1. ## define a relation...

i have a question and i am stuck on the first part was wondering if you could help.

Let R be the set of real numbers. Define a relation R(subscript1) on the set R as follows: m R(subscript1) n whenever n-m congruent to 0(mod24).

First off how do i define this relation?
I then have to proove R(subscript1) is an equivalence relation but i want to give this a bash first once i managed to define the relation.

Let R be the set of real numbers. Define a relation R(subscript1) on the set R as follows: m R(subscript1) n whenever n-m congruent to 0(mod24). First off how do i define this relation?

Is R a collection of ordered pairs? If so it is a relation on $\displaystyle \Re$.

Let R be the set of real numbers. Define a relation R(subscript1) on the set R as follows: m R(subscript1) n whenever n-m congruent to 0(mod24). Is it an equivalence relation.

You must show that R is reflexive, symmetric & transitive.

3. Originally Posted by Plato
Is R a collection of ordered pairs? If so it is a relation on $\displaystyle \Re$.

sorry but dont get this bit

You must show that R is reflexive, symmetric & transitive.

yeah get this bit
I just dont get that first bit Is R a collection of ordered pairs? If so it is a relation on $\displaystyle \Re$.

I just dont get that first bit Is R a collection of ordered pairs? If so it is a relation on $\displaystyle \Re$.
ordered pairs is what you would know as coordinates, sort of

like (m,n) is an ordered pair.

now, a relation is a set of such pairs. thus, saying $\displaystyle mR_1n$ is the same as saying $\displaystyle (m,n) \in R_1$, that is, the ordered pair (m,n) is an element of the (set) relation $\displaystyle R_1$. note that the relation is on the real numbers, this means that m and n are real numbers

the relation has been defined for you, you do not have to do that. simply prove it is an equivalence relation in the way Plato said. you do know the definitions of "reflexive", "symmetric" and "transitive", right?

I just dont get that first bit Is R a collection of ordered pairs? If so it is a relation on $\displaystyle \Re$.
Do you know the definition of relation?
A relation on $\displaystyle \Re$ is a subset of $\displaystyle \Re \times \Re$.
Does you R define a relation?

6. yeah pretty sure,

reflexive if for all x in R xRx
Symettric if for all x and y in R , xRy implies yRx
Transitive of for x,y&z in R, xRy and yRz implies xRZ.

i will have a look to see how apply it to this question

7. Originally Posted by Plato
Do you know the definition of relation?
A relation on $\displaystyle \Re$ is a subset of $\displaystyle \Re \times \Re$.
Does you R define a relation?
SO i dont actually need to define anything just go on to show its equivalant

yeah pretty sure,

reflexive if for all x in R xRx
Symettric if for all x and y in R , xRy implies yRx
Transitive of for x,y&z in R, xRy and yRz implies xRZ.

i will have a look to see how apply it to this question
yes, good.

now, remember how to interpret congruences

saying $\displaystyle m - n \equiv 0 \mod{24}$

is the same as saying $\displaystyle 24 \mid (m - n)$

which is the same as saying $\displaystyle m - n = 24k$ for some integer $\displaystyle k$

use any or all of these forms as they are convenient for you

SO i dont actually need to define anything just go on to show its equivalant
correct, they gave you the definition. they told you to define it "as follows.." meaning, they are saying, "this is how we want you to define it"

the definition is done, just state it and continue with the proof

10. cheers think i have got it now.

Let $\displaystyle R$ be the set of real numbers.

Define a relation $\displaystyle R_1$ on set $\displaystyle R$ as follows: .$\displaystyle m\:R_1\:n\text{ whenever }n-m \equiv 0\text{ (mod 24)}$

First off, how do i define this relation?

$\displaystyle n-m \equiv 0 \text{ (mod 24)}$ .means .$\displaystyle n-m\text{ is a multiple of 24.}$

That is: .$\displaystyle n - m \:=\:24a\text{ for some integer }a.$

I then have to prove $\displaystyle R_1$ is an equivalence relation.

You should be able to handle this part now . . .

12. thanks everybody problem solved

13. .