Logic - 2018-07-27

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user400188

03:57

@user21820 I see your points. The matrix formulation wont work under a probabilistic view unless the variables are independent. If they are not, associativity doesn't hold.

What I meant by "there are enough cases w[h]ere it will hold", was that you can often be sure events are independent or close enough to it, that this formulation would be of use.

user21820

Yea. There is this discrete feel to boolean algebra. It feel very special to me, because of the lattice structure. I don't have a problem with saying that we can look at independent probabilistic events and consider the probability of their conjunctions and disjunctions, but that is very far from logic, because practically none of the usual logic rules will hold.

In particular, it is true that P(A∧B) = P(A)·P(B) when A,B are independent, and P(A∨B) = 1−(1−P(A))·(1−P(B)) under the same condition, but that's about it.

Everything else has to still pass through the normal logic, so for example P(A∨¬A∧B) = P(A∨B).

Where "P" denotes probability.

Leaky Nun

Do you dream of waking up three times, and doubt reality when you finally wake up?

user21820

@LeakyNun Wow you have that dream too? I think in my whole life I had that kind of dream twice. In one occasion, I knew I was dreaming when I dreamed of having woken up, and so I was desperately trying to actually wake up. And I 'escaped' like 3 layers before finally waking up.

I didn't doubt reality though, because somehow I know when I am truly in the real world hahaha..

Leaky Nun

ain't your life simple

user21820

Lol... is there any distinction between your dreams and reality? Apparently some people dream only in black and white (especially those who grew up before colour TV).

user400188

04:11

I would expect only a portion of your brain does the dreaming. So you should be able to tell what is reality by receiving a signal from the other parts.

user21820

@user400188 Is this true for you? For some of my dreams I am unaware that it is a dream.

Until after I wake up.

But my dreams are never as sharp as reality, which is why I can recognize that they are dreams once I am awake. It is rare that I am cognizant of dreaming while still dreaming. Though I'm not sure it was a nice experience to know that when I was stuck in the waking dream...

Leaky Nun

I'm now thinking about this problem

if A is an n-by-n matrix with coefficients in the ring R

then det(tI_n - A) is a monic polynomial in R[t]

user21820

@LeakyNun: Did you ever observe water flowing in your dreams? I think my brain cannot render complicated fluid mechanics properly, so I can't dream it in high resolution.

Leaky Nun

lol

user21820

I used to be able to fly in my dreams though. It's been years since I last did. Kind of miss the weightless feeling.

@LeakyNun This sounds very familiar...

Leaky Nun

04:18

heh

user21820

But monicity is simple, isn't it? Only one term in the determinant can have degree n.

Leaky Nun

it's called the characteristic polynomial

@user21820 so basically by induction, expanding along the first row?

how can you prove that rigorously?

user21820

@LeakyNun No it's direct. I always use the definition of determinant as the sum over all permutations.

Leaky Nun

aha

user21820

Times the sign of the permutation of course.

Leaky Nun

04:19

ok that was easy

user21820

I find that it always makes things easy.

Leaky Nun

indeed it does

in Mathematics, 6 mins ago, by Leaky Nun

Let $R$ be a ring and $M$ a finitely generated module, say $M = \langle x_1, \cdots, x_n \rangle$. Let $\varphi \in \operatorname{End}_R(M)$. Let $\varphi(x_j) = \sum_i r_{ij} x_i$. Now we build a matrix $A = (a_{ij}) \in M_n(R[\varphi])$ such that $a_{ij} = \varphi \delta_{ij} - r_{ij}$. $\det A \in R[\varphi]$. Now, $M$ is an $R[\varphi]$-module. One can check that $A$ defines the zero transformation $M \to M$, so $\det A = 0$ by adjugate magic, and CH follows.

CH in one single-line message

user21820

@LeakyNun CH is Continuum Hypothesis. You proved it?? =P

Leaky Nun

lol

I suppose I'm in the logic room now

user21820

Wait are you trying to pull a fast one in the Math room?