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

1. Feb 13, 2010

### Dustinsfl

1. The problem statement, all variables and given/known data
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/$$\int g(x)dx$$ from -$$\infty$$ to $$\infty$$).

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.

2. Feb 13, 2010

### owlpride

~(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: Feb 13, 2010
3. Feb 13, 2010

### owlpride

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."