Need help on mathematical logic (and set theory) !?

[joseph1125]

I am totally stuck when I see this question. Please help!!!

 

[JeffM]

I am totally stuck when I see this question. Please help!!!

It is very hard to help someone with proofs because we do not know what theorems have previously been proved.

But if you want a suggestion, I might try a proof by contradiction.

 

[joseph1125]

It is very hard to help someone with proofs because we do not know what theorems have previously been proved.

But if you want a suggestion, I might try a proof by contradiction.

I use this kind of proof, is that okay?
Please challenge my proof if possible!

 

[JeffM]

I use this kind of proof, is that okay?
Please challenge my proof if possible!

It appears to me that you are assuming what is to be proven.

You say \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}\ and\ \{B = \emptyset\}\ and\ \{A \bigcap B = \emptyset\}.\)

But the first implication is exactly what you have been asked to prove. Maybe I misunderstand what you were trying to say.

Why not try a proof by contradiction?

 

[pka]

I am totally stuck when I see this question. Please help!!!

You should know for all sets \(\displaystyle A\) then \(\displaystyle A\subseteq A\cup B.\)

Proof by contradiction.
Suppose \(\displaystyle A\cup B=\emptyset~\&~A\ne\emptyset\) then \(\displaystyle \exists x\in A\) but that means \(\displaystyle x\in\emptyset\).

What is wrong with that picture?

 

[joseph1125]

It appears to me that you are assuming what is to be proven.

You say \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}\ and\ \{B = \emptyset\}\ and\ \{A \bigcap B = \emptyset\}.\)

But the first implication is exactly what you have been asked to prove. Maybe I misunderstand what you were trying to say.

Why not try a proof by contradiction?

Sorry, it should be "we assume that \(\displaystyle \{A \bigcup B = \emptyset\}\[/hex] is true"

Because my thinking was, in order to prove the whole thing is true. I have to determine \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}\[/hex]
is true or not, for implication like A implies B, if I want to make it has a possibility to show "false". A must be true to check if B=F. That was my workflow.\)\)

 

[JeffM]

Sorry, it should be "we assume that \(\displaystyle \{A \bigcup B = \emptyset\}\[/hex] is true" The theorem that you are asked to prove is conditional. It applies only when the conditional is true. The theorem simply does not apply if the conditional is false. There is no assumption that the conditional is always true.

Because my thinking was, in order to prove the whole thing is true. I have to determine \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}\ Yes, that is exactly what you have to demonstrate so you CANNOT assume it. That is called begging the question. It is a favorite trick of economists and an error of logic identified thousands of years ago.[/hex]
is true or not, for implication like A implies B, if I want to make it has a possibility to show "false". A must be true to check if B=F. That was my workflow.\)\)

\(\displaystyle \(\displaystyle .
Is English your native language? If it is, then you should realize that "I have to determine if X is true or not for implication like A implies B, if I want to make it has a possibility to show false" makes no grammatical sense whatsoever. If English is not your native language, then it is of course understandable that you are having difficulty translating your thought into English.

It is true that if the conditional statement \(\displaystyle A \bigcup B = \emptyset \implies A = \emptyset\),

is true, then it is also true that the conditional statement \(\displaystyle A \bigcup B = \emptyset \implies B = \emptyset\) is true.

But so what? The fact that both conditionals are true if one is true does not help you in the slightest prove that either of them is true.

I suggested that you try a proof by contradiction, and pka did 90% of the work of showing you how to do such a proof on the basis of what I presume is either an axiom or early theorem of set theory.\)\)

 

[joseph1125]

.
Is English your native language? If it is, then you should realize that "I have to determine if X is true or not for implication like A implies B, if I want to make it has a possibility to show false" makes no grammatical sense whatsoever. If English is not your native language, then it is of course understandable that you are having difficulty translating your thought into English.

It is true that if the conditional statement \(\displaystyle A \bigcup B = \emptyset \implies A = \emptyset\),

is true, then it is also true that the conditional statement \(\displaystyle A \bigcup B = \emptyset \implies B = \emptyset\) is true.

But so what? The fact that both conditionals are true if one is true does not help you in the slightest prove that either of them is true.

I suggested that you try a proof by contradiction, and pka did 90% of the work of showing you how to do such a proof on the basis of what I presume is either an axiom or early theorem of set theory.

I see what you mean, and you're right, English is not my native language
Okay, when I looked through the proof of pka, I can understand the beginning part, but doesn't understand the latter part

For Proof by contradiction, we have to assume that \(\displaystyle A \bigcup B = \emptyset \implies A =\emptyset\) = False
so \(\displaystyle A \bigcup B =\emptyset \) is true and \(\displaystyle A= \emptyset\) is false

However, "∃x∈A but that means x∈∅" I don't understand this part. Where is "∃x∈A" coming from?

 

[JeffM]

I see what you mean, and you're right, English is not my native language
Okay, when I looked through the proof of pka, I can understand the beginning part, but doesn't understand the latter part

For Proof by contradiction, we have to assume that \(\displaystyle A \bigcup B = \emptyset \implies A =\emptyset\) = False
so \(\displaystyle A \bigcup B =\emptyset \) is true and \(\displaystyle A= \emptyset\) is false

However, "∃x∈A but that means x∈∅" I don't understand this part. Where is "∃x∈A" coming from?

A proof by contradiction works as follows.

We want to prove X.

So we assume not X and deduce X, meaning that if not X is true, then X is also true. But that is nonsense.

Therefore not X is false.

THUS, X is true.

That is the general structure of a proof by contradiction.

In this case we want to prove that \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}\).

So we assume that \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A \ne \emptyset\}.\) This is the opposite of what we want to prove.

But if A is NOT empty, then it has at least one element, call it x.

\(\displaystyle So\ x \in A.\)

\(\displaystyle So\ x \in A \bigcup B\) because anything that is in A is also in the union of A and B.

\(\displaystyle So\ A \bigcup B \ne \emptyset\) because it has at least one element, namely x.

We have just proved that if \(\displaystyle A \bigcup B = \emptyset,\ then\ A \bigcup B \ne \emptyset,\) which is a nonsensical contradiction.

So the assumption is false.

Therefore, the opposite is true or

\(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}.

If you do not understand this, I am afraid you will need someone who speaks your native language well to explain it to you because this is as simple as I can make it.\)

 

[joseph1125]

A proof by contradiction works as follows.

We want to prove X.

So we assume not X and deduce X, meaning that if not X is true, then X is also true. But that is nonsense.

Therefore not X is false.

THUS, X is true.

That is the general structure of a proof by contradiction.

In this case we want to prove that \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}\).

So we assume that \(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A \ne \emptyset\}.\) This is the opposite of what we want to prove.

But if A is NOT empty, then it has at least one element, call it x.

\(\displaystyle So\ x \in A.\)

\(\displaystyle So\ x \in A \bigcup B\) because anything that is in A is also in the union of A and B.

\(\displaystyle So\ A \bigcup B \ne \emptyset\) because it has at least one element, namely x.

We have just proved that if \(\displaystyle A \bigcup B = \emptyset,\ then\ A \bigcup B \ne \emptyset,\) which is a nonsensical contradiction.

So the assumption is false.

Therefore, the opposite is true or

\(\displaystyle \{A \bigcup B = \emptyset\} \implies \{A = \emptyset\}.

If you do not understand this, I am afraid you will need someone who speaks your native language well to explain it to you because this is as simple as I can make it.\)

\(\displaystyle

Your explanation is much better than you thought!
Cheers!\)