PROOF EVALUATION (This type of exercise will appear occasionally): Each of the follow- ing a proposed "proof" of a "theorem". However the "theorem" may not be a true statement, and even if it is, the "proof" may not really be a proof. You should read each "theorem" and "proof" carefully and decide and state whether or not the "theorem" is true. Then: • If the "theorem" is false, find where the "proof" fails. (There has to be some error.) • If the "theorem" is true, decide and state whether or not the "proof" is correct. If it is not correct, find where the "proof" fails.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Proof Evaluation**

(This type of exercise will appear occasionally.)

Each of the following is a proposed "proof" of a "theorem". However, the "theorem" may not be a true statement, and even if it is, the "proof" may not really be a proof. You should read each "theorem" and "proof" carefully and decide and state whether or not the "theorem" is true. Then:

- If the "theorem" is false, find where the "proof" fails. (There has to be some error.)
- If the "theorem" is true, decide and state whether or not the "proof" is correct. If it is not correct, find where the "proof" fails.
Transcribed Image Text:**Proof Evaluation** (This type of exercise will appear occasionally.) Each of the following is a proposed "proof" of a "theorem". However, the "theorem" may not be a true statement, and even if it is, the "proof" may not really be a proof. You should read each "theorem" and "proof" carefully and decide and state whether or not the "theorem" is true. Then: - If the "theorem" is false, find where the "proof" fails. (There has to be some error.) - If the "theorem" is true, decide and state whether or not the "proof" is correct. If it is not correct, find where the "proof" fails.
**Theorem:**

For any sets \( A \) and \( B \), if \( \mathcal{P}(A) \cap \mathcal{P}(B) \neq \emptyset \), then \( A \cap B \neq \emptyset \).

**Proof:**

We will prove this by contradiction. Assume \( A \) and \( B \) are sets, and to get a contradiction, assume \( A \) and \( B \) are *not* disjoint. Then there is some element \( x \) in \( A \cap B \). So \( \{x\} \in \mathcal{P}(A) \cap \mathcal{P}(B) \), which means that \( \mathcal{P}(A) \cap \mathcal{P}(B) \) is not empty, and we are done.
Transcribed Image Text:**Theorem:** For any sets \( A \) and \( B \), if \( \mathcal{P}(A) \cap \mathcal{P}(B) \neq \emptyset \), then \( A \cap B \neq \emptyset \). **Proof:** We will prove this by contradiction. Assume \( A \) and \( B \) are sets, and to get a contradiction, assume \( A \) and \( B \) are *not* disjoint. Then there is some element \( x \) in \( A \cap B \). So \( \{x\} \in \mathcal{P}(A) \cap \mathcal{P}(B) \), which means that \( \mathcal{P}(A) \cap \mathcal{P}(B) \) is not empty, and we are done.
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