MHB How Do Sets W, Y, and Z Intersect?

  • Thread starter Thread starter rcs1
  • Start date Start date
  • Tags Tags
    Sets
rcs1
Messages
10
Reaction score
0
W = { x| 0< x < 3}
Y = { x| x > 2 }
Z = { x | 0 <= x < = 4}

then the problems:

1. (WUY) intersects Z =
2. (W intersects Y) intersects Z = do my propose answers below correct sir/mam?
1. 0 < x < = 4
2. 2< x < 3
hope you can help me on this
im using the line number ... but all i see in the intersection that lies in 1 to 2

im a little bit confused on this sir/mam

thank you much.
 
Physics news on Phys.org
rcs said:
W = { x| 0< x < 3}
Y = { x| x > 2 }
Z = { x | 0 <= x < = 4}

then the problems:

1. (WUY) intersects Z =

2. (W intersects Y) intersects Z = do my propose answers below correct sir/ma'am?

1. 0 < x < = 4 correct

2. 2< x < 3 correct

No further comment
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top