Bit_4

Suppose that you are told to toss a die until you have observed each of the six faces. What is the expected number of tosses required to complete your assignment? [Hint:If Y is the number of trials to complete the assignment, Y = Y_1 +Y_2 +Y_3 +Y_4 +Y_5 +Y_6 , where Y_1 is the trial on which the first face is tossed,Y_1 = 1, Y_2 is the number of additional tosses required to get a face different than the first, Y_3 is the number of additional tosses required to get a face different than the first two distinct faces, ...,Y_6 is the number of additional tosses to get the last remaining face after all other faces have been observed. Notice further that for i = 1,2,...,6, Y_i has a geome

I'm learning Italian and beside that damned "gli" sound I am having the most trouble with the r's. How do you roll or trill your r's? I've watched a few youtube tutorials but none of them really made sense to me and they contradict each other. Any advice?

There isn't a bijection between the empty set and the naturals, so does that mean it's technically uncountable, or am I thinking about this wrong?

How do you define a plane in R^n where n>3? I know how to do it in R^(3), but I'm not sure if you can just extend that to R^(n).

This is from Axler's Linear Algebra Done Right. My proof is below:

Let V be a finite-dimensional vector space.

Let a list (v_1,...,v_n) of vectors in V be linearly independent.

If (v_1,...,v_n) spans V, then we're done.

If it does not, then append a vector v_(n+1) not in the span of (v_1,...,v_n).

By the linear dependence lemma (a previous result in the text), the list (v_1,...,v_n,v_(n+1)) is linearly independent.

Repeat this process.

Since every linearly independent list has a length less than or equal to every spanning list (another previous result), the process must eventually terminate.

Once it does, we are left with a list that is lin

Theorem: In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

My version: If V is a finite-dimensional vector space, then for all lists t of vectors in V and for every spanning list p in V, if t is linearly independent, then len(t) ≤ len(p) (where len(x) is the length of x).

However, if you take the contrapositive, then you have: if len(t) > len(p), then t is linearly dependent. But since the quantifier is not negated as well, this is saying that the length of t is greater than the length of every spanning list, which I don't think makes sense. Did I go wrong somewhere?

I was working through the first exercises in Axler's Linear Algebra Done Right and I realized for one of them that I had invoked the statement "if u+v in U, then u,v in U". It seems intuitively true and it makes the exercise proof nice and simple, but it's not actually mentioned anywhere in what I have read so far (chapter 1).

Another way of stating the property of vector and subspaces is that there is a function that maps a pair of elements u and v from U to an element u+v in U. So for the converse to be true, that must mean that this function has an inverse, right? And for it to have an inverse, it must be bijective (right?). But I don't think it's injective because if it were then the

In his proof that Q is dense in R, he makes the claim

Hence there is an integer [;m;] (with [;-m_2 \leq m \leq m_1;]) such that [;m-1 \leq nx < m;].

Where [;nx;] is a real number and [;m_1,m_2;] are positive integers. Intuitively I can see how it makes sense but I'm not seeing how it's justified rigorously.

EDIT: This is from Principles of Mathematical Analysis a.k.a. Baby Rudin.

From our text, the Axiom of Archimedes is stated as "for all positive rational numbers a, there exists an integer n such that a > 1/n". I'm having trouble coming up with a proof of this just given the properties of the rational numbers, i.e. arithmetic, field axioms, and order.

So far I have that if a is expressed as a ratio of integers p/q, then pn > q for some n. After that I'm stuck.

EDIT: He actually just emailed me back and said he got the names backwards! Wayne is on the left and Wax is on the right.

Hello all, some of you may have seen my previous post asking which character was which on the cover of Alloy of Law. Well, I emailed Brandon himself and asked, and he told me that it is indeed Wax on the left and Wayne on the right, but he also said that Wax shouldn't have a gun. So that's that, in case anyone else wanted to know!

First of all, no spoilers please since I'm only half way through!

This question has been bugging since I'm pretty sure the guy on the left is Wax (bowler hat + gun), but is the other guy supposed to be Wayne? Or do I have them backwards, or am I just completely wrong? Again, since I haven't finished the book yet, just names or a simple yes/no to my questions will do. Thanks!

I'm reading through Gelfand's Lectures on Linear Algebra over winter break and I'm stuck on one of the early exercises. It reads:

Show that in an arbitrary basis [for the space of n-tuples]

e_1 = (a_11, a_12, ..., a_1n),

e_2 = (a_21, a_22, ..., a_2n),

.......................................

e_n = (a_n1, a_n2, ..., a_nn)

the coordinates N_1, N_2, ..., N_n of a vector x = (E_1, E_2, ..., E_n) are linear combinations of the numbers E_1, E_2, ..., E_n.

First of all, I'm confused how the scalar values N_1, N_2, ..., N_n can be linear combinations of anything since "linear combination" is defined in terms of vectors, not (just) scalars. Second, ignoring the first c