Proving: Sets Proofs - Techniques & Examples

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In summary: B\cap C$ if and only if $x\in A\cap B$ and $x\in A\cup C$. This is because $B\cap C=A\cup B\cup C$, which is a subset of $A$. Hence $x\in B\cap C$.
  • #1
paulmdrdo1
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prove the followinga. prove that if $A\cap B=\emptyset$, then $(A\times C)\cap (B\times C)=\emptyset$
b. $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$

i don't have any idea how i would start proving this.
can you give me some techniques on proofs.
 
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  • #2
paulmdrdo said:
prove the followinga. prove that if $A\cap B=\emptyset$, then $(A\times C)\cap (B\times C)=\emptyset$
b. $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$

i don't have any idea how i would start proving this.
can you give me some techniques on proofs.
a. Assume on the contrary that $(x,y)\in (A\times C)\cap (B\times C)$. Thus $x\in A\cap B$ and $y\in C$ (why?). Can you finish?

b. Let $x\in A\cup(B\cap C)$. Then $x\in A$ or $x\in B\cap C$.
Case 1. $x\in A$.
Here $x\in A\cup B$ and $x\in A\cup C$. (why?). Thus $x\in (A\cup B)\cap (A\cup C)$.

Case 2. $x\in B\cap C$.
Again, Here $x\in A\cup B$ and $x\in A\cup C$. (why?). Thus $x\in (A\cup B)\cap (A\cup C)$.

The above shows that any element of $A\cup (B\cap C)$ is in $(A\cup B)\cap(A\cup C)$. This means $A\cup (B\cap C)\subseteq (A\cup B)\cap(A\cup C)$. Can you show the reverse containment and finish?
 
  • #3
caffeinemachine i know nothing about formal way of proving. i can say why that is true in words or verbally but not in a generalize way. i want a formal presentation of proofs like what you are doing. can you give some tips. because every time i encounter this kind of problems i always feel discouraged.

but this is what i tried

$x\in A\cap B$ and $y\in C$ because we assume that $(x,y)\in (A\times C)\cap (B\times C) $

i don't know howfill in the other (whys)
 
  • #4
paulmdrdo said:
caffeinemachine i know nothing about formal way of proving. i can say why that is true in words or verbally but not in a generalize way. i want a formal presentation of proofs like what you are doing. can you give some tips. because every time i encounter this kind of problems i always feel discouraged.

but this is what i tried

$x\in A\cap B$ and $y\in C$ because we assume that $(x,y)\in (A\times C)\cap (B\times C) $

i don't know howfill in the other (whys)
The 'second why' is quite straightforward. If you know that $x\in A$, then $x$ also lies in any superset of $A$. Isn't it? Thus $x\in A$ gives $x\in A\cup B$ since $A\cup B$ is a superset of $A$. Similarly $x\in A$ gives $A\cup C$ too.

Now try to fill in for the third why.
 
  • #5


Sure, here are some techniques for proofs:

1. Direct proof: This is the most common type of proof, where you start with the given assumptions and use logical steps to arrive at the conclusion. In this type of proof, you may need to use definitions, theorems, and logical equivalences to show the validity of your argument.

2. Proof by contradiction: In this type of proof, you assume the opposite of what you are trying to prove and show that it leads to a contradiction. This proves that the original statement must be true.

3. Proof by induction: This technique is commonly used to prove statements about natural numbers. It involves proving a base case (usually n=1) and then showing that if the statement is true for n=k, it must also be true for n=k+1.

4. Proof by contrapositive: This is a type of proof where you prove the contrapositive of a statement instead of the original statement. The contrapositive of a statement "if A then B" is "if not B then not A". This technique can be useful when the direct proof is difficult.

5. Proof by cases: Sometimes, a statement can be proven by considering different cases and showing that the statement is true for each case. This technique can be useful when the statement is complex or involves multiple conditions.

Now, let's use some of these techniques to prove the statements given in the content:

a. To prove that if A∩B=∅, then (A×C)∩(B×C)=∅, we can use a direct proof. We start by assuming that A∩B=∅ and we want to show that (A×C)∩(B×C)=∅. Let (x,y) be an element of (A×C)∩(B×C). This means that x∈A and x∈B, which is a contradiction since we assumed that A∩B=∅. Therefore, (A×C)∩(B×C)=∅.

b. To prove that A∪(B∩C)=(A∪B)∩(A∪C), we can use a direct proof again. We start by assuming that x is an element of A∪(B∩C). This means that x∈A or x∈(B∩C). If
 

FAQ: Proving: Sets Proofs - Techniques & Examples

What is the purpose of proving sets?

Proving sets is a fundamental aspect of mathematics that allows us to establish the truth or validity of mathematical statements. It helps us to logically and rigorously demonstrate that a particular statement or relationship between sets is always true, regardless of specific examples or cases.

What are some common techniques used in proving sets?

There are several techniques that can be used in proving sets, including direct proof, proof by contradiction, proof by induction, and proof by contrapositive. Each technique has its own strengths and is chosen based on the nature of the statement being proved.

Can you provide an example of a direct proof for sets?

Sure, a direct proof involves starting with the given assumptions and using logical steps to arrive at the desired conclusion. For example, to prove that the sum of two even numbers is always even, we can start by assuming that a and b are even numbers. Then, we can express them as a = 2m and b = 2n, where m and n are integers. Thus, the sum a + b = 2m + 2n = 2(m + n), which is also an even number.

What is proof by contradiction and when is it used?

Proof by contradiction is a technique where we assume the opposite of what we want to prove and then show that it leads to a contradiction. This proves that the original statement must be true. This technique is often used when direct proof is not possible or when the statement is negative in nature.

How does proof by induction work?

Proof by induction is a technique commonly used for proving statements that involve a variable, such as n, where n is a positive integer. The proof involves showing that the statement is true for the base case, usually n = 1, and then assuming it is true for some arbitrary value of n. By using this assumption, we can show that the statement is also true for n+1. This proves that the statement is true for all positive integers n.

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