Recent content by cocoabeens

Whoops, sorry yes, G(n)=2^(n/3). So for the induction step, would it look like this- base case: for n=3, 2^(n/3)=2*R(n-3) left hand side: R(3)=2^(3/3) =2 right hand side: 2*R(3-3)=2 (r(0) must be 1?) inductive hypothesis: for some k that is divisible by 3, G(k)=2^(k+3/3) =2*G(k) holds true...

Okay, it seems I was confusing it up with properties of power sets. For the second part, I just work backwards, correct? I show that the element {x}∈P(Y) because of the given condition, and thus x∈Y. Because x∈X, therefore X⊆Y? Did I confuse up some symbols?

Say there's a 3 block/pixel/square shape in L- formation that can be rotated on a board of size 2 x n. G(n) is how many distinct ways the board can be tiled. I need to show that if n isn't divisible by 3, then G(n) is 0. Given a block of three squares fitting on a board of size 2xn, and k...

If I have finite sets X,Y, and need to prove that X ⊆ Y <=> P(X) ⊆ P(Y), where P() denotes the power set of a set. I started out saying that for infinite sets X,Y, x⊆X, and y⊆Y. Given that X⊆Y, we want to show that P(B)⊆P(Y). x⊆X, so through transitivity, x⊆Y (is this correct?). From here, I...

Thanks for the response! So for the first part, I was told that I need to deliver a formal proof, and that a table going over the cases was an insufficient proof. Would a formal proof be something like this- "Based on the table (of cases), we have shown that for all integers x,if x is even, it...

Hello, new to the forums here. I need to prove that for all integers x, x^2 = 0(mod4) or x^2 = 1(mod4). I started out by making a table of different cases. case 1: y=0 ->0mod4 case 2: y=1 ->1mod4 case 3: y=2 ->0mod4 ... case odd: y=2n ->0mod4 case even: y=2n+1 ->1mod4 From here, I'm not...