My difficulties in the text, How to Prove It 2e by Velleman

  • Thread starter Thread starter Remi Aure
  • Start date Start date
  • Tags Tags
    Difficulties Text
Click For Summary
SUMMARY

This discussion focuses on solving Exercise 1-(c) from "How to Prove It: A Structured Approach, 2nd edition" by Daniel J. Velleman. The problem requires analyzing the logical form of the statement \{n^2 + n + 1\ |\ n \in \mathbb{N}\} \subseteq \{2n + 1\ |\ n \in \mathbb{N}\} using specific logical symbols while avoiding certain set notation. The user initially struggled to derive the solution as presented in the appendix but ultimately arrived at a correct logical equivalence: ∀n ∈ ℕ ∃m ∈ ℕ (n^2 + n + 1 = 2m + 1). This solution aligns with the example provided on page 74 of the text.

PREREQUISITES
  • Understanding of logical symbols: ∈, ∉, =, ≠, ∧, ∨, →, ↔, ∀, ∃
  • Familiarity with set theory concepts and notation
  • Basic knowledge of mathematical logic and equivalences
  • Experience with self-study techniques in mathematical proofs
NEXT STEPS
  • Study the logical equivalences used in mathematical proofs
  • Review set theory definitions and notation, focusing on the symbols allowed in logical expressions
  • Practice deriving logical forms from set inclusion statements
  • Engage in discussions on platforms like MathHelpForum for collaborative problem-solving
USEFUL FOR

Students studying mathematical logic, educators teaching proof techniques, and anyone seeking to enhance their understanding of set theory and logical expressions in mathematics.

Remi Aure
Messages
2
Reaction score
0
I'm creating this thread for any major difficulties I come across in How to Prove It: A Structured Approach, 2nd edition by Daniel J. Velleman. This is a self-study (with any assistance if I can get it).

Homework Statement



The problem is to analyze the logical form of the given statement, using only the symbols \in, \notin, =, ≠, \wedge, \vee, \rightarrow, \leftrightarrow, \forall, and \exists in our answer, but not \subseteq, is-not-\subseteq, \wp, \cup, \cap, \, \{, \}, or \neg. (Thus, as the text goes on to say, we must write out the definitions of some set theory notation, and we must use equivalences to get rid of any occurrences of \neg.) This is Exercise 1-(c) on page 81.

Homework Equations



The statement is \{n^2 + n + 1\ |\ n \in \mathbb N\} \subseteq \{2n + 1\ |\ n \in \mathbb N\}.

The Attempt at a Solution



Although it seems correct to me, it's not exactly the solution given in the Appendix of solutions. His answer seems to be derived from facts that I can't find or discern from any of the previous discussion. What I ideally want is to see his solution derived in a way that I should've known from what's been explained in the text so far. If not that then one or two standard definitions or logical equivalences that might or might not have been introduced but nonetheless work to get me to the same result.

My answer:

\{n^2 + n + 1\ |\ n \in \mathbb N\} \subseteq \{2n + 1\ |\ n \in \mathbb N\}​
\equiv \forall x(x \in \{n^2 + n + 1\ |\ n \in \mathbb N\} \rightarrow x \in \{2n + 1\ |\ n \in \mathbb N\})​
\equiv \forall x(x \in \{y\ |\ \exists n \in \mathbb N (y = n^2 + n + 1)\} \rightarrow x \in \{y\ |\ \exists n \in \mathbb N (y = 2n + 1)\})​
\equiv \forall x(\exists n \in \mathbb N (x = n^2 + n + 1) \rightarrow \exists n \in \mathbb N (x = 2n + 1)).​

His answer:
. . . \equiv \forall n \in \mathbb N \exists m \in \mathbb N (n^2 + n + 1 = 2m + 1).​
 
Last edited:
Physics news on Phys.org
After a day, I made a parallel thread at mathhelpforum.com,

http://mathhelpforum.com/advanced-m...ext-_how-prove-structured-approach-2-ed_.html.

I eventually found a solution while interacting with that thread, but since Physics Forums is broader in scope and since these epic study threads are the only kinds I plan to make over a long period, I'll keep the developments of the cross-site threads congruous.

Critical or other feedback is always welcome.

New Solution to Exercise 1-(c) page 81

If one of the examples on page 74 says that the logical form of \{x_i\ |\ i \in I\} \subseteq A can be analyzed as \forall i \in I(x_i \in A), then the logical form of the statement \{n^2 + n + 1\ |\ n \in \mathbb{N}\} \subseteq \{2n + 1\ |\ n \in \mathbb{N}\} can be analyzed as

\equiv \forall n \in \mathbb{N} (n^2 + n + 1 \in \{2m + 1\ |\ m \in \mathbb{N}\})

\equiv \forall n \in \mathbb{N} (n^2 + n + 1 \in \{x\ |\ \exists m \in \mathbb{N} (x = 2m + 1)\})

\equiv \forall n \in \mathbb{N} (\exists m \in \mathbb{N} (n^2 + n + 1 = 2m + 1))

\equiv \forall n \in \mathbb{N} \exists m \in \mathbb{N} (n^2 + n + 1 = 2m + 1).​
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K