Recent content by Moo Of Doom

The important thing is that non-singular matrices represent invertible transformations. Thus, if B is non-singular, then Bu = v has a unique solution (i.e. u = B-1v). Since we know B0 = 0, there can be no other solutions to Bu = 0 unless B is singular.

While the space 3x3 matrices operate on is 3 dimensional, the space of all 3x3 matrices is 9-dimensional, one example of a basis being the set of nine matrices with one 1 entry, and the rest zeros: \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}\right)...

A collection of points such that no point is identical to any other.

It seems to me, that there is a misunderstanding here, because you guys appear to be working in different probability spaces. It appears Jarle is arguing in the probability space of five independent die rolls (here, there are 65 different outcomes, each occurring with probability 1/65), with...

1. Nope, that's not quite correct. You are adding the probabilities that the first card is X, ..., that the seventh card is X, but this only works if these are mutually exclusive events, which they clearly aren't in this case. A simple way to calculate the odds would be to calculate the odds of...

You might want to look at http://www.research.att.com/~njas/sequences/A000798 for some references.

Indeed, for size m > 2, this does not exist, because it would imply that Sm is a cyclic group (i.e. generated by one element), which it is not for m > 2.

Why is a1 = 3? Can't you solve it in two moves by moving the discs one after the other to the target rod? Since the two discs are "identical," this should be a solution, right? Or are the discs also labeled, and need to be stacked in the same exact order on the target rod? Also, I have no idea...

The operation in Z30 is addition, so your subgroup elements need to be generated by addition (for example, 25 + 25 = 20 (mod 30) must be in your subgroup).

Another equivalent definition is that it has finite diameter, where \mathrm{diam}(S) = \sup_{x, y \in S}(\mathrm{dist}(x, y)). This is applicable to any metric space (though the Heine-Borel theorem is not!).

Yes, in a direct proof one assumes the premises are true, and derives statements from that.

Of course. You can use a premise as often as you want. It remains true throughout the entire proof.

You were fine up to and including step 5. (1) ~P premise (2) P v Q premise (3) From disjunctive simplification we got: (P v Q) ^ (~P) -> Q (4) (Q->S) premise (5) From detachment i.e (Q->S) ^ Q -> S Then here your next steps should be something like (6) S -> (S v R) by disjunction introduction...

(~P) ^ (P v Q) implies Q, but it is not equivalent to Q (since Q says nothing about whether or not P holds). Thus, you cannot simply substitute it for (~P) ^ (P v Q) in an expression. The steps don't need to involve substituting the things into the original formula, but just noting what other...

"B if A" is the same as "if A, then B." If you read carefully, you'll notice dE_logics said the right thing (except with an incorrect value for the limit. I don't know where Halls got 0.5381 from...).