Recent content by Cyborg31

Didn't know you could move -I to the other side like normal algebra. Is the B^-2 a typo? Cause I don't know what that's supposed to represent.

Homework Statement A, B, I-Y^-1, X, Y are n x n invertible matrices (A(I-Y^-1))^-1 = YB Solve for Y Homework Equations The Attempt at a Solution (I - Y^-1)^-1A^-1 = YB (I - Y^-1)^-1A^-1B^-1 = Y A^-1B^-1 = (I - Y^-1)Y A^-1B^-1 = Y - I A^-1B^-1*(A^-1B^-1)^-1 = Y - I(A^-1B^-1)^-1 I = Y -...

a - c = (b - d)(mod m) would be (a - c) - (b - d) = mx for some x But its a = (b mod m)

Homework Statement Show that if a = (b mod m) and c = d(mod m) and m => 2, then a - c = (b - d)(mod m) Homework Equations c = d(mod m) <=> m|(c - d) d = c + xm The Attempt at a Solution I don't know how any equivalences for a = (b mod m), is there a way to get b from a = (b mod...

Uh ok... what's wrong with what I did? I did c|ac = a, before and that apparently that's wrong so I used bc/c = b, this time. How else do I cancel out the c? Should it be c|bc = c|acy => b = ay ? Is my solution correct or wrong?

For a|b <=> ac|bc a|b = x ac|bc = y b = ax bc = acy bc/c = acy/c => b = ay If b = ax and b = ay then x = y a|b = x <=> a|b = y therefore a|b <=> ac|bc Is this correct? <=> is equivalence, not <->.

Are the grades curved?

I know, that's what I meant "c divides ac". a|b = b/a right? Doesn't it follow that c|ac = ac/c?

Is Epistemology a science? Seems kinda similar... took an Intro Philosophy course before and a lot of it is about AIs, Turing Test, etc. and whether they can think or not. Lots of mentions of Computer Scientists, Cognitive Scientists, Neurologists, Behaviourists/Psychologists.

Homework Statement If a, b < c, and d are positive integers, prove the following inferences. 1. a|b \wedge c|d \rightarrow ac|bd 2. a|b <=> ac|bc Homework Equations The Attempt at a Solution 1. a|b = x, then b = ax c|d = y, then d = cy bd = axcy thus ac|bd =...

Homework Statement 1. Prove the following subset relationships: a. Constant \subseteq Logarithmic b. Logarithmic \subseteq Linear c. n log n \subseteq Polynomial Homework Equations O(1) Constant O(log n) Logarithmic O(n) Linear O(n log n) n log n O(n^m), m > 1 Polynomial of order m The...

Actually nm, I figured it out.

1. Let f, g: R -> R where f(x) = ax + b and g(x) = x^2 - x + 1. If (g o f)(x) = 9x^2 - 9x + 3, determine the values of a and b. So I got g(ax + b) = (ax + b)^2 - (ax + b) + 1 (a^2*x^2 + 2(a * b)x + b^2) - (ax + b) + 1 = 9x^2 - 9x + 3 (9x^2 + 2(3 * b)x + b^2) - (3x + b) + 1 (9x^2 + 2(3 *...

Homework Statement Let h: Z -> Z be defined as follows: h(n) = 0 if n is even and 1 if n is odd Determine h^2, h^3, and h^500 Homework Equations The Attempt at a Solution If n is even, h(n) = 0 then h(h(n)) = h(0) = 0 If n is odd, h(n) = 1 then h(h(n)) = h(1) = 1 If n is even h(n) = 0...

Because of the brackets, x element of C and by the Union of (A n B) and C, x is an element of (A n B) u C. (A n (B u C)), has the brackets shifted over. So if x element of C, Union of C and B means x element of (B u C). I know x element of (B u C) but not A, but by the definition of...