Equivalence Relations

**Tazsweet19** · September 21st 2008, 05:16 PM

To be as concrete as possible we can consider “free vectors” and “bound vectors” on the real line R (the usual number line which is in one-one correspondence with R and is, in fact, coordinatized by R.) [/FONT]In this case a bound vector is defined to be any ordered pair of real number (x1[/SIZE][/FONT], x2), e.g. (4, 5). The equivalence relation on bound vectors is (x1, x2) ~ (y1, y2) if and only if x1- x2 = y1 - y2.

Verify that “~” is an equivalence relation.

x~x = reflexive -> 4 ~ 4
x~y <=> y~x = symmetric -> 4 ~ 5 <=> 5 ~ 4
x~y and y~z => x~z = transitive -> 4 ~ 5 and 5 ~ z => 4 ~ z

Is it correct?

**Jhevon** · September 21st 2008, 05:26 PM

Originally Posted by Tazsweet19

In calculus the notion of “free vector” is defined by an equivalence relation on the set of “bound vectors”. To be as concrete as possible we can consider “free vectors” and “bound vectors” on the real line R (the usual number line which is in one-one correspondence with R and is, in fact, coordinatized by R.) In this case a bound vector is defined to be any ordered pair of real number (x1, x2), e.g. (4, 5). The equivalence relation on bound vectors is (x1, x2) ~ (y1, y2) if and only if x1- x2 = y1 - y2.

Verify that “~” is an equivalence relation.

x~x = reflexive -> 4 ~ 4
x~y <=> y~x = symmetric -> 4 ~ 5 <=> 5 ~ 4
x~y and y~z => x~z = transitive -> 4 ~ 5 and 5 ~ z => 4 ~ z

Is it correct?

no, it is not correct. review the definition of an "equivalence relation"

you have to show:

(a) ~ is reflexive: $(x_1,y_1)$ ~ $(x_1, y_1)$ for all $(x_1,y_1) \in \mathbb{R} \times \mathbb{R}$

(b) ~ is symmetric: $(x_1,y_1)$ ~ $(x_2,y_2)$ implies that $(x_2,y_2)$ ~ $(x_1,y_1)$ for $(x_1,y_1),(x_2,y_2) \in \mathbb{R} \times \mathbb{R}$

(c) ~ is transitive: $(x_1,y_1)$ ~ $(x_2,y_2)$ and $(x_2,y_2)$ ~ $(x_3,y_3)$ implies that $(x_1,y_1)$ ~ $(x_3,y_3)$ for $(x_1,y_1),(x_2,y_2),(x_3,y_3) \in \mathbb{R} \times \mathbb{R}$

of course you show the relations based on the equation you were given

**Tazsweet19** · September 21st 2008, 06:49 PM

I try to figure out the question:

In particular, describe the partition of the set of all ordered pairs {(x1, x2)} which corresponds to ~. In words, explain what is a free vector?

[SIZE=3]A x B := {(a,b) | a E A, b E B} -> {(x1,x2) | x1 e A, x2 e B}

Is it correct?

**wisterville** · September 21st 2008, 07:20 PM

Hello,

Maybe you want to understand the equivalence relation as a subset V of $A\times A$ , $A$ the set of "bound vectors." So, we are talking about ordered pairs of ordered pairs.

$A\times A=\{((x_1, x_2), (y_1, y_2))|(x_1, x_2)\in\mathbb{R}\times\mathbb{R}, (y_1, y_2)\in\mathbb{R}\times\mathbb{R}\}.$
You have to show
(a) $((x_1, x_2), (x_1, x_2))\in V$ for all $(x_1, x_2)\in\mathbb{R}\times\mathbb{R}$ ,
(b) $((x_1, x_2), (y_1, y_2))\in V$ implies $((y_1, y_2), (x_1, x_2))\in V$ for all blablabla...etc.
where $V=\{((x_1, x_2), (y_1, y_2))|((x_1, x_2), (y_1, y_2))\in A\times A, x_1-x_2=y_1-y_2\}$ .

Bye.

**wisterville** · September 21st 2008, 08:05 PM

Hello,

The equivalence class of a bound vector $(x_1, x_2)$ is $\overline{(x_1, x_2)}=\{(y_1, y_2)|y_1\in\mathbb{R}, y_2\in\mathbb{R}, x_1-x_2=y_1-y_2\}$ .
It corresponds to the real number x_1-x_2.
(Show that any real number has its counterpart as an equivalence class of a bound vector.)

Bye.

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