Building a set notation part II

[reenmachine]

Homework Statement

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

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

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!

 

[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:

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

 

[Dick]

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?

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.

 

[HallsofIvy]

Homework Statement

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

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

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!

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.

 

[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.

 

[Dick]

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.

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?

 

[reenmachine]

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

 

[reenmachine]

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?

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!

 

[Fredrik]

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.

Homework Statement

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

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

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!

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?

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?

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.

edit:

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

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.

 

[Dick]

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

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

 

[Fredrik]

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.

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.

 

[reenmachine]

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

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.

 

[Dick]

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.

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.