What is the equivalence class [3] in a relation defined by powers of 2?

[knowLittle]

Homework Statement

Let $H = \{ 2^{m} : m \in Z\}$
A relation R defined in $Q^{+}$ by $aRb$, if $\frac{a}{b} \in H$

a.) Show that R is an equivalence Relation

b.) Describe the elements in the equivalence class [3].

The Attempt at a Solution

For part a, I think I am able to solve it, tell me what you think of my solution:
Assume:
$2^{m} R 2^{m} , m \in Z$
Since, $2^{0} \equiv 1 \equiv \frac{2^{m} }{2^{m} } \in H$
So, R is reflexive, for any m in Z.

Assume:
$2^{m} R 2^{n}$, then $\frac{ 2^{m}} { 2^{m} }, m , n \in Z$
If $m<0 \frac{1}{2^{m} 2^{n}}$ , then it still satisfies $2^{0} , 2^{m+n} \in H$, since m+n is in Z as well.
Without loss of generality for n<0 or both m and n <0.
Thus, $2^{n} R 2^{m}$ and R is symmetric.

Assume:
$2^{m} R 2^{n}$ and $2^{n} R 2^{d}$, since n, m, and d are in Z , they are transitive.

I have problems in part b.
What would an equivalence class [3] mean in this powers of 2 relation?

 

[PeroK]

You've gone about this the wrong way. You must prove R is an equivalence relation on Q+, not on H.

So, you need to show the properties hold for elements of Q+.

 

[pasmith]

I have problems in part b.
What would an equivalence class [3] mean in this powers of 2 relation?

3 is a positive rational number. [3] is the equivalence class of 3, ie. the set of all positive rational $a$ such that $aR3$.

 

[knowLittle]

3 is a positive rational number. [3] is the equivalence class of 3, ie. the set of all positive rational $a$ such that $aR3$.

I know that a/b or 3/1 is in $Q^{+}$.
Would be multiples of 3 or numbers that can be divided and reduced into 3/1?

 

[knowLittle]

You've gone about this the wrong way. You must prove R is an equivalence relation on Q+, not on H.

So, you need to show the properties hold for elements of Q+.

So, the purpose of mentioning H is to say that whatever a/b is in the form $2^{m}, m \in Z$?
It will always be in Q+.
Could you give one or two examples of elements in this relation class?

 

[PeroK]

So, the purpose of mentioning H is to say that whatever a/b is in the form $2^{m}, m \in Z$?
It will always be in Q+.
Could you give one or two examples of elements in this relation class?

You've still got things the wrong way round. You would start, for symmetry say:

Let a, b be in Q+ and aRb, then a/b is in H, so $∃ m \in Z \ s.t. \ a/b = 2^{m}$...

 

[knowLittle]

You've still got things the wrong way round. You would start, for symmetry say:

Let a, b be in Q+ and aRb, then a/b is in H, so $∃ m \in Z \ s.t. \ a/b = 2^{m}$...

This is what I don't understand, what if a/b =7. This cannot be explained by $2^{m}$

 

[PeroK]

This is what I don't understand, what if a/b =7. This cannot be explained by $2^{m}$

If a/b = 7 then $a \ notR b$

 

[vela]

This is what I don't understand, what if a/b =7. This cannot be explained by $2^{m}$

In addition to what PeroK said, if a=7 and b=14, neither of which is of the form $2^n$, you'd have aRb because a/b = 1/2 = 2^-1.

 

[pasmith]

I know that a/b or 3/1 is in $Q^{+}$.
Would be multiples of 3 or numbers that can be divided and reduced into 3/1?

In the definition of $aRb$, both $a$ and $b$ are strictly positive rationals, and $aRb$ if and only if there exists $m \in \mathbb{Z}$ such that $ab^{-1} \in H = \{2^m : m \in \mathbb{Z}\}$.

Thus, for example, 4/7 and 5/41 are not equivalent, because
$$\frac 47 \div \frac 5{41} = \frac{164}{35} \notin H.$$
On the other hand, 5/8 and 5/16 are equivalent, because
$$\frac 58 \div \frac {5}{16} = 2 \in H.$$

 

[knowLittle]

So, to correct my 'proof' as Perok mentioned:

Homework Statement

Let $H = \{ 2^{m} : m \in Z\}$
A relation R defined in $Q^{+}$ by $aRb$, if $\frac{a}{b} \in H$

a.) Show that R is an equivalence Relation

b.) Describe the elements in the equivalence class [3].

The Attempt at a Solution

For part a, I think I am able to solve it, tell me what you think of my solution:
Assume:
$2^{m} R 2^{m} , m \in Z$
Since, $2^{0} \equiv 1 \equiv \frac{2^{m} }{2^{m} } \in Q^{+}$
So, R is reflexive, for any m in Z.
Is this correct?

Assume:
$2^{m} R 2^{n}$, then $\frac{ 2^{m}} { 2^{n} }, m , n \in Z$ and the fractions are in Q+.
If $2^{m} R 2^{n}$, then $\frac{ 2^{m}} { 2^{n} }$ is in Q+ and m,n in Z.
Assume $\frac{ 2^{n} }{ 2^{ m} }$, then $2^{n}R 2^{m}$ and R is symmetric.
Thus, $2^{n} R 2^{m}$ and R is symmetric.

Assume:
$2^{m} R 2^{n}$ and $2^{n} R 2^{d}$, since n, m, and d are in Z and
$\frac{ 2^{m-n}}{1}$ and $\frac{ 2^{n-d}}{1} ,$ are in $Q^{+}$ and $m-n, n-d \in Z$,
then $\frac{2^{m}}{2^{n} } \times \frac{ 2^{n} }{2^{d}} = \frac{2^{m}}{2^{d} } \in Q+$, then
they are transitive.
Is this correct?

 

[LCKurtz]

Well, that's a new one. You have made it impossible to quote your revised post. How the hell did you do that?

In any case, it is still messed up, even on reflexive. You have to start with $a\in Q^+$. You aren't given that $a$ is some power of $2$.

If $a\in Q^+$ please state what you have to prove to show $aRa$. Do it in response to this post, not by editing something previous.

 

[knowLittle]

Well, that's a new one. You have made it impossible to quote your revised post. How the hell did you do that?

In any case, it is still messed up, even on reflexive. You have to start with $a\in Q^+$. You aren't given that $a$ is some power of $2$.

If $a\in Q^+$ please state what you have to prove to show $aRa$. Do it in response to this post, not by editing something previous.

Reflexive:
Assume $a \in Q^+$.
If $a \in Q^+$ and $\frac{a}{a} \in \{2^{m}: m\in Z\}$, then $aRa$.

Since, $\frac{a}{a} \equiv 1 \equiv {2^0}, 0 \in Z$, then $aRa$.

Is this right?

 

[PeroK]

Yes, you're getting there! But, let me tidy things up for you, as you're still thinking a little back to front. And, there's no need for equivalent signs rather than equal signs:

$$Let \ a \in Q^+$$
$$\frac{a}{a} = 1 = 2^0 \in H \ as\ 0 \in \mathbb{Z}$$
$$∴ \ aRa$$
$$∴ \ R \ is \ reflexive$$

 

[knowLittle]

Thanks, Perok and LCKurtz.
Now, I want to prove symmetry.
Let $a,b \in Q^+$

$\frac{a}{b} = 2^m \in H; \frac{b}{a} = 2^n \in H$ for some m,n $\in \mathbb{Z}$

Then, $aRb, bRa$

and R is symmetric.

Is this correct?

 

[PeroK]

I'm afraid not, I'm sorry to say. To prove symmetry you need to show that:

For a, b in Q+, aRb => bRa

In other words, if aRb, then bRa.

This stuff's hard until you get your head round what the hell is going on!

 

[knowLittle]

Ok.
$a,b \in Q^+$

If $aRb , \frac{a}{b} \in H$ for some $2^m , m \in Z$

or $\frac{a}{b} = 2^m$

Solve for b/a:

$\frac{1}{2^m} = \frac{2^0}{2^m} = 2^{-m} = \frac{b}{a}$

Then, $\frac{b}{a} = 2^{-m} \in H$

So, $aRb$ implies $bRa$

Correct

 

[PeroK]

Yes, you've got it!