ T$, but cannot prove the transitivity of $S\circ T$. Any suggestions?
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Given a transitive set$$a,$$ I can prove that $$\bigcup a$$ is also transitive, but I don't quite like my method because I must first prove $$a \subseteq \mathcal{P}(\bigcup a)$$ to get at $$\bigcup a\subseteq \mathcal{P}(\bigcup a).$$ Could you please show me a smarter proof?
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Please give me feedback on my answer to this question. Question:  For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;  $x\equiv y$   if $x^{2}=y^{2}$  on $R$  To show $\equiv$   is reflexive, we need to show that ,  $\forall x\in R:X\equiv x.$  Let $x\in R$,   $X\equiv x$   if $X^{2}=x^{2}$  , which is obvious.  $[x]\triangleq:\{y\in R/x\equiv y\}$  $[0]=[y\in R/0\equiv y\},$  when $y=0$,  Then $y^{2}=0^{2}=0$  Thus; there are not many infinity many equivalence classes.
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Is checking transitivity of a digraph not easier than (in terms of asymptotic complexity) taking the transitive closure of the digraph? Do we know any lower bound better than $\Omega(n^2)$ to determine if a digraph is transitive or not?
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If $G$ is a monoid, $e$ is its identity, if $ab=e$ and $ac=e$, can you give me a counterexample such that $b\neq c$?   If not, please prove $b=c$.  Thanks a lot.
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For all $x,y\in\Bbb{R}$ define that $x\equiv y$ if $x^2=y^2$ . Then $\equiv$ is an equivalence relation on $\Bbb{R}$ , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.  Solution  True. To show that $\equiv$ is reflexive we need to show that $\forall x\in\mathbb{R} :x\equiv x$. Let $x\in\mathbb{R} , x\equiv x \mbox{ if } x^{2}=x^{2}$, which is obvious.  $[x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0}$. Hence $y^{2} =0^{2} =0$ which implies that $y=0$.  Can anyone please give me correct answer to this question.
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The generalize rule for a transitive relation is   a -&gt; b b -&gt; c therefor a -&gt; c   If an element has less than 3 elements, can it still be transitive? If so, does that provide any useful information?
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Hei, guys!  I'm having some trouble with the next problem:  Let $A$ and $B$ be subgroups of $G$. Show that $\sim$ is an equivalence relation when it is defined as follows: $g\sim g'\Leftrightarrow g' = agb$ for some $a$ from $A$, $b$ from $B$.  Should I prove this is valid for any a and b? It says for some elements a and b. But I might as well take the unit element.   Could you give me a clue on how to solve it?  Cheers!
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If $g \circ f$ is injective, so is $g$   I don't think this is true. I think that $f$ has to be surjective.  So I am going to try to prove that:     If $g \circ f$ is injective, and $f$ is surjective, then $g$ is injective.   First off, $g \circ f$ means that $(g \circ f)(a) = (g \circ f)(b)$ then $a=b$.  And $f$ being surjective means that $\forall b \in B, \exists a\in A$ such that $f(a)=b$  Proof  Suppose that $g$ was not injective.  $g(f(a))= g(f'(a))$ and $f(a)\ne f'(a)$  But $g(f(a)) = g(f(b)) \to a=b$ $g(f(a))= g(f(b)) \to f(a)=f(b) \to a=b$ Hence contradiction, since $g$ is not injective.  So $g$ is injective.  Is this an acceptable proof, I think my logic is iffy around the second last paragraph.
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Let $S$ be the relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)S(c,d)$ if and only if $ad=bc$. Prove this is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.  I think I've found this to be reflexive and symmetric, but I'm stuck on transitivity. Can someone check my work so far and assist with testing transitivity?  Reflexive: Let $(x,y)S(x,y)$. Then $xy = yx$. So $S$ is reflexive. Symmetric: Suppose $(a,b)S(c,d)$. Then, $ad = bc$. Therefore, $da = cb$ and $cb = da$. Therefore, $(c,d)S(a,b)$. Thus, $S$ is symmetric.  Transitive: Suppose $(a,b)S(b,c)$. Then $ac = bb$.   This is where I'm stuck. Any ideas?
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Problem:  If a set $A$ has $n$ elements in it, how many reflexive relations can be defined on it?  My solution  Is the answer  summation of (n^2 - n)C(i)  for i=0 to n^2 -n  $$\sum_{i=0}^{n^2-n} C(n^2-n,i) = \sum_{i=0}^{n^2-n} \binom{n^2-n}i$$  How? well if i make a matrix $n\times n$  now the diagonal elements have to be selected,out of remaining $n^2-n$ any number of elements can be selected.
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"Consider a language with a single object constant a, a single unary function constant s, and two unary relation constants p and q. We start with the premises shown below. We know that p is true of s(a) and only s(a). We know that q is also true of s(a), but we do not know whether it is true of anything else.  ¬p(a) p(s(a)) ∀x:¬p(s(s(x))) q(s(a)) Prove ∀x.(p(x) ⇒ q(x)). Hint: Break the problem into two parts - first prove the result for s(x), and then use that intermediate conclusion to prove the overall result."  I know I have to reach AX:(p(s(X)) => q(s(X))) and p(a) => q(a) first, but I'm struggling getting started. I also don't know if my first steps are correct. Any help? Thank you.
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If $X,Y$ are reflexive, symmetric, and transitive, then $X \times Y$ is an equivalence relation where ${(a,b):a\in X, b\in Y}$.    I am trying to self learn these topics. I do know what an equivalence relation is. I wanted to know how would someone prove that $X\times Y$ be an equivalence relation?