Recent content by NastyAccident

Anybody?

Okay, so for the basis case just say that there is one move left and not worry about the month and day?

I guess that's where I am really bloody confused. I know there are twelve dates that guarantee victory. 1/20, 2/21, 3/22, 4/23, 5/24, 6/25, 7/26, 8/27, 9/28, 10/29, 11/30, 12/31. In the beginning, I choose that as the base case since I am trying to present something where no matter what the...

Homework Statement (The December 31 Game) Two players alternately name dates. On each move, a player can increase the month or the day of the month, but not both. The starting position is January 1, and the player who names December 31 wins. According to the rules, the first player can start...

Ahh, so in this case in point there was simply no intersection between the set with A being removed and the set with B being removed?

Homework Statement In any group of (n) people, if one person has brown hair, then everyone has the same hair color. Suppose: For any group of people everyone has the same hair color. Case 1: In any group of 1 person, everyone has the same hair color. Case 2: Now, with a group of k+1...

Okay, so my counterexample for No. 2 is: f(x)=|x| {-3,...,-1} = A and {1,...,200} = B There is no intersection between A ∩ B. However, there is an intersection with f(A) ∩ f(B) that gives the set {1,3}. Thus, {null} != {1,3}. Figured out the wording that I was missing =) Just need...

Well, rest easy. I just checked your work and it is definitely 2arcsin(sqrt(x)) + C . The answer that was given by the book is an unfortunate typo. Also, u = sqrt(x) is another way to use substitution.

Homework Statement Suppose f is a function with sets A and B. 1. Show that: I_{f} \left(A \cap B\right) = I_{f} \left(A\right) \cap I_{f} \left(B\right) Inverse Image of F (A intersects B) = Inverse Image of F (A) intersects Inverse Image of B. 2. Show by giving a counter example that...

First, I apologize for the (x-3)(x+1)>0 statement. It should have been (x-3)(x+1) < 0. I hit the wrong key. I now see and understand why the split is occurring like that since you are trying to find terms that create a negative which would yield something that is less than zero. Now, this...

Okay, so I think I follow but I want to make sure that my train of thought is correct. We have one case: (x-y)^2/(xy)>=0 with two subcases: xy<0 xy>0 There is no >= in the sub cases since zero no matter what cannot be in the denominator. When you introduce a negative x or negative y...

Okay, the way of factoring originally brought up confused me for a second since it appeared to be completing a square to a problem that could of already been factored. After reading your original suggestion, I had the thought of breaking it down to (x-3)(x+1) >0. Though, after that I was...

Going your way: x^2-2x+1-4 < 0 (x-1)^2-4<0 (x-1)^2<4 -1<x<3 0<x+1<4 x+1<(x-1)^2 x+1<x^2-2x+1 0<x^2-3x 0<x(x-3) x=0 or x=3? Somewhere near that?

Homework Statement (!) Assuming only arithmetic (not the quadratic formula or calculus), prove that \left\{x \in\Re: x^2-2x-3 < 0\right\} = \left\{x \in\Re: -1 < x < 3\right\} Homework Equations \left\{x \in\Re: x^2-2x-3 < 0\right\} = \left\{x \in\Re: -1 < x < 3\right\}...

So, cases: 1.) xy <= 0 2.) xy >= 0 x/y + y/x >= 2 x^2/xy + y^2/xy >=2xy/xy (x^2+y^2-2xy)/(xy) >= 0 (x-y)^2/(xy)>=0 (x-y)^2 >= 0 The first case contains all real numbers in the first and third quadrants so results like (1,1), (-3,-1), et cetera work. The second case contains all...