Hi SSequence:
Yes.
If ##M## is a transitive model, then yes. For general models, the answer is no, see this stackexchange thread: https://math.stackexchange.com/questions/2118514/two-models-of-zfc-such-that-there-is-a-isomorphism-between-their-ordinals
Generally, forcing starts with a countable transitive model M of ZFC. Since it is countable, M will not even be close to containing all subsets of the naturals, so it is possible to add a new generic set G to M. G will not be in the L of M[G], since L^{M[G]}=L^M \subseteq M \subset M[G], but may...
Hmm... you and I have very different intuitions about what the most insightful solution is. I would consider mentioning that column rank = row rank to be far more insightful than offering a formula for the solution, since it tells us why there is a solution in the first place. Additionally, it...
The ##w_i##'s exist because each ##v_i## is in the range of ##X##. Saying that there is some ##w## such that ##Xw=v## is simply what it means for v to be in the range of ##X##. I'm really not sure what more I can say on that point.
As for ##Z##, I'm just using the fact that every linear...
All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since \sin (\theta) ~ \theta for small \theta) at the origin, the...
This limit does not exist. Note that the function isn't even defined on the path x=-y^2, and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.
Sure. Both players will almost surely be wearing at least one white hat. For i=1,2, let n_i be the position of the first white hat on player i's head. Then have player 1 point at hat n_2 and player 2 point at hat n_1. It is easy to see that both players win iff n_1 = n_2, which happens with...
No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but...
Yes, but not just those. We also have products involving a^{-1} and b^{-1}. So a list of all elements generated by a and b would include:
1,\ a,\ b,\ a^{-1},\ b^{-1},\ a^2,\ ab,\ ab^{-1},\ ba,\ b^2,\ ba^{-1},\ a^{-1}b,\ a^{-2},\ a^{-1}b^{-1},\ b^{-1}a,\ b^{-1}a^{-1},\ b^{-2},\ \ldots
But this...
Remember that the symmetric group is generated by the set of all transpositions, so it suffices to show that all transpositions can be generated by these two elements. Hint: first try to show that those two elements generate the set {(12), (23), (34), ..., (n-1 n)}. Then show that all...
If you just want the field to exist, certainly. If you want the multiplication on the field to be efficiently computable, probably not. Suppose that G is a group with a hard discrete logarithm. Note that for a discrete log to be well-defined over G, G must be cyclic. The only cyclic groups which...
This is incorrect. For i>4, i^(-3) < 1/4 i^(-2)
Edit: Ah, I think I see where you're confused. You seem to have missed the ellipses. The condition must hold for all natural numbers, not just for the first three.
I'm not sure. Certainly the theory PA + "all Goodstein sequences terminate" proves Goodstein's theorem without mentioning transfinite numbers, but it's probably not what you would consider a "natural" axiom system. I can state with certainty that adding axioms for exponents would not enable you...
Sorry for just posting the solution earlier, I was late to dinner. Simon, that's pretty creative :) I actually used a different (and somewhat more boring) method though:
Yeah, it's true. The proof is pretty easy, simply note that \{x \in X: P(x)\} is nonempty, open, and closed, so since X is connected, it must be all of X. This fact is the most commonly used way of proving that something is true on a connected domain.
Actually, this statement is a little bit weaker than induction. Induction states that (P(1) \wedge \forall n (P(n) \Rightarrow P(n+1))) \Rightarrow \forall n P(n). Your statement is equivalent to (P(1) \wedge \forall n (P(n) \Leftrightarrow P(n+1))) \Rightarrow \forall n P(n). Note the iff...
Such a dividing curve need not exist. Consider the following: Let A\subseteq \mathbb{R}^2 be defined by:
A = \{(x, \sin(\frac{1}{x})): x\in [-\frac{1}{\pi}, 0)\cup (0,\frac{1}{\pi})\} \cup \{(x, 0): x \in [-\frac{\pi}{2}, -\frac{1}{\pi}] \cup [\frac{1}{\pi}, \frac{\pi}{2}]\} \cup \{(0, y)...
First, it follows easily by induction on n that for each n, a_n = 2^{n-1}(p+1) - 1. By Euler's theorem 2^{p-1} \equiv 1 \mod p, and so a_p = 2^{p-1}(p+1) - 1 \equiv p+1 - 1 \equiv 0 \mod p, so p \vert a_p, but p \neq a_p, so a_p is not prime. Q.E.D.
I don't want to give away the answer, since I found it by googling this exact problem, so I'd just be copying someone else, but I will give you a couple of hints. First, the prime you should choose is one large enough that all of the numbers have distinct residues modulo that prime. Second, the...