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Stream: deprecated: mathematics

Topic: Simplifying Set Expressions

Jacob Zelko (May 25 2023 at 16:54):

Hi folks!

Happily working through How To Prove It and ran into an interesting struggle on simplifying set statements. I was given this problem:

Simplify the statement. Which variables are free and which are bound? If the statement has no free variables, say wheter it is true or false:

$5 \not\in \{x \in \R | 13 - 2x > c\}$

and the following is my attempt:

$5 \not\in \{x \in \R | x < 6 - \frac{c}{2} \}$ and x is a bound variable and c is a free variable

Does that look like I simplified it about right? The part that I am still a bit confused about is the part of $x \in \R$ and whether I can simplify that further. Otherwise I think that is all I can do from what I can tell.

Thanks!

Jacob Zelko (May 25 2023 at 16:56):

P.S. I meant to title this post “Simplifying Set Expressions” but am unsure how to fix the post title. :frown:

EDIT: Figured it out!

Jean-Baptiste Vienney (May 25 2023 at 17:02):

Jacob Zelko said:

and the following is my attempt:

$5 \not\in \{x \in \R | x < 6 - \frac{c}{2} \}$ and x is a bound variable and c is a free variable

Does that look like I simplified it about right? The part that I am still a bit confused about is the part of $x \in \R$ and whether I can simplify that further. Otherwise I think that is all I can do from what I can tell.

Thanks!

It's almost all good except for the $6$, how did you get that $6$?

Jacob Zelko (May 25 2023 at 17:05):

Additionally, another problem that I ran into which I found very challenging was this one (it has the same requirements in trying to determine an answer):

$4 \in \{x \in \{y | \text{y is a prime number} \} | 13 - 2x > 1 \}$

I took a few steps to figure this one out as best as possible. What I first did is say that $P(x)$ represents the statement x is a prime number. Then I simplified this first to $4 \in \{P(x) | 13 -2x > 1\}$ which I then finally simplified to $4 \in \{P(x) | x < 6\}$ and said that the statement is false and that $x$ and $y$ are bound variables. Does that seem to make sense? My question more was about if I did my simplification step of using a statement correctly for the statement about primes.

Jacob Zelko (May 25 2023 at 17:07):

@Jean-Baptiste Vienney — sorry about that, just saw your response while I was still typing up my challenges! Let me get you that piece of work. Hang on just a moment…

Jacob Zelko (May 25 2023 at 17:08):

AH you’re right I’m crazy. I did a small arithmetic error. That part should be $\frac{13}{2}$

Jacob Zelko (May 25 2023 at 17:09):

Not $6$

Jean-Baptiste Vienney (May 25 2023 at 17:15):

Jacob Zelko said:

Additionally, another problem that I ran into which I found very challenging was this one (it has the same requirements in trying to determine an answer):

$4 \in \{x \in \{y | \text{y is a prime number} \} | 13 - 2x > 1 \}$

I took a few steps to figure this one out as best as possible. What I first did is say that $P(x)$ represents the statement x is a prime number. Then I simplified this first to $4 \in \{P(x) | 13 -2x > 1\}$

There is a small issue here (although the general idea is good), does it make sense to say that $5 \in \{Q(x)\}$ where $Q(x)$ is "x > 4"
for instance?

Ralph Sarkis (May 25 2023 at 17:19):

Jacob Zelko said:

Simplify the statement. Which variables are free and which are bound? If the statement has no free variables, say wheter it is true or false:

$5 \not\in \{x \in \R | 13 - 2x > c\}$

I am not sure what the book is trying to get you to do, but if someone asked me to simplify this, I would write $-7 \leq c$, and I couldn't say whether it is true or false because that depends ont the value of $c$ which is free as you said. I don't know if your attempt counts as a simplified expression.

For the other one :

$4 \in \{x \in \{y | \text{y is a prime number} \} | 13 - 2x > 1 \}$

I would say "$4$ is a prime number and $5 > 1$" and this is false. because $4$ is not a prime number.

Additionally, writing $\{P(x) \mid x < 6\}$ is not great use of the set-builder notation imo. It is understandable, but usually the left part of set builder is only to say which set we are starting from and what notation will we use for elements of that set. I would prefer something like: Let $\mathbb{P}$ be the set of prime numbers, then consider the set $\{x \in \mathbb{P} \mid x< 6\}$.

Jacob Zelko (May 25 2023 at 17:20):

Hm. I see what you mean @Jean-Baptiste Vienney . That doesn't make sense because that would just be the expression $x > 4$ as an element in some set. Would it make more sense then to say $\{x \in Q(x)\}$ using your definition of $Q$? I would read the latter formulation I presented as the set of all $x$'s greater than $4$ and that $5 \in \{x \in Q(x) \}$ is true.

Jacob Zelko (May 25 2023 at 17:24):

Oh @Ralph Sarkis this is super helpful! Yea, the textbook confused me on what exactly the left and right parts of set builder notation precisely are used for. They may’ve been abusing the notation slightly in an effort to provide greater challenge but i really like the presentation you laid out there.

Jean-Baptiste Vienney (May 25 2023 at 17:24):

Jacob Zelko said:

Hm. I see what you mean Jean-Baptiste Vienney . That doesn't make sense because that would just be the expression $x > 4$ as an element in some set. Would it make more sense then to say $\{x \in Q(x)\}$ using your definition of $Q$? I would read the latter formulation I presented as the set of all $x$'s greater than $4$ and that $5 \in \{x \in Q(x) \}$ is true.

Yeah, that's exactly the problem. The usual notation is $\{x | Q(x)\}$, so you can say that $5 \in \{x| Q(x)\}$ is true.

Jean-Baptiste Vienney (May 25 2023 at 17:25):

or better that $5 \in \{x \in \mathbb{R}|Q(x)\}$ is true.

Jacob Zelko (May 25 2023 at 17:28):

I’m interested to know how you got to $-7 \leq c$ in your formulation of the element hood test in the first problem I showed @Ralph Sarkis . How did you get to that?

Ralph Sarkis (May 25 2023 at 17:38):

I am unrolling all symbols one by one to get something that I find as simple as I want. We start with the statement
$5 \not\in \{x \in \R | 13 - 2x > c\}$
This is the same thing as (by looking at the definition of $\not\in$:
$\mathsf{NOT} \left(5 \in \{x \in \R | 13 - 2x > c\}\right)$
I can simplify what is inside the NOT by replacing $x$ by $5$ in the right part (because that is what set-builder notation means):
$\mathsf{NOT} \left( 13 - 2\cdot 5 > c\right)$
A bit of algebra gives me:
$\mathsf{NOT} \left( 3 > c\right)$
Finally, I put the NOT back insider: (EDIT: I fixed my arithmetic mistake noted by Jean-Baptiste)
$3 \leq c$

Jean-Baptiste Vienney (May 25 2023 at 17:42):

I think you wanted to write $\mathsf{NOT} \left(3 > c\right)$

Jacob Zelko (May 25 2023 at 17:52):

Oh wow. There were a few examples on how to do what you just did but I was struggling to follow them. This was super clarifying. Thanks @Ralph Sarkis. Imma go back to my problems and retry this simplification process Better now that it is more clear to me

Jacob Zelko (May 25 2023 at 17:52):

Thanks for the help you both!!! :clap: