~(p->q) of part of A Probabilistic Proof of Wallis's Formula for pi

  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Formula Pi Proof
Click For Summary
SUMMARY

The discussion centers on the negation of a theorem related to continuous functions and probability distributions, specifically addressing the statement that if g(x) is a nonnegative continuous function with a finite integral, then a*g(x) can be a continuous probability distribution. The negation of the implication P implies Q is established as not P or Q, where P is defined as "g(x) is a non-negative continuous function whose integral is finite." The correct negation of P is articulated as "g(x) is a function which is not continuous or assumes negative values or whose integral diverges." This clarification resolves confusion regarding the necessity of negating both parts of the implication.

PREREQUISITES
  • Understanding of continuous functions in calculus
  • Familiarity with probability distributions
  • Knowledge of logical implications and negations
  • Basic concepts of integrals and their properties
NEXT STEPS
  • Study the properties of nonnegative continuous functions in probability theory
  • Learn about the implications of the negation of logical statements in mathematical proofs
  • Explore the relationship between integrals and probability distributions
  • Investigate Wallis's Formula and its applications in probability and statistics
USEFUL FOR

Mathematics students, educators, and anyone interested in advanced calculus, probability theory, and logical reasoning in mathematical proofs.

Dustinsfl
Messages
2,217
Reaction score
5

Homework Statement


Write in symbols the negation of the theorem stated in part a.

part a:We immediately see that if g(x) is a nonnegative continuous function whose integral is finite, then there exists an a>0 such that a*g(x) is a continuous probability distribution (take a=1/[tex]\int g(x)dx[/tex] from -[tex]\infty[/tex] to [tex]\infty[/tex]).

The negation of P implies Q is equivalent to not P or Q.

The issue is with the P part.

Is ~P= g(x) isn't a nonnegative continuous fuction whose integral isn't finite?

I am not sure if both is parts need to be negated or just one.

Then for the Q part it is just itself starting with the existential quantifier.
 
Physics news on Phys.org
Dustinsfl said:
The negation of P implies Q is equivalent to not P or Q.

~(P => Q) is equivalent to "P and not Q."

Since you don't need the negation of P, does that solve your problem?
 
Last edited:
Just for completeness sake:

I suppose P is the statement "g(x) is a non-negative continuous function whose integral is finite." I would re-write this as "g(x) is a function which is non-negative and continuous and whose integral is finite" to emphasize the internal logical structure. Since ~(a and b and c) = ~a or ~b or ~c, the negation of P is:

"g(x) is a function which is not continuous or assumes negative values or whose integral diverges."
 

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
945
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
9
Views
6K
  • · Replies 100 ·
4
Replies
100
Views
13K
  • · Replies 10 ·
Replies
10
Views
2K