Homework Statement
So, I have a set
S = (x, y) \in \mathbb{R} | (x and y \in \mathbb{Q}) or (x and y \notin \mathbb{Q})
I want to find
T = \mathbb{R} minus S
so I am negating this and get
T = (x, y) \in \mathbb{R} | (x or y \in \mathbb{Q}) and (x or y \notin \mathbb{Q})...
Looks like you'll probably have to use induction on n.
1. Show that the recurrence relation is true for n = 1. (base step)
2. Assume that, for some n in the natural numbers, your recurrence relation holds. Try to show that n + 1 holds as well. (inductive step).
It's probably going to...
Perhaps you should try solving directly from the formula for Newton's Law of Cooling.
i.e. dT/dt = -k(T - Tambient)
where T = temp, t = time, k is a positive constant.
if you are considering the codomain to be (1/2, 1], then since your image is equal to the codomain it is both open AND closed by the very definition of a topology (the null set and the whole set are both open and closed)... but we aren't considering them in R, we are considering only (-1, 1) and...
f(V) = (sqrt(3)/2, 1]
f(V)c = {x in [-1, 1] | x not in f(V)}.
sqrt(3)/2 is not in f(V), so it IS in f(V)c. Thus, [-1, sqrt(3)/2].
If that were the complement, then f(V) union f(V)c would not be the entire interval... which is a basic complement law. right?
Call U the universe of discourse...
So, what you really have is [A|B], where B is the column vector of solutions, right? So when you multiply by the matrix you have by its inverse what you really get is [I|A-1B]. Remember when you right it in the form [A|B], it is really just shorthand saying Ax = B. When A is invertible, you have...
Well... lim log(n)/sqrt(n) = lim log(e^n)/sqrt(e^n), since in either case, e^n and n both go to infinity (one just does it faster!)
There are a lot of sequence theorems involved with sequences, functions, and combinations of both but I don't know which you have learned or are expected to know...
I was mistaken about f(V) before, here is what we have:
x takes values in R, but f(X) = [-1, 1].
So, f(V) = (sqrt[3]/2, 1]. But that is still open in [-1, 1].
so I am still missing something I think...