Recent content by ayusuf

Homework Statement If L is a language over \Sigma then lim n -> inf of Ln = \Sigma* iff (\Sigma\cup{\lambda})\subseteq L Also Lk = {x1x2...xk | x1, x2, ...xk \in L} Homework Equations The Attempt at a Solution I started by saying there is an w is an element of sigma then it is...

Okay for the first part kind of what I was saying that c = dk. It's like saying 12 = 6*k where k = 2 right? I know it's pretty clear but I just want to make sure I fully digest this stuff. Okay now for p divides b2 because according to Euclid's lemma it must divide one of the factors right...

Okay I kind of get it. Thanks!

The reason your saying b2 is a multiple of p is because c/d = k and if we isolate c we get c = dk. So in this c is b2, d is p and k is c2q right? also how do go from p dividing b2 to p dividing b? Thanks.

Right so from that example it would be wrong to say that A \subseteq B but rather we should say A is a proper \subset of B because A \neq B.

Okay now I understand what you meant by multiple but where are you getting that famous fact from? Okay so now if we isolate b2 in this equation we will have b2 = pc2q and by taking the square root of both sides we'll have b = sqrt(pc2q) = csqrt(pq). Why would p divide b?

But everytime A \subseteq B that must mean A = B right? If not please give me an example. Thanks.

So from a2q = b2p which implies that p divides a2. I understand up to there but when you say write a as a multiple of p I get confused. Could you elaborate a little bit more on that? Thanks.

Yes exactly so if A \subseteq B then every element in A must be in B and if A does not equal B then A is a proper \subset of B.

If A \subseteq B does that mean A = B which means B = A because if A is a proper \subset of B then A does not equal B right. I am wrong right?

Prove, if p and q are distinct prime numbers, then sqrt(p/q) is irrational. I know how to prove that if p is a distinct prime number, then sqrt(p) is irrational. From there let sqrt(p) = q/r and then prove but for this I'm stuck. Do we let sqrt(p/q) = (a/b)/(c/d).Thanks.

Cool, Thanks! :D

Okay I kind of get. By showing that for k>=3 we know (k-1)^2 >= 2 which originally means 2k^2 >= (k+1)^2 for k>=4 but by proving the first inequality we have proved the previous inequality and thus by proving the previous inequality this means we have proved 2^k >= k^2 for k >= 4. Am I right?

It is 3 because (3-1)^2 >= 2 because 2^2 >= 2 which is 4 >= 2 but what does that prove then?

Okay so from k^2 - 2k - 1 >= 0 for k>= 4 we can rewrite it as (k-1)^2 >= 0 for k>= 4. Yes I'm sorry but I still don't see it.