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An equivalence relation T on $\mathbb{N}$ is defined for all $x,y\in\mathbb{N}$ by

$$xRy \to x=2^ny\;\;\;\text{or}\;\;\; y=2^nx \;\;\; \text{for some non-nagtive integer}\;\;\; n$$

Write down the equivalence class [1] using any set notation.

MMy attempt

Since $1R\frac{1}{2^n}$, but $y$ must be a natural number so we can rule this out.

Since $1R2^n$, by $y=2^nx$, the answer must be

$$[1]=\{2^n, \;\;\;\text{for some non-negative integer}\;\;\; n\}$$

However, my answer shows $$[1]=\{2^n|n\in\mathbb{Z}, n\geq 0\}$$

I would like to clarify, do my answer mean the same thing? I was thinking more of, since the $n$ is fixed at the beginning by the word "for some", and hence, there is only 1 element in the equivalence class.

But it seems the answer is saying that there can be infinitely many elements, like $(2,4,8,16,...)$

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Is this a question of wording? By putting the "$\text{for some non-negative integer }n$" inside the brackets, you are indicating that "for some" should be interpreted as "for all," effectively meaning that each non-negative integer $n$ gives us one element of $[1]$. This is the correct answer. If the statement was outside of the brackets, you would be saying that there is only one element, which would be wrong. –  rayradjr Nov 22 '12 at 20:53
    
I am explicity saying there is only 1 element. –  Yellow Skies Nov 22 '12 at 20:58
    
@SingaporeanDude: If you say there is only one element, you should specify which one that is (the notaion $[1]$ should designate one well defined set). As you did not do this, most people would read your expression as a clumsy and not entirely correct way of expressing what the second expression says more clearly. If that is not what you meant, then what you did mean makes no sense. –  Marc van Leeuwen Nov 24 '12 at 11:07

1 Answer 1

up vote 2 down vote accepted

The way the equivalence relation is defined, given natural numbers $x$ and $y$, they are equivalent if and only if such an $n$ exists, but $n$ can vary between pairs in the same equivalence class.

The two answers you gave are not equivalent. Or rather I should say that the first answer is not well defined. The second one is correct.

Did this answer your questions?

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Yeah thanks!!!!! This part helped! "n can vary between pairs in the same equivalence class." –  Yellow Skies Nov 22 '12 at 20:59
    
Hey espen I'm sorry one last question. I realized that according to the answer's definition of the equivalence classes, if I do the same for [2], i will get $[2]={2^n*2|n\in Z,z \geq 0}$ Dosen't that means [1] contains the element 2, and [2] also contains the element 2, and hence they are not equivalence classes? –  Yellow Skies Nov 22 '12 at 21:11
1  
No, $[1]=[2]$. By the definition of the eq. rel., $[2]=\{2\cdot 2^{n} | n\in \mathbb{Z}, n\geq -1\}$. Remember, $xRy$ iff $x=2^n y$ or $y=2^n x$ for some $n$. If two equivalence classes share an element, they are equal. This is true for any equivalence relation. –  espen180 Nov 22 '12 at 21:21

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