Question about proof from a guy with a highschool education

[reenmachine]

What is the definition of $(1,2)$ and the other couples?

Definition of a couple? Not sure , my guess would be that the first componant is an element of $A$ and the second is an element of $B$.

What if you take the union?

What do you mean by "take the union"?

 

[micromass]

Definition of a couple? Not sure , my guess would be that the first componant is an element of $A$ and the second is an element of $B$.

https://www.physicsforums.com/showpost.php?p=4335608&postcount=60

 

[Fredrik]

So does $(1,2)\cup (1,3)\cup (2,2)\cup (2,3) = \{1,2,3\}$ ?

Definition of a couple? Not sure

"Couple" isn't the right word in English. It's "pair". And when we say "pair", we mean "ordered pair".

What do you mean by "take the union"?

Use the definition of $\cup$ to rewrite $(1,2)\cup (1,3)\cup (2,2)\cup (2,3)$ in the form {something,something,...,something}.

Then use that you know what it means for two sets to be equal. X=Y if and only if...

 

[reenmachine]

You're making it complicated by choosing a set with ordered pairs as elements, but maybe you meant { when you wrote (? I will consider the example with { instead.

If $A=\{\{0,1\},\{1,2\}\}$, then $\bigcup A=\{0,1\}\cup\{1,2\}=\{0,1,2\}$.

This is the post that confused me yesterday , in MY quote I used ordered pairs but you didn't but I got mixed up for some reasons.So I thought you could take all the components of the ordered pairs to form the new $\cup$ set.

 

[reenmachine]

Use the definition of $\cup$ to rewrite $(1,2)\cup (1,3)\cup (2,2)\cup (2,3)$ in the form {something,something,...,something}.

Then use that you know what it means for two sets to be equal. X=Y if and only if...

That would be $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$

X=Y if and only if all elements of X are in Y and all elements of Y are in X.

 

[micromass]

That would be $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$

Right. So

\bigcup (A\times B) = \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}

Now, what about

\bigcup \bigcup (A\times B) = \bigcup \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}

 

[reenmachine]

Right. So

\bigcup (A\times B) = \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}

Now, what about

\bigcup \bigcup (A\times B) = \bigcup \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}

Since the elements aren't ordered pairs anymore , then now I guess $\bigcup \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \} = \{1,2,3\}$

So basically my hypothesis that you need a ''higher'' level of U to achieve this result was true?

 

[micromass]

Since the elements aren't ordered pairs anymore , then now I guess $\bigcup \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \} = \{1,2,3\}$

Yes.

So basically my hypothesis that you need a ''higher'' level of U to achieve this result was true?

Correct.

 

[reenmachine]

Yes.
Correct.

Finally I get something right , getting rare these days

thanks guys!

 

[reenmachine]

It still leaves me with the problem $$\bigcup_{i \in N}R × [i , i + 1]$$.

If elements are going to ressemble something like $\{π\} , \{π , 4.3565\}$(with $i = 4$) (from the ordered pair $(π,4.3565)$) , but $\{π\}$ isn't $π$ so how do I proceed to describe them?

$\{π\}$ isn't in $R$ right?

But $π$ is.

Edit: What I do know is that $$\bigcup(\bigcup_{i \in N}R × [i , i + 1]) = R$$

So $$\bigcup_{i \in N}R × [i , i + 1]$$ will be the set of all element $\{a\}$ with $a \in R$ and all elements $\{a,d\}$ with $d \in R \ \ 1 ≤ d$.

 

[jibbles]

I'm back!

I want something to be crystal clear in my head:

$A=\{1,2\}$
$B=\{2,3\}$
$A \cup B = \{1,2,3\}$
$A \cap B = \{2\}$
$A × B = \{(1,2) , (1,3) , (2,2) , (2,3)\}$

Then what exactly is $$\bigcup(A × B)$$ ?

Is it $\{1,2,3\}$?

Basically , when you have to unionize a cartesian product set , you take all the elements inside the ordered pairs to make the set?

So any notation like $$\bigcup(A × B)$$ with a cartesian product will be equal to $A \cup B$?

Formally, the ordered set (a,b) denotes the set {{a}, {a,b}}. Hence

$\bigcup{(A{\times}B)}=\bigcup{\{(1,2),(1,3),(2,2),(2,3)\}}$

$=\bigcup{ \{\{\{1\},\{1,2\}\},\{\{1\},\{1,3\}\},\{\{2\},\{2,2\}\},\{\{2\},\{2,3\}\}\} }$

$=\bigcup{ \{\{\{1\},\{1,2\}\},\{\{1\},\{1,3\}\},\{\{2\},\{2\}\},\{\{2\},\{2,3\}\}\} }$

$=\bigcup{ \{\{\{1\},\{1,2\}\},\{\{1\},\{1,3\}\},\{\{2\}\},\{\{2\},\{2,3\}\}\} }$

$=\{\{1\},\{1,2\},\{1\},\{1,3\},\{2\},\{2\},\{2,3\}\}$

$=\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,3\}\}.$

Hope this helps.

 

[reenmachine]

About $$\bigcup_{i \in N}R × [i , i + 1]$$

Since this is probably more of a conceptual question I'll ask it here.

My last and final attempt at denoting this set : $\{\{x\},\{x,y\} |\ x,y \in R \ \ y ≥ 1 \}$

I checked the book's solution , which is $\{(x,y) : x,y \in R \ \ y ≥ 1\}$

I find it weird that the dummy variables would be put between "()" like it's an ordered pair to present the set such as "the set of all (x,y) such that...".Anybody can enlighten me on using my method versus the one in the book concerning the left side of the notation?

If the elements of the set I'm trying to denote will be between brackets such as $\{1\} , \{1,2\}, \{2\} , \{2,1\}$ for the set $(1,2) \cup (2,1)$ then why would I present the dummy variables between "()" in the notation instead of {}?

 

[Fredrik]

About $$\bigcup_{i \in N}R × [i , i + 1]$$

Since this is probably more of a conceptual question I'll ask it here.

My last and final attempt at denoting this set : $\{\{x\},\{x,y\} |\ x,y \in R \ \ y ≥ 1 \}$

I checked the book's solution , which is $\{(x,y) : x,y \in R \ \ y ≥ 1\}$

I find it weird that the dummy variables would be put between "()" like it's an ordered pair to present the set such as "the set of all (x,y) such that...".Anybody can enlighten me on using my method versus the one in the book concerning the left side of the notation?

If the elements of the set I'm trying to denote will be between brackets such as $\{1\} , \{1,2\}, \{2\} , \{2,1\}$ for the set $(1,2) \cup (2,1)$ then why would I present the dummy variables between "()" in the notation instead of {}?

The elements of $\bigcup_{i \in N}\mathbb R × [i , i + 1]$ are all ordered pairs, so it makes a lot of sense to describe the set in the form "the set of all ordered pairs such that..." You could write {{x},{x,y}} instead of (x,y), but you wrote {x},{x,y}. Apart from that it's fine, but I would prefer to see a comma or a $\land$ between the two properties. Without a symbol there, a reader might think that you forgot to put one there, and might be unsure if you meant $\land$, $\lor$ or something else.

Most people prefer to write (x,y) because it's simpler. Some (e.g. physics students) do it because they don't even know that there's a definition of (x,y). They just know that ordered pairs have the property that (a,b)=(c,d) if and only if a=c and b=d.

The set can also be written as
$$\{(x,y)\in\mathbb R\times\mathbb R : y\geq 1\}$$ or as
$$\{z\in\mathbb R\times\mathbb R:\exists x,y~~(z=(x,y)~\land~y\geq 1)\}.$$

 

[reenmachine]

The elements of $\bigcup_{i \in N}\mathbb R × [i , i + 1]$ are all ordered pairs, so it makes a lot of sense to describe the set in the form "the set of all ordered pairs such that..." You could write {{x},{x,y}} instead of (x,y), but you wrote {x},{x,y}. Apart from that it's fine, but I would prefer to see a comma or a $\land$ between the two properties. Without a symbol there, a reader might think that you forgot to put one there, and might be unsure if you meant $\land$, $\lor$ or something else.

Most people prefer to write (x,y) because it's simpler. Some (e.g. physics students) do it because they don't even know that there's a definition of (x,y). They just know that ordered pairs have the property that (a,b)=(c,d) if and only if a=c and b=d.

The set can also be written as
$$\{(x,y)\in\mathbb R\times\mathbb R : y\geq 1\}$$ or as
$$\{z\in\mathbb R\times\mathbb R:\exists x,y~~(z=(x,y)~\land~y\geq 1)\}.$$

Now I'm lost again.

You ask me to re-write $(1,2)\cup (1,3)\cup (2,2)\cup (2,3)$ in the form of $\{something,something...\}$.

I then responded with $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ to which micromass responded that I was correct.But in $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ , the ordered pairs aren't between brackets like for example: $\{ \{\{1\} , \{1,2\}\} , \{\{1\},\{1,3\}\} , \{\{2\} , \{2,2\}\} ,\{\{2\} \{2,3\}\} \}$.

Micromass then asked me to find the elements of \bigcup \bigcup (A\times B) = \bigcup \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}

To which I responded $\{1,2,3\}$.

But if the previous set was to be $\{ \{\{1\} , \{1,2\}\} , \{\{1\},\{1,3\}\} , \{\{2\} , \{2,2\}\} ,\{\{2\} \{2,3\}\} \}$ instead of $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ , then the double-unionized set would have been $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ instead of $\{1,2,3\}$ no?

What am I missing?

 

[Fredrik]

You ask me to re-write $(1,2)\cup (1,3)\cup (2,2)\cup (2,3)$ in the form of $\{something,something...\}$.

You asked if $(1,2)\cup (1,3)\cup (2,2)\cup (2,3) = \{1,2,3\}$, so I asked you to rewrite the left-hand side in the form {something,something,...} so you can check if this equality holds or not.

I then responded with $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ to which micromass responded that I was correct.

\begin{align}&(1,2)\cup (1,3)\cup (2,2)\cup (2,3)\\
&=\{\{1\},\{1,2\}\} \cup\{\{1\},\{1,3\}\} \cup\{\{2\},\{2,2\}\} \cup\{\{2\},\{2,3\}\}\\
&=\{ \{1\},\{1,2\},\{1\},\{1,3\},\{2\},\{2,2\},\{2\},\{2,3\} \}\\
&=\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,2\},\{2,3\}\}
\end{align} Now it's trivial to see that the equality you asked about doesn't hold. You just have to see that {3} is an element of the set on the right, but not an element of the set on the left.

But in $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ , the ordered pairs aren't between brackets

None of the elements of this set is an ordered pair. But some of its subsets are, e.g. $\{\{1\},\{1,3\}\}$ and $\{\{2\},\{2,3\}\}$.

like for example: $\{ \{\{1\} , \{1,2\}\} , \{\{1\},\{1,3\}\} , \{\{2\} , \{2,2\}\} ,\{\{2\} \{2,3\}\} \}$.

This set is equal to {(1,2),(1,3),(2,2),(2,3)}.

Micromass then asked me to find the elements of \bigcup \bigcup (A\times B) = \bigcup \{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}

To which I responded $\{1,2,3\}$.

Good.

But if the previous set was to be $\{ \{\{1\} , \{1,2\}\} , \{\{1\},\{1,3\}\} , \{\{2\} , \{2,2\}\} ,\{\{2\} \{2,3\}\} \}$ instead of $\{ \{1\} , \{1,2\} , \{1,3\} , \{2\} , \{2,2\} , \{2,3\} \}$ ,

It wasn't. That first set is equal to {(1,2),(1,3),(2,2),(2,3)}, which is equal to $A\times B$. The second is equal to $\bigcup A\times B$.

 

[reenmachine]

\begin{align}&(1,2)\cup (1,3)\cup (2,2)\cup (2,3)\\
&=\{\{1\},\{1,2\}\} \cup\{\{1\},\{1,3\}\} \cup\{\{2\},\{2,2\}\} \cup\{\{2\},\{2,3\}\}\\
&=\{ \{1\},\{1,2\},\{1\},\{1,3\},\{2\},\{2,2\},\{2\},\{2,3\} \}\\
&=\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,2\},\{2,3\}\}
\end{align} Now it's trivial to see that the equality you asked about doesn't hold. You just have to see that {3} is an element of the set on the right, but not an element of the set on the left.

Hmm yeah that makes more sense.

None of the elements of this set is an ordered pair. But some of its subsets are, e.g. $\{\{1\},\{1,3\}\}$ and $\{\{2\},\{2,3\}\}$.

The elements of $\bigcup_{i \in N}\mathbb R × [i , i + 1]$ are all ordered pairs

Ok , but since $\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,2\},\{2,3\}\} = \bigcup A × B$ if $A=\{1,2\}$ and $B=\{2,3\}$ and you're saying that none of the elements of this set is an ordered pair , why would the elements of the set $\bigcup_{i \in N}R × [i , i +1]$ be ordered pairs? Isn't this the same situation as $\bigcup A × B$? This is probably the key part that I'm not sure to understand.What is the difference? The very exemple we were talking about is the reason I wrote $\{x\},\{x,y\}$ instead of $(x,y)$.

Basically , what I think is that elements of $R × [i , i +1]$ are ordered pairs but that elements of $\bigcup_{i \in N}R × [i , i +1]$ will come out similarly to the elements of the set $\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,2\},\{2,3\}\}$ , which you told me weren't ordered pairs.

thanks man! Hope you are having a nice week-end!

 

[Fredrik]

Ok , but since $\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,2\},\{2,3\}\} = \bigcup A × B$ if $A=\{1,2\}$ and $B=\{2,3\}$ and you're saying that none of the elements of this set is an ordered pair , why would the elements of the set $\bigcup_{i \in N}R × [i , i +1]$ be ordered pairs? Isn't this the same situation as $\bigcup A × B$? This is probably the key part that I'm not sure to understand.What is the difference? The very exemple we were talking about is the reason I wrote $\{x\},\{x,y\}$ instead of $(x,y)$.

Basically , what I think is that elements of $R × [i , i +1]$ are ordered pairs but that elements of $\bigcup_{i \in N}R × [i , i +1]$ will come out similarly to the elements of the set $\{\{1\},\{1,2\},\{1,3\},\{2\},\{2,2\},\{2,3\}\}$ , which you told me weren't ordered pairs.

thanks man! Hope you are having a nice week-end!

All you have to do to see that the elements of $\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$ are ordered pairs is to note that every element of $\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$ is an element of at least one of the $\mathbb R\times[i,i+1]$ sets.

The source of the confusion seems to be the similarity between the notations
$$\bigcup A\times B$$ and
$$\bigcup_{i\in\mathbb N}A\times B_i.$$ The former denotes the union of the elements of $A\times B$, but the latter denotes the union of the sets $A\times B_i$, not their elements.
$$\bigcup_{i\in\mathbb N}A\times B_i=(A\times B_1)\cup (A\times B_2)\cup\cdots.$$ I guess this sort of thing is why micromass said that he doesn't like the $\bigcup S$ notation.

 

[reenmachine]

I will have to agree with micromass there.

This is a pretty disgusting way to denote all of this

Thanks , it finally start to make sense (well except for that choice of notation).

 

[reenmachine]

This also means that $$\bigcap_{i \in N}R × [i , i +1] = \varnothing $$ right?

 

[Fredrik]

That equality holds since no $(x,y)\in\mathbb R\times\mathbb R$ is an element of all of those sets. In fact, a given (x,y) can belong to at most two of those sets.

 

[reenmachine]

That equality holds since no $(x,y)\in\mathbb R\times\mathbb R$ is an element of all of those sets. In fact, a given (x,y) can belong to at most two of those sets.

How could this happen?

 

[Fredrik]

For example, (0,2) is an element of ℝ×[1,2] and an element of ℝ×[2,3].

 

[reenmachine]

So if we unionize this set , like

$$\bigcup\bigcup_{i \in N}R × [i , i +1]$$

do we now get $\{x\}$ and $\{x,y\}$ type of elements? (that aren't ordered pairs anymore)

 

[Fredrik]

That's right. Since e.g. (2,3)={{2},{2,3}} is an element of $\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$, we know that {2} and {2,3} are elements of $\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$.

So the elements of $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$ will be real numbers. (Edit: The previous sentence was corrected after reenmachine's reply below). So we have $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]=\mathbb R$.

 

[reenmachine]

That's right. Since e.g. (2,3)={{2},{2,3}} is an element of $\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$, we know that {2} and {2,3} are elements of $\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$.

So the elements of $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$ will all be singleton sets whose elements are real numbers. So we have $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]=\mathbb R$.

Thank you!

When you mean singleton sets , you mean that that elements like {2},{2,3} will finish as {2},{3}.Only point I'm still wondering is I thought $\{\{2\},\{3\}\}$ didn't equal $\{2,3\}$ since the elements of the former are sets containing real numbers and the elements of the latter are real numbers.I thought R would only be the latter.So basically 4 unions instead of 3.

Or did you just meant that in $\{...2,3,4...\}$ the numbers are all singletons even if they arent between brackets?

 

[reenmachine]

Here's an interesting exercise from the book of proof:

$$\bigcup_{x \in p(N)}X = ?$$

"p" being powerset here.

My answer would be $N$.

So when you unionize the elements of a powerset , you always end up with the original set.

The only thing I'm wondering is what happens with $\varnothing$

$\varnothing$ would become an element of $\bigcup X$ but since I'm saying $\bigcup X = N$ is the presence of $\varnothing$ destroying the whole logic?

 

[Fredrik]

Thank you!

When you mean singleton sets , you mean that that elements like {2},{2,3} will finish as {2},{3}.Only point I'm still wondering is I thought $\{\{2\},\{3\}\}$ didn't equal $\{2,3\}$ since the elements of the former are sets containing real numbers and the elements of the latter are real numbers.I thought R would only be the latter.So basically 4 unions instead of 3.

Or did you just meant that in $\{...2,3,4...\}$ the numbers are all singletons even if they arent between brackets?

If {2} and {2,3} are elements of a set S, then $\{2\}\cup\{2,3\}=\{2,3\}$ is a subset of $\bigcup S$. This makes 2 and 3 elements of $\bigcup S$.

What I said is that since {2} and {2,3} are elements of $\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$, the elements of $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$ will be things like {2} and {3} (singleton sets whose elements are real numbers). I see now that this is wrong. $\{2\}\cup\{2,3\}=\{2,3\}$ will be a subset of $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$, so the elements of $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]$ will be things like 2 and 3. This implies that $\bigcup\bigcup\bigcup_{i\in\mathbb N}\mathbb R\times[i,i+1]=\mathbb R$. Somehow I got the right final result even though made a blunder. Maybe I just made a second one that compensated for the first one.

 

[Fredrik]

Here's an interesting exercise from the book of proof:

$$\bigcup_{x \in p(N)}X = ?$$

"p" being powerset here.

My answer would be $N$.

So when you unionize the elements of a powerset , you always end up with the original set.

The only thing I'm wondering is what happens with $\varnothing$

$\varnothing$ would become an element of $\bigcup X$ but since I'm saying $\bigcup X = N$ is the presence of $\varnothing$ destroying the whole logic?

You're confusing the two notations again. We have
$$\bigcup X\neq \bigcup_{X\in\mathcal P(N)} X=\bigcup\mathcal P(N) =N.$$

 

[reenmachine]

You're confusing the two notations again. We have
$$\bigcup X\neq \bigcup_{X\in\mathcal P(N)} X=\bigcup\mathcal P(N) =N.$$

I still don't get what happens to $\varnothing$.

If $\varnothing$ is an element of P(N) , how could he be an element of N?

Is it $\{P(1) \cup P(2) \cup P(3)...\}$ ? Those notations are horrendous.

Is it the union of all elements of the powerset of N? This is where I was coming from with $\varnothing$ being an element of the union of all elements of the powerset of N.

I'm sorry if there's some endless confusions , I'm struggling more with the notations than with the logic they are describing.Thank you!

 

[Fredrik]

I don't understand what ∅ has to do with this. $\bigcup_{X\in\mathcal P(N)} X$ and $\bigcup\mathcal P(N)$ both denote the union of all subsets of N, and this is clearly N (regardless of what set N is).

$\varnothing$ is an element of $\mathcal P(N)$ (regardless of what N is), but not an element of $N$ (unless you have specifically chosen N to be a set that has $\varnothing$ as an element).

By the way, X is a dummy variable in the question you asked in post #631, so the notation $\bigcup X$ doesn't even make sense there.

 

[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.