Recent content by avec_holl

Swell! I agree that my earlier proofs weren't worded very well and that probably created some confusion. I'm sorry about that. So, as it stands, is my logic reasonable? I also realize that there are better solutions but I wanted to make sure I got everything cleared up with my logic before I...

Couple thoughts that might be a bit coherent. We know that: f(a) = C + D and f(-a) = C - D for some real numbers C and D. I define the function E such that is always has the property that E(a) = C and E(-a) = C; hence E is even. I define the function O such that O(a) = D and O(-a) = -D...

The proof should address the first question. To address your second criticism, I'll revise the proof . . . Suppose that f is a function whose domain is the set of real numbers, then f(a) and f(-a) both exist. Given any two real numbers f(a) and f(-a) there exist two real numbers C and D...

I'm still struggling with this. Given any two real numbers f(a) and f(-a) there exists two real numbers c and [/itex]d[/itex] such that f(a) = c + d and f(-a) = c - d. The number c has the property that if E(a) = c then E(-a) = c and the number d has the property that O(a) = d and O(-a) = -d...

Normally, I probably would use the case by case method but I'd already proved the lemma I used to derive the contradiction. So, using the lemma that I provided, do you think that (b) is a valid proof? Also, have you read through (a)? I recognize that the point of contention is where I...

I stated in the quoted bit that O cannot be the zero function which means that f cannot be even. The point of this lemma is: In (b) we have that the sum of an odd function and an even function is odd. The lemma proves that this can only be true if E is the zero function - if it isn't, we...

Alright, I'm a bit confused at the moment so I'll post my original argument with all my thoughts included . . . (a) Suppose that f is some function whose domain is the set of real numbers. This implies that both f(a) and f(-a) exist. Since, given any two real numbers c and d there exist two...

Here's what I thought: Given any two numbers c and d, there exist two numbers a and b such that a + b = c and a - b = d. However, I guess this doesn't apply (maybe it's a consequence of this theorem?). Maybe this is incorrect in general. I guess I'll work on this a bit longer then! Could...

How does this pertain to the current health care reform bill? The current bill would allow Joe, Sarah, Bob, Emily, Frank and Amy to keep private health care if they desired and does not cut funding from medicaid or medicare.

Homework Statement 13. (a) Prove that any function f with domain \mathbb{R} can be written f = O + E where E is even and O is odd. (b) Prove that this way of writing f is unique. Homework Equations N/A The Attempt at a Solution (a) Suppose that f is some function with domain...

Thanks for your reply! Sorry about the misnomer, my textbook refers to this result as the Remainder Theorem but I guess it isn't.

Homework Statement Prove that for any polynomial function f and number a, there exists a polynomial function g and number b such that: f(x) = (x-a)g(x) + b Homework Equations N/A The Attempt at a Solution Proof: Let P(n) be the statement that for some natural number n, f(x) =...

I just looked into Vector Calculus, Linear Algebra, and Differential Forms and it looks like an excellent textbook - very well reviewed. Thanks for the suggestion!

Does anyone have some suggestions for a good multi-variable/vector calculus book? I have a fairly reasonable math background - managed to self-teach myself calculus through Micahael Spivak's text Calculus pretty successfully - and I'm looking for something that's fairly rigorous. One of my...

Homework Statement Let S be a set of Natural numbers with the property that every even number in S is divisible by 5. Which of the following must be true. a. 2 is not in S b. 5 is not in S c. S contains all multiples of 10 d. Every even number in S is divisible by 10 e. S contains no...