Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was given this question (there were prior questions defining the equivelence relation ~)

let E be the set of equivelence classes of ~
 define f from E to Reals
    f:={([(x,y)],r)|x=yr,\: x,y,r in Reals}

I'm trying to understand what it means; is the following correct?

f:[(x,y)] mapsto r

E=(Reals X {Reals/0})/~ 

ie the Quitient set of (Reals X {Reals/0}) by ~ ((x,y) was ealier defined as an element of (Reals X {Reals/0}) ))

Am I understanding this right?

My function takes in a equivlence class, [(x,y)] and reaturns a real value r, furfilling the condition that x=yr?

share|improve this question
    
You may want to check your spelling! But I think your basic understanding is more-or-less okay - you'll need to do some work to show that f is well-defined and that depends on the definition of ~. –  yatima2975 Aug 26 '10 at 14:35
add comment

2 Answers 2

up vote 2 down vote accepted

Yes; the function is defined to take equivalence classes as inputs, and return reals as values. If you put in the class of $(x,y)$, which is $[(x,y)] = \{(a,b)\in \mathbb{R}\times\mathbb{R}^{\star}\,|\, (x,y)\sim (a,b)\}$, it returns a real number $r$. Which real number? The real number $r$ such that $x=yr$. (I'm using $\mathbb{R}^{\star}$ to denote the nonzero real numbers) The function takes equivalence classes as inputs, and has real numbers as outputs.

You would need to make sure that this function is well defined, though; that is, the definition of the function depends on the name you give the equivalence class; but the class $[(x,y)]$ has many different "names"; it can also be called $[(a,b)]$ for some other real numbers $a$ and $b$. So you need to make sure that if $[(x,y)]=[(a,b)]$ (that is, if $(x,y)\sim (a,b)$) then the real number $r$ such that $x=yr$ is the same as the real number $s$ such that $a=bs$. Otherwise, what you describe would not be a function.

share|improve this answer
    
Good. this is what i've done –  oxinabox Aug 27 '10 at 0:34
add comment

That looks right to me. Recall that a function $f : X \to Y$ is formally a subset of $X \times Y$, i.e. a set of ordered pairs $(x,y), x \in X, y \in Y$, with the property that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$.

Here $X$ is the set $E$ of equivalence classes, and $Y = \mathbb{R}$. You are told that $f$ is the set of ordered pairs $([(x,y)],r)$ for which $x = yr$. One has to check that this is well-defined with respect to the equivalence relation ~: if $x=yr$ and $(x,y)~(x',y')$, then $x'=y'r$. Then $f$ makes sense as a set. To check that it is a function, one should check that if $([(x,y)], r) \in f$ and $([(x,y)], r') \in f$ (i.e. $x=yr$ and $x=yr'$) then $r=r'$ (which should follow since $y \ne 0$ by assumption).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.