Equivalence class Definition and 23 Threads

So, my approach and solution are as follows: $$ [x * y] = \{ z \in M : z \sim (x * y) \} $$ Since we know that ##a * b \sim a^{\prime} * b^{\prime}## we can rewrite ##z## as ## x^{\prime} * y^{\prime} ##. Plugging this in yields: $$ [x * y] = \{ x^{\prime}, y^{\prime} \in M : x^{\prime} *...

I'm following this video on how to establish an equivalence relation to define the tensor product space of Hilbert spaces: ##\mathcal{H1} \otimes\mathcal{H2}={T}\big/{\sim}## The definition for the equivalence relation is given in the lecture vidoe as ##(\sum_{j=1}^{J}c_j\psi_j...

Dear Everyone, $\newcommand{\R}{\mathbb{R}}$ I am struck in writing the equivalence classes. And the problem is this: Let ${\R}^{2}= \R \times \R$. Consider the relation $\sim$ on ${\R}^{2}$ that is given by $({x}_{1},{y}_{1}) \sim ({x}_{2},{y}_{2})$ whenever...

Let ##d_1## and ##d_2## be two metrics on the same set ##X##. We say that ##d_1## and ##d_2## are equivalent if the identity map from ##(X,d_1)## to ##(X,d_2)## and its inverse are continuous. We say that they’re uniformly equivalent if the identity map and its inverse are uniformly...

Homework Statement I am looking at the quotient group G = Z/3Z which is additive and abelian. The equivalence classes are: [0] = {...,0,3,6,...} [1] = {...,1,4,7,...} [2] = {...,2,5,8,...} I want to prove [0] is a normal subgroup, N, by showing gng-1 = n' ∈ N for g ∈ G and n ∈ N. Since G...

Please refer to the video at 37:02 from the link above. I am struggling with the notation "=" of the property (a) which concludes to [a]=[m]. shouldn't it be [a]⊆[m], just like [m]⊆M.

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown > Question: Let ~ be the equivalence relation on the set ℤ of integers defined by a~b if a-b is divisible by 5. Let k ∈ Em belong to the equivalence class of m, and l ∈ En belong to the equivalence...

Homework Statement Show that every partition of X naturally determines an equivalence relation whose equivalence classes match the subsets from the partition. Homework Equations ( 1 ) we know that equivalence sets on X can either be disjoint or equal The Attempt at a Solution Let Ai be a...

Consider this self-evident proposition: "If object A has the same mass as object B and object C separately, then object B has the same mass as object C." Why isn't this stated as a law, but the zeroth law of thermodynamics is? Is there a physical quantity u such that the u of A is equal to the...

Hello! :) I have to find an equivalence class $[g] \in \mathbb{Z_{15}}^{*}$ so that each equivalence class $\in \mathbb{Z}^{*}_{15}$ is a power of $[g]$. $\mathbb{Z}^{*}_{15}=\{[1],[2],[4],[7],[8],[11],[13],[14]\}$ I tried several powers of the above classes,and I think that there is no...

I am given that the relation ~ is defined on the set of real numbers by \(x\)~\(y\) iff \(x^2=y^2\). First part of the problem said to prove ~ is an equivalence relation, that wasn't bad. The second part asks to "Describe the equivalence classes". This just seems really vague to me. Is this a...

Homework Statement ~ is a equivalence relation on integers defined as: a~b if and only if 2a+3b is divisible by 5 What is the equivalence class of 0 Homework Equations The Attempt at a Solution [0] = {0, 5n} n is an integer My reasoning for choosing 0 is that if a=0...

Hi all, So the equivalence class X/\sim is the set of all equivalences classes [x]. I was wondering if there was a way of writing it in terms of the usual quotient operation: G/N=\{gN\ |\ g\in G\}? From what I've read, it would be something like X/\sim = X/[e]. But, since I'm looking at the de...

Why in equivalence class of N of even number and odd number, the even number are taken as related to 0 and odd number are related as 1 i.e [0] and [1]. Instead of [0], even number can also be related to [2] or [4]? Or [2] or [4] could also be taken, as it is just an convention. Thanks.

i have two relations given to me which are both defined on the integers Z by relation 1: x~y if 3x^2 -y^2 is divisibale by 2 and relation 2: x~y if 3x^2 -y^2 ≥0 I have used three properties to figure out that relation 1 is eqivalence relation as it stands for all three properties i.e...

(x1, y1)Υ(x2, y2) ⇔ x1 × y2 = x2 × y1 for all x1, x2 ∈ Z and y1, y2 ∈ Z+ have been shown to be an equivalence relation in tutorial. Specify the equivalence class [(2; 3)] as induced by Υ. i don't understand what it means by 'Specify the equivalence class [(2; 3)] as induced by Υ.'...

Homework Statement I'm trying to prove that "if R is an equivalence relation on a set A, prove that if s and t are elements of A then either [s] intersect [t] = empty set, or, [s]=[t]" Homework Equations The Attempt at a Solution I know that if you were to start trying to solve...

Homework Statement Definition: If A is a set and if ~ is an equivalence relation on A, then the equivalence class of a\inA is the set {x\inA l a~x}. We write it as cl(a)Let S be the set of all integer. Given a,b \in S, define a~b if a-b is an even integer. so, the equivalent class of a...

Homework Statement Suppose [d], [b] \in Z sub n.

Homework Statement Find the equivalence class [2] for the following equivalence relations: a) R: Z <-> Z, where xRy, iff |x| = |y| b) T: N <-> N, where xTy, iff xmod4 = ymod4 N means natural numbers etc...there wasnt the correct symbols in the latex reference Homework Equations ?? The...

Homework Statement Prove that if (a1, b1) ~ (a2, b2) and (c1, d1) ~ (c2, d2), then (a1, b1) + (c1, d1) ~ (a2, b2) + (c2, d2) and (a1, b1) \bullet (c1, d1) ~ (a2, b2)\bullet (c2, d2). Let [a, b] denote the equivalence class with respect to ~ of (a, b) in Z x (Z-{0}), and define Q to be the...

I read the textbook 5 times now and I can't seem to figure out what the equivalence class suppose to be and how to find it, and i don't understand quotient set either (more importantly how to find it). I'm not familiar with any Equivalences at all if anyone can help me with it that would be...

Ok, I'm having trouble understanding the path-homotopy equivalence class. It's kind of blurry when they apply the operation... [f]*[g] = [f*g] ...where [f] is the path-homotopy equivalence class of f. I can see that an element in [f]*[g] is in [f*g], but not the other way around. For...