# Show proofs

• Aug 12th 2007, 05:48 AM
ArchReimann
Show proofs
Can anyone help me with these proofs:

1. Prove that for any set A and B, the
$P(A \cap B) = P(A) \cap P(B)$
where P(X) = power set of X.

2. Prove that for every real number x, there is a real number y such that
$(x+y = xy) \leftrightarrow x \neq 1$

3. Suppose F and G are nonempty family of sets and every element of F is disjoint from some element of G. Prove that the union of all sets in F and the intersection of all sets in G are disjoint.

THANKS! MY MATH 109 GRADE DEPENDS ON THIS!
• Aug 12th 2007, 06:16 AM
ThePerfectHacker
Quote:

Originally Posted by ArchReimann
2. Prove that for every real number x, there is a real number y such that
$(x+y = xy) leftrightarrow x neq 1$

For any $x\not = 1$ choose $y=\frac{x}{x-1}$.
• Aug 12th 2007, 06:20 AM
ArchReimann
Sorry for the messed up statements.

Um, can you post the step by step proof please? I'm kind of cnfused about the structure of formal proofs. Thanks a lot!
• Aug 12th 2007, 09:04 AM
Rebesques
Quote:

Suppose F and G are nonempty family of sets and every element of F is disjoint from some element of G. Prove that the union of all sets in F and the intersection of all sets in G are disjoint.

Let the families be $\Im, \ \wp$ respectively. We have that there is some element $\{g_0\}\in \wp$ such that, $(\forall \{f\}\in\Im) \ \{f\}\cap\{g_0\}=\emptyset$. (1)

Suppose now the sentence "the union of all sets in $\Im$ and the intersection of all sets in $\wp$ are disjoint" is false. This means "the union of all sets in $\Im$ and the intersection of all sets in $\wp$ have a common element". Call this $f$. Certainly, $\{f\}\in \Im$ and $f\in \cap\wp$ (the intersection of this family). Now since $f\in \cap\wp$, we must have that $\{f\}$ intersects all elements of the family: $\{f\}\cap\{g\}\neq\emptyset, \ \forall \{g\}\in\wp$. This and (1) contradict each other.
• Aug 12th 2007, 09:09 AM
ArchReimann
OMG that's exactly what I wanted. My professor told us that he will give double points on #3 if we use prrof by contradiction. Thanks thanks thanks.

Now the other two..hmmm.
• Aug 12th 2007, 06:47 PM
Rebesques
Quote:

Originally Posted by ArchReimann
Can anyone help me with these proofs:

1. Prove that for any set A and B, the
$P(A \cap B) = P(A) \cap P(B)$
where P(X) = power set of X.

Double inclusion: We show $P(A \cap B)\subset P(A) \cap P(B)$ and $P(A) \cap P(B)\subset P(A \cap B)$.

For the first inclusion, let $s\in P(A \cap B)$. We will prove $s\in P(A) \cap P(B)$.

We have $s\subset A \cap B$, so $(s\subset A$ and $s\subset B)$. Then $(s\in P(A)$ and $s\in P(B))$, which gives $s\in P(A) \cap P(B)$. qed

The other inclusion is up to you :)

Quote:

2. Prove that for every real number x, there is a real number y such that
$(x+y = xy) \leftrightarrow x \neq 1$

Double implication. We must show $(x+y = xy) \rightarrow x \neq 1$ and $(x+y = xy) \leftarrow x \neq 1$.

For the first one. Suppose by contradiction $x+y = xy$ and $x=1$. Then $1+y = y$, a contradiction.

For the second one. Suppose $x\neq1$. Then choose for $y$what Hacker has proposed. The proof is now complete.