# Homework Help: Building a set notation part II

1. May 9, 2013

### reenmachine

1. The problem statement, all variables and given/known data

As an exercise , the book I'm reading ask me to build a set notation for the following set:

$\{... \ , \frac{1}{27} \ , \frac{1}{9} \ , \frac{1}{3} \ , 1 \ , 3 \ , 9 \ , 27 , \ ...\}$

3. The attempt at a solution

After playing with the numbers a couple of minutes , I came with this result:

$\{ x \in R : \exists y \in N \ \ 3^y = x \ \ \ 1/(3^y) = x\}$

Here I'm wondering if the right side is correct.The reason for my doubts is the fact I used two properties of x instead of one.

thanks!

Last edited: May 9, 2013
2. May 9, 2013

### reenmachine

Attempt at describing the set using another road:

$\{ x \in \{3^y\}\cup\{1/(3^y)\} : y \in N \}$

Is this accepted? Or is making a statement about y on the right side instead of x disqualifies this method?

edit:

$\{ x \in R : \exists y \in N \ \ x \in \{3^y\}\cup\{1/(3^y)\}\}$

Last edited: May 9, 2013
3. May 9, 2013

### Dick

1 is also supposed to be an element of your set. Is it? And you still aren't describing it as directly as you could.

4. May 9, 2013

### HallsofIvy

Almost! The (modern) standard definition of "N" is that it is the set of positive integers and so does not include 0 which means your set does not include $$3^0= 1$$.

Think about using "I" (the set of all integers) instead of N.

5. May 9, 2013

### reenmachine

oops , forget about it , brain cramp

I was unaware that $3^0 = 1$ to be honest.

I thought it gave us 0 and that $1/(3^y) = 1$ if $y=0$.

But I guess the fact that I thought it would give us 0 was not good since 0 isn't part of the set.

Last edited: May 9, 2013
6. May 9, 2013

### Dick

Once you've got the powers straightened out, why don't you try to use Z instead of N, like in the last post. Remember 3^(-k)=1/3^k?

7. May 9, 2013

### reenmachine

$\{ x \in R : \exists y \in Z \ \ 3^y=x \}$

Last edited: May 9, 2013
8. May 9, 2013

### reenmachine

Yeah this is what I did but I had to refresh my mind about negative exponants.This made me see the problem more complicated than it was.

thanks man!

9. May 9, 2013

### Fredrik

Staff Emeritus
Edit: I clicked the quote button and then left the computer for a while. Didn't see that a lot had happened since then.

Edit 2: Oh, and I also overlooked the issue with 0 that HallsofIvy mentioned.

If you want to use two properties P(x) and Q(x), you need to make it clear if you mean "P(x) and Q(x)", "P(x) or Q(x)", or something else.

In this case, you could type \text{ or }. But a better approach is to replace $\mathbb N$ with $\mathbb Z$. You know that $x^{-y}=1/x^y$ for all $x,y\in\mathbb R$, right?

When you read it as "the set of all x in $\{3^y\}\cup\{1/(3^y)\}$ such that..." you should see that this only makes sense if y is some specific number defined earlier. In that case, your notation defines a subset of a set with only two elements.

This is logically correct, because $x \in \{3^y\}\cup\{1/(3^y)\}$ means $$x=3^y\text{ or }x=1/3^y,$$ but the notation is kind of ugly.

Last edited: May 9, 2013
10. May 9, 2013

### Dick

Better, but still overcomplicated. What's wrong with $\{3^y : y \in Z \}$?

11. May 9, 2013

### Fredrik

Staff Emeritus
The book reenmachine has been using defines $\mathbb N$ that way, but I don't think there's a standard. I know I prefer to include 0.

12. May 9, 2013

### reenmachine

Not sure.

It's just that when I'm saying it verbally , like ''the set of all $3^y$ such that $y \in Z$'' sounds a little bit weirder than ''the set of all $x \in R$ such that $3^y = x$ if there exist a $y \in Z''$.

But $\{3^y : y \in Z \}$ is the shortest answer so you're right.

Last edited: May 9, 2013
13. May 9, 2013

### Dick

Your way isn't wrong and the shortest way isn't necessarily right, but once you get used to it, I think shorter way is easier to read.