Question about proof from a guy with a highschool education

[reenmachine]

Ok I think I see where I screwed up.

Even if $\varnothing$ is part of the union of all subsets of $N$ , with suppose $N=\{1,2\}$ , then the set will be $\{1\} \cup \{1,2\} \cup \{2\} \cup \varnothing$ which gives us $\{1,2\}$ anyway.

Basically $\varnothing \cup Z = Z$ so $\varnothing$ plays no role (unless between brackets as a set containing an empty set) in an union.

Is that a correct understanding?

 

[Fredrik]

Yes, that sounds good.

 

[reenmachine]

Yes, that sounds good.

Thank you!

 

[reenmachine]

There's something I'm wondering about a specific notation.

An exercise ask me to write all the elements of the set $\{x \in Z : |x| < 5\}$.

$|x|$ seemed to be a dummy variable representing any elements of the set.So my logic was that this set was the set of all elements of $Z$ that were inferior to $5$.So something like $\{... , -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4\}$.

When I checked the solution , it was $\{-4,-3,-2,-1,0,1,2,3,4\}$.

Why is $|x|< 5$ relevant to negative numbers?

Why isn't it something like $-5 < x < 5$ ?

I encountered another exercise with a similar problem when I checked the solution compared to my answer.

thanks!

 

[micromass]

http://en.wikipedia.org/wiki/Absolute_value

 

[reenmachine]

http://en.wikipedia.org/wiki/Absolute_value

I see , I didn't read the whole thing because there's some concepts that are probably above my head the deeper you go into the page , but from what I understood , basically it's like you draw two lines (left and right) from 0 on the x-plane that are equal to the absolute value , and elements of that set will be on those lines and in Z in that case.

Thank you!

 

[Fredrik]

...basically it's like you draw two lines (left and right) from 0 on the x-plane that are equal to the absolute value , and elements of that set will be on those lines and in Z in that case.

I'm sorry, but that didn't make much sense.

For all $x\in\mathbb R$, the absolute value of $x$, denoted by $\left|x\right|$, is defined by
$$\left|x\right|=\begin{cases}x &\text{ if }x\geq 0\\ -x &\text{ if }x<0.\end{cases}$$ This is equivalent to saying that $\left|x\right|=\sqrt{x^2}$.

If you prefer to think of it in terms of "the real line", then |x| is the distance between 0 and x. (Distances are never negative). This should not be thought of as a definition. It's just a way to think about absolute values that's good enough to not cause many problems.

Dumbed down: The absolute value removes the minus sign, if there is one. For example, we have |5|=|-5|=5.

 

[reenmachine]

I'm sorry, but that didn't make much sense.

For all $x\in\mathbb R$, the absolute value of $x$, denoted by $\left|x\right|$, is defined by
$$\left|x\right|=\begin{cases}x &\text{ if }x\geq 0\\ -x &\text{ if }x<0.\end{cases}$$ This is equivalent to saying that $\left|x\right|=\sqrt{x^2}$.

If you prefer to think of it in terms of "the real line", then |x| is the distance between 0 and x. (Distances are never negative). This should not be thought of as a definition. It's just a way to think about absolute values that's good enough to not cause many problems.

Dumbed down: The absolute value removes the minus sign, if there is one. For example, we have |5|=|-5|=5.

Yeah I tried to make the distance analogy but didn't do that good of a job expressing myself very clearly

thank you!

 

[CompuChip]

An exercise ask me to write all the elements of the set $\{x \in \mathbb{Z} : |x| < 5\}$.

$|x|$ seemed to be a dummy variable representing any elements of the set.

Actually the dummy variable is x, as you know since we have discussed the formal notation extensively a while back.

(Have a $\mathbb{Z}$ or $\mathbf{Z}$ by the way, so we won't have to guess whether Z really means the same).

 

[reenmachine]

Actually the dummy variable is x, as you know since we have discussed the formal notation extensively a while back.

Yes this is what I meant.Thank you!

(Have a $\mathbb{Z}$ or $\mathbf{Z}$ by the way, so we won't have to guess whether Z really means the same).

Sorry about that , I will try to use those from now on.

 

[reenmachine]

Maybe I'm just having a good day but I just quickly re-read the whole first chapter of the book of proof and I feel like I am ready to move on to the next one.The material seemed clearer than ever.I did many exercises and the only ones that I'm having some problems with are the one involving sin and cos in set notations but since I know I'll cover some trigonometry in school very soon I'm not worried about that (I already did some trigonometry way back but it was the very basics).That's not to say I'm above making many blunders though

Thanks a lot guys.

 

[reenmachine]

I'm reading the logic chapter without much difficulty (at least for now) , but I still would like to see how good my understanding is.

Is there a better way to write truth table than using code and trying to be symmetric on the forum?

 

[Fredrik]

I don't know really. I'm sure you can do something with the table and tabular environments, I don't know how to use them exactly. Edit: The array environment is very good for this. See micromass' post here.

A simple option is to use this site, where you can e.g. write a & (a > b) to get the truth table for $a\land(a\to b)$. Unfortunately, the result isn't in the form of LaTeX or an image you can link to. So you will have to link to the site and post the input.

Of course, that site is probably a more reliable way to check your results than to ask me, so you may not have to post your attempts here. You can just use the generator to find the correct answer.

 

[reenmachine]

I don't know really. I'm sure you can do something with the table and tabular environments, I don't know how to use them exactly.

A simple option is to use this site, where you can e.g. write a & (a > b) to get the truth table for $a\land(a\to b)$. Unfortunately, the result isn't in the form of LaTeX or an image you can link to. So you will have to link to the site and post the input.

Of course, that site is probably a more reliable way to check your results than to ask me, so you may not have to post your attempts here. You can just use the generator to find the correct answer.

thanks a lot! I will try it and finish the chapter on logic and see if I have some conceptual questions :)

 

[reenmachine]

In the book of proof , they say that I should internalized and mesmorized the truth tables.How true is that? I guess internalizing them means to understand their logic and be able to quickly do it in your head while mesmorizing them would mean just knowing them even if you don't necessarily think about the logic while recalling it.Of course I agree that internalizing them is important , but mesmorizing them? Isn't this the kind of thing that will just happen out of nowhere eventually after playing with truth tables for long enough?

 

[Fredrik]

Do you mean "memorize"? The only other similar word I know is "mesmerize", which means to hypnotize.

You should definitely commit the basic ones to memory and make sure that you understand why they're defined the way they are, but how you do it is up to you. There aren't that many details that you have to force yourself to remember. It's just stuff like:

$p\lor q$ is true when p and q are both true.
$p\Rightarrow q$ is always true when p is false.

That's pretty much it. Everything else in the most basic truth tables is obvious.

 

[reenmachine]

Do you mean "memorize"? The only other similar word I know is "mesmerize", which means to hypnotize.

You should definitely commit the basic ones to memory and make sure that you understand why they're defined the way they are, but how you do it is up to you. There aren't that many details that you have to force yourself to remember. It's just stuff like:

$p\lor q$ is true when p and q are both true.
$p\Rightarrow q$ is always true when p is false.

That's pretty much it. Everything else in the most basic truth tables is obvious.

Thank you! Yeah , I'm comfortable with the two statements above.Of course in the beginning I got mixed up like most people but once you try a couple of real statements instead of p and q you see why it's like that pretty quickly.

Sorry about the typo , thought it needed an "s".

 

[reenmachine]

Here's a little something they ask me to do in the logic section.

Write the following as an english sentence and say if the statement is true:

$\forall n \in N , \exists X \in p(N) , |X| < n$

For all natural number n , there exist a subset X of N such that |X|< n.

The statement is true because of the subset $\varnothing$

$\forall n \in Z , \exists m \in Z , m = n + 5$

For all integer n , there exist an integer m such that m = n + 5

The statement is clearly true because every integer + 5 results in another integer.

 

[Fredrik]

Both answers look good to me.

You should post questions like these (questions about textbook-style problems) in the homework section.

 

[reenmachine]

Both answers look good to me.

You should post questions like these (questions about textbook-style problems) in the homework section.

oops completely forgot in the moment , I will.

 

[reenmachine]

I think this is conceptual , so I will post it here.

I'm thinking about the difference between $\exists \forall$ and $\forall \exists$.

I just wanted to know if I'm on the right track.

With $\exists \forall$ , you could say $\exists y \in Z$ such that $\forall x \in Z$ , $y - 1 = x$.That would mean that a single $y \in Z$ minus one would result in all $x \in Z$ , which is obviously false.

With $\forall \exists$ , you could say $\forall x \in Z$ , $\exists y \in Z$ such that $y - 1 = x$.This means that for all the $x \in Z$ , there is a $y \in Z$ that equals x if you substract one to it.This is true.

thanks!

 

[CompuChip]

Correct!
Usually \exists\forall describes a special element of your set, e.g. a more sensible use would be something like:
\exists y \in \mathbb{Z} \text{ such that } \forall x \in \mathbb{Z}, x + y = x
(This statement asserts the existence of an additive identity, otherwise known as 0).

 

[reenmachine]

Correct!
Usually \exists\forall describes a special element of your set, e.g. a more sensible use would be something like:
\exists y \in \mathbb{Z} \text{ such that } \forall x \in \mathbb{Z}, x + y = x
(This statement asserts the existence of an additive identity, otherwise known as 0).

Thank you!

 

[CompuChip]

And as a practical example of the other possibility, consider

\forall y \in \mathbb{Z}, \exists x \in \mathbb{Z}, x + y = 0

which claims the existence of an additive inverse to all y.

 

[reenmachine]

And as a practical example of the other possibility, consider

\forall y \in \mathbb{Z}, \exists x \in \mathbb{Z}, x + y = 0

which claims the existence of an additive inverse to all y.

Okay so to demonstrate that there's a -1 for 1 and etc...

thank you!

 

[CompuChip]

Exactly, but without fixing the notation.
As long as we're talking about the integers, -1, -2, ... is great notation but it doesn't always work.
For example, the statement also holds in the set { 0, 1, 2 } with addition defined modulo 3. It does not contain any negative numbers, but still 0 + 0 = 0, 1 + 2 = 0 and 2 + 1 = 0.

So the statement really says something about (an operation on) our set. I stated it here for $\mathbb{Z}$, but I can state it for any set and then it may be an axiom, or it may be provable from axioms, or it may be false.

 

[reenmachine]

For example, the statement also holds in the set { 0, 1, 2 } with addition defined modulo 3. It does not contain any negative numbers, but still 0 + 0 = 0, 1 + 2 = 0 and 2 + 1 = 0.

I don't understand what you're saying :(

Why 1+2=0 and 2+1=0 in this set?

 

[Fredrik]

I don't understand what you're saying :(

Why 1+2=0 and 2+1=0 in this set?

He's talking about letting the + symbol denote "addition modulo 3" instead of the usual addition operation. The dumbed down explanation of the definition is that when we "should" get to 3, we start over with 0 instead. So e.g. 2+1=0 and 2+2=1.

 

[reenmachine]

He's talking about letting the + symbol denote "addition modulo 3" instead of the usual addition operation. The dumbed down explanation of the definition is that when we "should" get to 3, we start over with 0 instead. So e.g. 2+1=0 and 2+2=2.

I'm not sure I understand , why isn't it 2+2 = 1?

If you restart at 0 when you reach 3 , then why isn't 4 = 1 and 5 = 2 and 6 = 0 and so on?

 

[Fredrik]

I'm not sure I understand , why isn't it 2+2 = 1?

If you restart at 0 when you reach 3 , then why isn't 4 = 1 and 5 = 2 and 6 = 0 and so on?

It is. I typed that wrong. I fixed it in an edit after you posted this, but somehow I didn't see that you had made this post.