Question about proof from a guy with a highschool education

[reenmachine]

{ and } are reserved symbols in LaTeX. If you want to display them, you must type \{ and \}. If you want text to be interpreted as text and not as a product of variables, you must write \text{something}. LaTeX Guide. Yes, the first few days of using LaTeX, you may spend as much time on how to write things as on the math, but it's just the first few days.

A

3. That's right.
7. The equation has two solutions in ℝ, and 3 isn't one of them.
8. The equation has three solution in ℝ, and -2 is just one of them.
9. What you need to know is that $\sin 0=0$ and that $\sin(t+2\pi)=\sin t$ for all $t\in\mathbb R$.
15. You seem to have forgotten how to read the notation $\{x\,|\,\varphi(x)\}$. (The author writes a colon where I write a vertical bar). It's read as "the set of all x with property $\varphi$". Can you translate $\{5a+2b\,|\,a,b\in\mathbb Z\}$ to words in this way?

7.Meant -3.I have no clue how to find the other one.Seems I suffer from memory loss , how do you deal with variables with exposants in algebra to find the value of the variable? This is highly embarrassing :shy:

8.same question as 7.

9.What does t mean? why 2(3.14)?

15.Not sure how to do it.A try: {x : ''there exist'' a,b in Z such that 5a + 2b = x } ?

 

[reenmachine]

Hint: 2^n.

Hint: 2^n.

By the way, I think it's slightly more popular these days to define the "natural numbers" as including 0. This author doesn't, and that's fine too. You just need to know that different authors have different conventions.C

31. The question is asking you how many elements the set {{{1},{2,{3,4}},∅}} has.

Do you mean I can use formulas to the nth - level in notations?

And here I am going backward.I'm now confused about what is an element or not.What the hell is going on

In {{{1},{2,{3,4}},∅}} , you could say 1, 2 ,3 ,4 are elements.Unless this is a powerset , which would make {∅} an element and many more.The multiples brackets are confusing the hell out of me.

This is highly frustrating as it seem I'm regressing instead of progressing today , now I'm confusing subsets and elements again , thought that was behind me

The answer to 31 would be 4?

edit: okay , it doesn't matter how many subsets of a set you have to dig to find an element , if the element is there it's still an element of the big set.

 

[reenmachine]

17 If N={1,2,...}, your answer means {2,4,6,8,10,...}, so this is incorrect. Hint: 2^n.

I'm close to give up

I've tried many things , such as ( x in N : there exist y in N such that x = 2y ) to make sure x is an even integers to operate with them but it still doesn't work out.

EDIT:Think I got it , $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ , is that it?

This would mean if y=3 , then x = 2^3 = 8 , if y=5 , then x = 2^5 = 32

BOOYAH!

sorry my blood pressure is rising , don't want to sound like I'm pissed but math can do that to me

thanks for the patience

 

[Fredrik]

7.Meant -3.I have no clue how to find the other one.Seems I suffer from memory loss , how do you deal with variables with exposants in algebra to find the value of the variable? This is highly embarrassing :shy:

You understand the formula $(a+b)^2=a^2+b^2+2ab$, right? Suppose that you encounter an equation of the form $x^2+ax+b=0$. There's a clever trick involving that formula that gets x alone on one side of the equality sign:
\begin{align}0 &=x^2+ax+b=x^2+\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2+ax+b =\left(x+\frac a 2\right)^2-\left(\frac{a}{2}\right)^2+b\\
\left(x+\frac a 2\right)^2 &=\left(\frac{a}{2}\right)^2-b\\
x+\frac a 2 &=\pm\sqrt{\left(\frac{a}{2}\right)^2-b}.\\
x &=-\frac a 2 \pm\sqrt{\left(\frac{a}{2}\right)^2-b}.
\end{align} You should memorize both this formula and its derivation. Use the formula to find the solutions.

9.What does t mean?

Did you not see that the sentence ended with "for all $t\in\mathbb R$"? I don't mean to add to your frustration, but if you're still wondering about this after taking the whole sentence into account, then you should go back and read what micromass and I said about dummy variables again.

why 2(3.14)?

Because sin is defined to ensure that this picture tells the truth for all $t\in\mathbb R$.

If you're wondering specifically about the value $2\pi$, then check out this picture, and read the Wikipedia article on "radians" if you need an explanation.

15.Not sure how to do it.A try: {x : ''there exist'' a,b in Z such that 5a + 2b = x } ?

It's "the set of all 5a+2b such that a and b are integers". Edit: I guess I should have looked more closely at what you wrote, because it's the same set as mine. There is however a much simpler way to state the result. Can you think of a set that contains all the sums 5a+2b such that a and b are integers, and doesn't contain anything else?

Do you mean I can use formulas to the nth - level in notations?

I meant e.g. that $2^5=2\cdot 2\cdot 2\cdot 2\cdot 2=32$.

And here I am going backward.I'm now confused about what is an element or not.What the hell is going on

In {{{1},{2,{3,4}},∅}} , you could say 1, 2 ,3 ,4 are elements.Unless this is a powerset , which would make {∅} an element and many more.The multiples brackets are confusing the hell out of me.

This is highly frustrating as it seem I'm regressing instead of progressing today , now I'm confusing subsets and elements again , thought that was behind me

edit: okay , it doesn't matter how many subsets of a set you have to dig to find an element , if the element is there it's still an element of the big set.

That's not it. For example, {x} has only one element, which is denoted by x. This is true even if x={y,z}, so that {x}={{y,z}}. A trick you could use here is to replace matching pairs of curly brackets and the stuff between them with a letter, like x, and then work your way out.

{{{1},{2,{3,4}},∅}} Set a={1} and b={3,4}.
{{a,{2,b},∅}} Set c={2,b}.
{{a,c,∅}} Set d={a,c,∅}.
{d}

This set clearly has only one element. Actually to see this, you only had to make the observation that the expression started with {{ and ended with }}. This made everything between irrelevant. Edit: Oops, not really. See post #258 (written by jbriggs444) for an example that shows that what I just said is incorrect.

EDIT:Think I got it , $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ , is that it?

This would mean if y=3 , then x = 2^3 = 8 , if y=5 , then x = 2^5 = 32

BOOYAH!

sorry my blood pressure is rising , don't want to sound like I'm pissed but math can do that to me

Yes, that's correct.

There is however a simpler way to write the answer: $\{2^n\,|\,n\in\mathbb N\}$.

 

[reenmachine]

That's not it. For example, {x} has only one element, which is denoted by x. This is true even if x={y,z}, so that {x}={{y,z}}. A trick you could use here is to replace matching pairs of curly brackets and the stuff between them with a letter, like x, and then work your way out.

{{{1},{2,{3,4}},∅}} Set a={1} and b={3,4}.
{{a,{2,b},∅}} Set c={2,b}.
{{a,c,∅}} Set d={a,c,∅}.
{d}

This set clearly has only one element. Actually to see this, you only had to make the observation that the expression started with {{ and ended with }}. This made everything between irrelevant.

Just to comment right away on that part of the post , the worst thing is I actually wrote this but edited it.I knew I had to eliminate all the brackets that were inside brackets (not counting the brackets immediately between the brackets of the set we're searching the cardinality of) but I didn't do it for some reasons.

The answer 1 was the first that got to my mind so this is good news.I noticed I've been getting a little bit emotionnal the last 2 or 3 days.Hopefully I can learn from these frustrations and rise up to the challenge.

Will get back at you about the other part of your post soon enough.

thanks a lot

 

[micromass]

Thanks for the explanation. According to Wikipedia, the axiom of foundation says that every non-empty set x has an element y that's disjoint from x. This clearly rules out x={x}, but it's not obvious how it rules out the existence of disjoint x,y such that x={x,y}.

You should apply the axiom to $\{x\}$. So the axiom tells you that $\{x\}$ has an element disjoint from it. So $x$ and $\{x\}$ are disjoint. But if $x=\{x,y\}$, then $\{x,y\}$ and $\{x\}$ are not disjoint.

Sorry reen, for this off-topic discussion.

 

[Fredrik]

You should apply the axiom to $\{x\}$. So the axiom tells you that $\{x\}$ has an element disjoint from it. So $x$ and $\{x\}$ are disjoint. But if $x=\{x,y\}$, then $\{x,y\}$ and $\{x\}$ are not disjoint.

Ah, of course. So the proof of the proposition that no set in ZFC set theory is an element of itself goes like this: Suppose that there's a set x such that $x\in x$. Then {x} is a set. (There are many ways to see that; one is to use the assumption and the power set axiom). The foundation axiom says that there's a y in {x} that's disjoint from x. Since the only element of {x} is x, this means that x is disjoint from x. This implies that x=∅. This contradicts that $x\in x$.

Sorry reen, for this off-topic discussion.

I suspect that Reenmachine is just relieved to see that I too can fail to find proofs that look pretty simple when you have them in front of you.

 

[jbriggs444]

This set clearly has only one element. Actually to see this, you only had to make the observation that the expression started with {{ and ended with }}. This made everything between irrelevant.

Nit pick.

{{1},{2}} starts with {{ and ends with }} but has two elements.

It is often helpful to decorate expressions like these with white space to make them easier for the reader to parse by eye.

 

[Fredrik]

Agreed. Thanks for the correction. I have added a comment about it to the post where I said that.

 

[reenmachine]

You understand the formula $(a+b)^2=a^2+b^2+2ab$, right?

Well , yeah , I understand it very clearly.

Suppose a=2 and b=3:

The left side is (2+3)(2+3) , the 2×2 and 3×3 on the left side are equal to 2^2 and 3^3 on the right side.Then on the left side you are left with a×b and b×a , which means the same thing , so 2×3 and 3×2 , which is really 2(2×3) denoted by 2ab on the right side.

Suppose that you encounter an equation of the form $x^2+ax+b=0$. There's a clever trick involving that formula that gets x alone on one side of the equality sign:
\begin{align}0 &=x^2+ax+b=x^2+\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2+ax+b =\left(x+\frac a 2\right)^2-\left(\frac{a}{2}\right)^2+b\\
\left(x+\frac a 2\right)^2 &=\left(\frac{a}{2}\right)^2-b\\
x+\frac a 2 &=\pm\sqrt{\left(\frac{a}{2}\right)^2-b}.\\
x &=-\frac a 2 \pm\sqrt{\left(\frac{a}{2}\right)^2-b}.
\end{align} You should memorize both this formula and its derivation. Use the formula to find the solutions.

Wow , that's pretty good.But while I understand it on an operating level (like if x=3 , a=4 and b=-21 for exemple) , I would like to understand a little bit more in depth.

Take the way I presented the formula $(a+b)^2=a^2+b^2+2ab$ at the top of the post , I think I made it clear that I connected all the dots as to why this formula was an equation.With this one , not so much.In fact , I guess the key part where I'm not sure what's happening is when ax disappears and (x+a/2)^2 makes his entrance.

Suppose you take my exemple of x=3 , a=4 and b= -21.

This means that 3^2 + 4(3) -21 = 0.

Then the formula introduces (a/2)^2 , but since they both add and substract it it's not that confusing for the moment and it has no immediate impact on the formula.

It still gives 9 + (4/2)^2 - (4/2)^2 +12 - 21 = 9 + 4 - 4 + 12 - 21 = 0.

Then this is where I'm not sure what is happening ''abstractly''.

You have (3 + 4/2)^2 - (4/2)^2 -21 = 0.What happened to 12? Why does it work? I thought about it for 15-20 minutes and couldn't figure it out.

The end of the formula is pretty clear and I understand WHY you need to introduce some new elements in the formula in order to isolate the x on an operating level.That I have no problem of understand , it's just this specific case which has me scratching my head as to why it all work out.

Just a minor question about the end of the formula , why the ± sign? And when you have the -a/2 ahead of the ± sign , what difference does it make on -a/2 compared to the rest of the square root?

Anyway , with my exemple of x=3 , a=4 and b= -21 it all works out in the end and 3 is indeed isolated.

thanks!


Then you have

 

[Fredrik]

I think it's much harder to see what's going on when you assign values to the variables and work with the numbers instead.

The statement

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

is an easy to prove theorem about real numbers. This theorem implies that for all $a,x\in\mathbb R$, we have
$$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$ What I did was to rewrite $x^2+ax+b$ as a sum of five terms, three of which are the ones on the right-hand side of the equality above. So then I could just apply the result above to those three terms.

This trick is called "completing the square".

What happened to 12?

12 is one of the three terms whose sum you rewrote as (3 + 4/2)^2.

Just a minor question about the end of the formula , why the ± sign?

$x=\pm a$ really means $x\in\{-a,a\}$. When you see ± in a formula, you should think of the formula as representing two different formulas, one with a plus sign and one with a minus sign. Each of the formulas is a solution to the equation.

 

[reenmachine]

I think it's much harder to see what's going on when you assign values to the variables and work with the numbers instead.

The statement

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

is an easy to prove theorem about real numbers. This theorem implies that for all $a,x\in\mathbb R$, we have
$$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$ What I did was to rewrite $x^2+ax+b$ as a sum of five terms, three of which are the ones on the right-hand side of the equality above. So then I could just apply the result above to those three terms.

This trick is called "completing the square".

Sorry I'm a bit confused here , do you mean that the two formulas are related? because I didn't think so.

Also , in $$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$

Shouldn't it be 2ax? If the a/2 is what makes the 2 in 2ax disappear , then why isn't the x divided by 2 also? (I understand in practice it wouldn't make sense if we tried it with numbers but I'm trying to understand the logic).

 

[reenmachine]

I suspect that Reenmachine is just relieved to see that I too can fail to find proofs that look pretty simple when you have them in front of you.

ahahah I don't know if I'm relieved but I feel less lonely

Though I don't mind being the lonely student of many mentors , you certainly won't find that in the school system.Maybe it is my insecurity , but I'm sometimes scared that if I don't ''get it'' fast enough people will get bored so I put pressure on myself to move forward.I think it can be both good and bad , good because I force myself to a higher standard and bad because it can get frustrating when I don't meet that high standard.

In other words , I assume it's funnier to teach someone when you can see his progress (even though I have no clue how to teach so that's just an assumption) , so I want to show progress for the help I'm receiving.

 

[reenmachine]

Sorry reen, for this off-topic discussion.

No problem , I'm honored to have undirectly generated some mathematical discussions

 

[Digitalism]

Shouldn't it be 2ax? If the a/2 is what makes the 2 in 2ax disappear , then why isn't the x divided by 2 also? (I understand in practice it wouldn't make sense if we tried it with numbers but I'm trying to understand the logic).

When you multiply out (x + a/2)^2 you get x^2 + ax/2 + ax/2 + (a/2)^2 the two middle terms add up to ax which shows the equivalence between both sides of the equation you quoted

 

[Fredrik]

Sorry I'm a bit confused here , do you mean that the two formulas are related? because I didn't think so.

Not related? They are essentially the same statement.

Edit: On second thought, I'm not sure what you meant by "the two formulas". The two that I was referring to when I said that they are "essentially the same statement" are the ones in this quote:

The statement

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

is an easy to prove theorem about real numbers. This theorem implies that for all $a,x\in\mathbb R$, we have
$$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$

Edit 2: OK, maybe I see what you meant. When I said "There's a clever trick involving that formula that gets x alone on one side of the equality sign", it wasn't clear enough that what I meant by "that formula" was $(a+b)^2=a^2+b^2+2ab$.

Also , in $$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$

Shouldn't it be 2ax? If the a/2 is what makes the 2 in 2ax disappear , then why isn't the x divided by 2 also? (I understand in practice it wouldn't make sense if we tried it with numbers but I'm trying to understand the logic).

Don't replace the variables with numbers. Just use the theorem

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

to evaluate
$$\left(x+\frac a 2\right)^2,$$
or perhaps even better, forget about the theorem and just use that multiplication is distributive over addition. Don't replace the variables with numbers.

Maybe it is my insecurity , but I'm sometimes scared that if I don't ''get it'' fast enough people will get bored so I put pressure on myself to move forward.I think it can be both good and bad , good because I force myself to a higher standard and bad because it can get frustrating when I don't meet that high standard.

In other words , I assume it's funnier to teach someone when you can see his progress (even though I have no clue how to teach so that's just an assumption) , so I want to show progress for the help I'm receiving.

You clearly have made some progress, and I think that the rate at which you make progress will increase significantly once you have reached the level where you understand the basics. We just have to get you there.

I am getting a little bit tired, but I can certainly handle one more bunch of exercises, like the bunch we're discussing now. After that, I think you should start using the homework forum for exercises. Keep using this thread for discussions about definitions and theorems about sets, relations and functions.

 

[Fredrik]

I should also have said that this is a fun thread and a fun experiment. These are interesting topics, and it's also interesting to me to see how this works out. These things are usually taught to people with a lot more experience of math than you have, so I had no idea if you would find them easy or hard. That's why this is a bit of an "experiment".

I think you would have found it much easier to understand things like set-builder notation, dummy variables, and the formal definitions of relations and functions if you had had more experience with basic algebra, 1st and 2nd degree polynomial equations, the informal description of a function from X into Y as "a rule that associates exactly one element of Y with each element of X", plotting graphs of functions, etc.

I guess there's a reason why these things aren't taught at the high school level. I still think this is doable. You just have to be prepared to do a lot of silly-looking exercises, and maybe also to take some time to refresh your memory about some high school stuff that you don't know as well as you will need to. You will also have to be willing to try and try again when you don't see the solution to a problem. Stubbornness pays off. Think of what you did to find the expression $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ for one of the problems discussed above. When you finally had it in front of you, you knew that it was right before I had confirmed it. This is a good way to learn stuff. The downside is that it takes a lot of time.

 

[reenmachine]

You clearly have made some progress, and I think that the rate at which you make progress will increase significantly once you have reached the level where you understand the basics. We just have to get you there.

I am getting a little bit tired, but I can certainly handle one more bunch of exercises, like the bunch we're discussing now. After that, I think you should start using the homework forum for exercises. Keep using this thread for discussions about definitions and theorems about sets, relations and functions.

I think I've also had a rough week on a mental energy level compared to the two other weeks I've spend in this thread.Do not worry , I won't come here with countless exercises everyday , I might use one here and there just to ensure that I understood correctly by presenting something I've worked on , but I will try to keep this thread for definitions and concepts.

 

[reenmachine]

I should also have said that this is a fun thread and a fun experiment. These are interesting topics, and it's also interesting to me to see how this works out. These things are usually taught to people with a lot more experience of math than you have, so I had no idea if you would find them easy or hard. That's why this is a bit of an "experiment".

This is an experiment for me as well.I'm well aware that I lack some basic stuff that would make my life easier , but at this point I'm in so I have to fight without these basics.If I do end up understanding most of what is in the book of proof , I might become a weird case in the sense that I'll know some deep math all the while lacking some basics which would normally be illogical in a student that followed a normal path.

I think you would have found it much easier to understand things like set-builder notation, dummy variables, and the formal definitions of relations and functions if you had had more experience with basic algebra, 1st and 2nd degree polynomial equations, the informal description of a function from X into Y as "a rule that associates exactly one element of Y with each element of X", plotting graphs of functions, etc.

I guess there's a reason why these things aren't taught at the high school level. I still think this is doable. You just have to be prepared to do a lot of silly-looking exercises, and maybe also to take some time to refresh your memory about some high school stuff that you don't know as well as you will need to. You will also have to be willing to try and try again when you don't see the solution to a problem. Stubbornness pays off. Think of what you did to find the expression $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ for one of the problems discussed above. When you finally had it in front of you, you knew that it was right before I had confirmed it. This is a good way to learn stuff. The downside is that it takes a lot of time.

I do have an history of self-learning something successfully (at least I think) , the english language! Three years ago , I wasn't capable of writing a single sentence nevermind have conversions about advanced mathematics.At the beginning it was painful , it was hard to keep reading and trying to have discussions when I struggled to understand half of what was said , but eventually it got easier and one day without realizing it I could think in english in my head.

Thanks a lot Fredrik , your feedback is always greatly appreciated and can't thank you enough for the time you've spend helping me.I know I'm repeating myself , but I want you to know that I'm sincerely grateful.

 

[reenmachine]

$\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$.

About this , would it be better to write $\{∀x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ ?

 

[micromass]

About this , would it be better to write $\{∀x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ ?

No. The well-formed formula is

$\{x\in A~\vert~\varphi(x)\}$

and not

$\{\forall x\in A~\varphi(x)\}$

That's just notation though. We chose one form above the other because we wanted our notations to be the same in every case.

 

[Fredrik]

Since $\{x\in A\,|\,\varphi(x)\}$ is to be read "the set of all $x\in A$ such that $\varphi(x)$", I see your point, but no one writes a $\forall$ there. As micromass said, this is just a notational convention, so it doesn't require any deeper analysis.

However, if we denote this set by B, then we have
$$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right),$$ where $\Leftrightarrow$ is read as "if and only if", and $\land$ is read as "and". (These logical symbols are defined by truth tables that I assume are included in the logic section of the Book of Proof). This "sentence" can be thought of as the definition of the notation $B=\{x\in A\,|\,\varphi(x)\}$.

The symbol $\leftrightarrow$ is often used instead of $\Leftrightarrow$.

Edit: I actually messed up that sentence in the language of set theory when I posted this, but I have changed it, so it should be OK now.

 

[reenmachine]

Since $\{x\in A\,|\,\varphi(x)\}$ is to be read "the set of all $x\in A$ such that $\varphi(x)$", I see your point, but no one writes a $\forall$ there. As micromass said, this is just a notational convention, so it doesn't require any deeper analysis.

However, if we denote this set by B, then we have
$$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right),$$ where $\Leftrightarrow$ is read as "if and only if", and $\land$ is read as "and". (These logical symbols are defined by truth tables that I assume are included in the logic section of the Book of Proof). This "sentence" can be thought of as the definition of the notation $B=\{x\in A\,|\,\varphi(x)\}$.

The symbol $\leftrightarrow$ is often used instead of $\Leftrightarrow$.

Edit: I actually messed up that sentence in the language of set theory when I posted this, but I have changed it, so it should be OK now.

Extremely clear! Thanks a lot!

Just a minor question , why the () instead of the {}? Or more precisely , why isn't the sentence $$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right),$$ between {}?

 

[Fredrik]

Because what I did there defines the meaning of the curly brackets.

The alphabet of the formal language of set theory includes only a bare minimum of symbols. Typically, we would define it as containing the symbols $\land$, $\lnot$ (read as "and" and "not") and only a few others. Then we use the fundamental symbols to define new ones, to make things easier for ourselves. For example, it would be convenient to have a symbol that we can think of as representing the word "or", so we make the following definition: For arbitrary statements p and q, we define the expression $p\lor q$ to mean $\lnot(\lnot p \land \lnot q)$. Now the symbol $\lor$ can be read as "or".

This will make more sense when you have studied truth tables. The only motivation I can give you right now is that the statement "p or q" is true if and only if it's not true that p and q are both false.

The set-builder notation is introduced the same way. For all sets A,B and all properties $\varphi$, we define the notation
$$B=\left\{x\in A\,|\,\varphi(x)\right\}$$ to mean
$$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right).$$ This too can be viewed as short notation for a significantly longer expression, where the $\Leftrightarrow$ is expressed using $\land$ and $\lnot$.

 

[reenmachine]

Because what I did there defines the meaning of the curly brackets.

The alphabet of the formal language of set theory includes only a bare minimum of symbols. Typically, we would define it as containing the symbols $\land$, $\lnot$ (read as "and" and "not") and only a few others. Then we use the fundamental symbols to define new ones, to make things easier for ourselves. For example, it would be convenient to have a symbol that we can think of as representing the word "or", so we make the following definition: For arbitrary statements p and q, we define the expression $p\lor q$ to mean $\lnot(\lnot p \land \lnot q)$. Now the symbol $\lor$ can be read as "or".

This will make more sense when you have studied truth tables. The only motivation I can give you right now is that the statement "p or q" is true if and only if it's not true that p and q are both false.

The set-builder notation is introduced the same way. For all sets A,B and all properties $\varphi$, we define the notation
$$B=\left\{x\in A\,|\,\varphi(x)\right\}$$ to mean
$$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right).$$ This too can be viewed as short notation for a significantly longer expression, where the $\Leftrightarrow$ is expressed using $\land$ and $\lnot$.

Thanks a lot , I actually knew a little bit about $\lnot$ and these kind of statements from when I red a little bit about formal logic.

EDIT:

Suppose that F=It's friday , A=I'm alone , B=I'm at home and C=I'm playing baseball:

In chronological order of knowledge:

1)B→A↔¬F
2)F→(B∧¬A)∨(¬B∧(A∨¬A))
3)C↔F∧(¬B∧¬A)
4)¬B∧F → C∧¬A
5)F→ ¬B ↔C

Therefore with the additionnal statements we can conclude that the bolded red in the 2nd statement F→(B∧¬A)∨(¬B∧(A∨¬A)) is impossible but was a possibility before knowing the next statements.

That would mean:
If I'm at home , then I'm alone if and only if it's not friday
If it's friday , then either (I'm at home and not alone) or (not at home and alone or not alone).
I'm playing baseball if and only if it's friday and I'm not at home and not alone.
If I'm not at home on friday , then I'm playing baseball.
If it's friday , then I'm not at home if and only if I play baseball.

We could also conclude that I'll never be alone on friday , whether I'm at home or playing baseball , which are my only two options.

 

[reenmachine]

$$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right).$$ This too can be viewed as short notation for a significantly longer expression, where the $\Leftrightarrow$ is expressed using $\land$ and $\lnot$.

Giving it a try just for fun:

$$\forall x \left(x\in B \Leftrightarrow \left(x\in A\ \land\ \varphi(x)\right)\right).$$

∀x (x ∈ B → ¬(¬x ∈ A) ∧ ¬(¬p(x)))

?

 

[Fredrik]

What I had in mind was to first rewrite $p\leftrightarrow q$ as
$$(p\land q)\lor (\lnot p\land\lnot q).$$ Then you can use the definition of $\lor$ I mentioned earlier. An alternative is to rewrite $p\leftrightarrow q$ as
$$(p\rightarrow q)\land(q\rightarrow p).$$ Then you can use that $p\rightarrow q$ is defined as $\lnot(p\land\lnot q)$.

What you did was just to replace the $\leftrightarrow$ with $\rightarrow$, and insert $\lnot\lnot$ in two places.

 

[micromass]

Edit: Is my first rewrite even correct? I have deja vu. Feels like this is a mistake I've made before. I need to think about it.

The truth table seems to say that it's correct, so I think it is.

 

[Fredrik]

The truth table seems to say that it's correct, so I think it is.

Yes, I just checked it myself. I saw your post just after I deleted the comment where I questioned it.

 

[reenmachine]

Edit: Is my first rewrite even correct? I have deja vu. Feels like this is a mistake I've made before. I need to think about it.

I think it is.

Suppose P= I love her and Q= She loves me

Then I love her if and only if she loves me

-

I love her and she loves me OR I don't love her and she don't loves me

-

If I love her she loves me AND If she loves me I love her

And finally that:

It's not true that I love her and she don't loves me.

 

[reenmachine]

Another try

∀x ( ¬(x ∈ B ∧ ¬(x ∈ A ∧ p(x))) ∧ ¬((x ∈ A ∧ p(x)) ∧ ¬x ∈ B) )

(big spaces for clearness)

Just to be sure , is there a difference between ¬(R) and ¬R ? (I don't think there is)

Take note that I used the alternative method you've proposed in your earlier post.I skipped the part with the arrows and directly defined p → q with ¬(p∧¬q).

 

[Fredrik]

That looks correct, except for a missing ) at the end. There's no difference between ¬(R) and ¬R. The "she loves me" argument is fairly convincing, but if you want to know for sure, you must verify that the truth tables are the same. 1=true, 0=false.

p     q     p ↔ q
1     1       1[/color]
1     0       0[/color]
0     1       0[/color]
0     0       1[/color]

p     q     (p∧q) ∨ (¬p ∧ ¬q)
1     1       1   1[/color]  0  0  0    
1     0       0   0[/color]  0  0  1
0     1       0   0[/color]  1  0  0
0     0       0   1[/color]  1  1  1

 

[reenmachine]

That looks correct, except for a missing ) at the end. There's no difference between ¬(R) and ¬R. The "she loves me" argument is fairly convincing, but if you want to know for sure, you must verify that the truth tables are the same. 1=true, 0=false.

p     q     (p∧q) ∨ (¬p ∧ ¬q)
1     1       1   1[/color]  0  0  0    
1     0       0   0[/color]  0  0  1
0     1       0   0[/color]  1  0  0
0     0       0   1[/color]  1  1  1

Good , finally I got something right

About this part of the code in the quote , both the left and right side of the ∨ has to be completely different for it to be true is that correct? either 0 or 111 /or/ 1 or 000 for this specific case.

 

[Fredrik]

About this part of the code in the quote , both the left and right side of the ∨ has to be completely different for it to be true is that correct? either 0 or 111 /or/ 1 or 000 for this specific case.

The ∨ operation has the following truth table:

p     q     p∨q
1     1      1
1     0      1
0     1      1
0     0      0

Don't confuse "p or q" with "either p or q", which has 0 1 1 0 in the final column.

Maybe I should have used more colors in the big truth table.

p     q     (p∧q) ∨ (¬p ∧ ¬q)
1     1       1[/color]   1[/color]  0[/color]  0[/color]  0[/color]    
1     0       0[/color]   0[/color]  0[/color]  0[/color]  1[/color]
0     1       0[/color]   0[/color]  1[/color]  0[/color]  0[/color]
0     0       0[/color]   1[/color]  1[/color]  1[/color]  1[/color]

First you write down all possible truth values of p and q. That's the first two columns. Then you use the black columns to fill in the green columns. Then you use the green and black to fill in the brown columns. Then you use the brown to fill in the red. The brown and green columns are just intermediate steps. The actual truth table consists of the black and red columns.

Can you use truth tables to show that "if p then q" is equivalent to "if not q then not p"?

 

[reenmachine]

Can you use truth tables to show that "if p then q" is equivalent to "if not q then not p"?

Let's verify:

p     q      v   ¬q   ¬p
[U][B]1     1      1    0    0[/B][/U] 
1     0      0    1    0
0     1      0    0    1
0     0      1    1    1

?

Or did you wanted me to show the proof based on your tables?

 

[reenmachine]

I made a mistake that basically switched all the numbers in the last colums but that's because I originally misplaced both the ¬q and ¬p and replaced them without changing the 1s and 0s.It wasn't a logical mistake.

fixed

 

[Fredrik]

I meant that you should write down the truth table for p→q, and then write down the truth table for ¬q→¬p. They should be the same.

I don't understand the table you did.

Edit: Here's what I wanted you to do. Fill in numbers instead of the question marks in the tables below. In the second one, the first and third column of question marks should contain the truth values of ¬q and ¬p respectively, and the second column should contain the truth values of ¬q→¬p. The first and third are not part of the actual truth table, but it will be easier for you to write down the correct truth values for ¬q→¬p if you first write down the truth values of ¬p and ¬q.

p     q     p→q
1     1      ?
1     0      ?
0     1      ?
0     0      ?

p     q    ¬q → ¬p
1     1    ?  ? ?
1     0    ?  ? ?
0     1    ?  ? ?
0     0    ?  ? ?

 

[reenmachine]

p     q     p→q
1     1      1
1     0      0
0     1      0
0     0      1

p     q    ¬q → ¬p
1     1    0  1 0
1     0    1  0 0
0     1    0  0 1
0     0    1  1 1

?

 

[Fredrik]

The first one is 75% right. The final column should be 1 0 1 1. What you wrote down are the truth values of p↔q.

In the second one, you got the negations right. But the implication has the same problem as in the first table.

 

[reenmachine]

The first one is 75% right. The final column should be 1 0 1 1. What you wrote down are the truth values of p↔q.

In the second one, you got the negations right. But the implication has the same problem as in the first table.

? confusing here

p     q     p→q
1     1      1
1     0      0
0     1      1
0     0      1

Not so sure about the second one.

p     q    ¬q → ¬p
1     1    0  1 0
1     0    1  0 0
0     1    0  1 1
0     0    1  1 1

So ''if p , then q'' doesn't exclude that ''if not p , then q'' ? Therefore only the statement ''if p , then not q'' will be considered false?

 

[Fredrik]

Yes, that's right.

 

[CompuChip]

Yes and that is really by definition. Consider p being the statement "it is Friday" and q "I am alone". Let's consider the statement S: $p \implies q$.

If you are alone because it is Friday then the statement is clearly true. If it is Friday and your house is filled with friends, it is clearly false. This matches our intuition about what "implies" means.

Now suppose it's not Friday. You may be alone, or not. Basically, S has nothing to say about this. Still, we have to assign either "true" or "false" to it. The choice is somewhat arbitrary, I think looking at the intuitive meaning an argument can be made for both. So we need to make a choice, and we define S to be true if p is false, whatever the value of q.
Note that if we defined it to be true, then $p \implies q$ would have had exactly the same truth table as $p \land q$. If we defined it to be true if p is false and q is false, it would have had the same truth table as $p \Leftrightarrow q$. And if we had defined it to be true if p is false and q is true we would get the same truth table as for $q$.

 

[Fredrik]

So ''if p , then q'' doesn't exclude that ''if not p , then q'' ? Therefore only the statement ''if p , then not q'' will be considered false?

I'm not sure I understand this question, which you seem to have edited in seconds after I replied "yes, that's right". If you're asking why p→q is considered true when p is false (regardless of whether q is true or false), then CompuChip already gave you a good reply: If we make the final column 1 0 0 0, we get the truth table of ∧. If we make it 1 0 0 1, we get the truth table of ↔. If we make it 1 0 1 0, we get the truth table of q.

I will contribute the following example: Your friend is sure that his team is going to win the big game, so he says "if my team loses tonight, I will wear a dress to work tomorrow". His team wins. Is there any kind of clothing he can wear at work the next day that will force you to conclude that he lied to you? (If the final column of the truth table doesn't end with 1 1, the answer would be yes).

 

[reenmachine]

Yes and that is really by definition. Consider p being the statement "it is Friday" and q "I am alone". Let's consider the statement S: $p \implies q$.

If you are alone because it is Friday then the statement is clearly true. If it is Friday and your house is filled with friends, it is clearly false. This matches our intuition about what "implies" means.

Now suppose it's not Friday. You may be alone, or not. Basically, S has nothing to say about this. Still, we have to assign either "true" or "false" to it. The choice is somewhat arbitrary, I think looking at the intuitive meaning an argument can be made for both. So we need to make a choice, and we define S to be true if p is false, whatever the value of q.
Note that if we defined it to be true, then $p \implies q$ would have had exactly the same truth table as $p \land q$. If we defined it to be true if p is false and q is false, it would have had the same truth table as $p \Leftrightarrow q$. And if we had defined it to be true if p is false and q is true we would get the same truth table as for $q$.

Yeah this is what I thought , basically anything that is possible is true and the rest if false.

thanks man!

 

[reenmachine]

Is there any kind of clothing he can wear at work the next day that will force you to conclude that he lied to you?

The answer would be no.

I like these logical statements exercises , so much fun.

thanks a lot !

 

[Fredrik]

The answer would be no.

Right. My point was that this example provides the motivation for the choice to define → this way, with 1 1 in the final two positions of the truth table.

I like these logical statements exercises , so much fun.

Section 2.6 of the Book of Proof has more of them. So you can either do a few of those, or just prove a few of the the equivalences (2.1)-(2.5) using truth tables.

Do you understand all the exercises from post #249 now, including how to solve 2nd-degree polynomial equations, and how to derive the formula for the solutions of a 2nd-degree polynomial equation?

 

[reenmachine]

Do you understand all the exercises from post #249 now, including how to solve 2nd-degree polynomial equations, and how to derive the formula for the solutions of a 2nd-degree polynomial equation?

Here's the link for the exercises: http://www.people.vcu.edu/~rhammack/BookOfProof/Sets.pdf

I'm going to try a couple of random ones , if you think there's some in particular that I should attempt please tell me so.

Solutions:

1.{...,-21,-16,-11,-4,-1,4,9,...}
3.{-2,-1,0,1,2,3,4,5,6}
13.{...,-2,-1,0}
15.{...,-2,-1,0,1,2,3,...}
21.$\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=y^2\}$
27.$\{x\in\mathbb R\,|\,\exists y\in\mathbb R~~ x=yπ/2\}$

?

thanks!

 

[micromass]

Here's the link for the exercises: http://www.people.vcu.edu/~rhammack/BookOfProof/Sets.pdf

I'm going to try a couple of random ones , if you think there's some in particular that I should attempt please tell me so.

Solutions:

1.{...,-21,-16,-11,-4,-1,4,9,...}

Why -4?

3.{-2,-1,0,1,2,3,4,5,6}

OK

13.{...,-2,-1,0}

OK

15.{...,-2,-1,0,1,2,3,...}

This is true, but I'm curious how you found it.

21.$\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=y^2\}$

OK

27.$\{x\in\mathbb R\,|\,\exists y\in\mathbb R~~ x=yπ/2\}$

Why do you take $y\in \mathbb{R}$? In that case, shouldn't things like $\frac{\sqrt{2}\pi}{2}$ also be in the set?

 

[reenmachine]

Why -4?

Guess it's -6 , just a brain cramp.

This is true, but I'm curious how you found it.

I had no clue how to solve at first but then I thought to myself , it's not 5a=2b or 5a+2b=something , it's just 5a+2b , so basically all numbers in Z could be inserted there.

Why do you take $y\in \mathbb{R}$? In that case, shouldn't things like $\frac{\sqrt{2}\pi}{2}$ also be in the set?

$\{x\in\mathbb R\,| xπ/2\}$ ? I'm confused about what to do.

EDIT:$\{x\in\mathbb R\,|\,\exists y\in\mathbb Z~~ x=yπ/2\}$ ?

 

[Fredrik]

15.

As I said before, it's the set of all 5a+2b such that a and b are integers. It's pretty obvious that this is the set of all integers, $\mathbb Z$, as you have now concluded. (Edit: Ohh...after reading micromass' reply below, I see it's not as obvious as I thought. It's only obvious that this set will be a subset of $\mathbb Z$). If you want to prove that this "guess" is correct, you need to rely on the axiom that says that two sets are equal if and only if they have the same elements. Define $Z=\{5a+2b\,|\,a,b\in\mathbb Z\}$. We want to prove that $Z=\mathbb Z$.

Let $x\in Z$ be arbitrary. The definition of Z tells us that there exist $a,b\in\mathbb Z$ such that $5a+2b=x$. Since $\mathbb Z$ is closed under addition and multiplication, this implies that $x\in\mathbb Z$.

Let $x\in\mathbb Z$ be arbitrary. Define $a,b\in\mathbb Z$ by $a=x$ and $b=-2x$. We have $a,b\in\mathbb Z$ and $5a+2b=5x+2(-2x)=5x-4x=x$. This implies that $x\in Z$.

21. Note that your result can be simplified to $\left\{n^2\,|\,n\in\mathbb N\right\}$. The proof of that is similar to the proof of 15. (Most people use n and m for integers rather than x and y, but it's not necessary to do this. n is a dummy variable here, so you can use x instead).

27. The answer in your edit is correct, and can be simplified to $\left\{\frac{n\pi}{2}\,\big|\,n\in\mathbb Z\right\}$.