@tb by their comment, I guess I answered the OP's question.

@JonasTeuwen Doing homework is quite hard. Some of the questions are absolute nonsense. Like questions about pointwise convergence.

@robjohn Maybe. At least one of the two.

@N3buchadnezzar That's what the people say flunk, yes.

Since the question is not about pointwise convergence.

@JonasTeuwen I am still jealous.

But maybe you have got your share in the past? ;-). I probably can only do this for a couple of years, so I'll try to maximize that :-).

I am taking 5 subjects this semester

and the next =)

@N3buchadnezzar what subjects?

Number theory, Calculus 1, Linear Algebra, Mechanics, Matlab

Matlab is a course??

Holy cow.

haha

Its called "Introduction to programming and IT"

Proprietary software as a course...

80% of the exam is pure matlab

The computer engineers have python instead =(

Yeah, that's way better.

Next semester: Calculus 2, Linear algebra 2, electromagnetics, philosophy and science history, Calculus 3.

Those courses (except perhaps electromagnetics) give me the willies.

But if it doesn't require calc 2 as a prerequisite it would probably also give me the willies.

Good luck with those.

^^

Normal is to take 4 subjects a semester.

Thinking about taking Algebra, or Geometry for the fall.

Not sure what branch of mathematics I want to study deeper.

@AsafKaragila ayt?

@N3buchadnezzar Analysis.

Yeah, during next fall I will be taking an introductory course to Real Analysis.

Great.

No rudin or baby

I despise "baby" as a name for that book.

Depends on how you define powers, rite?

I read some about dedikins cuts, and so on.

But I guess I qill learn some of this in Real Analysis, right?

Why would you need Dedekind cuts?

No, probably not. Maybe the definition of $a^x$ when $x$ is real.

so far, for me stuff like

$$ \large 2^{\pi} $$

Makes little sense, except if you consider the taylor serie of $\pi$

But still then, you are deppendent on the rules stated above to evaluate the expression.

@Matt ?

@AsafKaragila Sorry, internet was broken.

Askew.

Oh it's fine.

I capped out for the day, awesome.

Or do a barrel roll.

@AsafKaragila Can I ask you a question?

Sure!

What is the forcing poset?

Oh yeah, it's there. Sorry, I've had a beer and I have a big glass of arak with me here now. (It should only interrupt my immediate understanding, not my math skills.)

@AsafKaragila $(M_\mathcal{U}, \leq)$. Now I'd like to show that $m_G$ is an infinite subset of $\omega$.

$\mathcal U$ is a free ultrafilter, I'd guess?

I've not heard free before but it's an ultrafilter.

Free means that there is no $x$ such that $\{x\}$ is in the ultrafilter.

Looks like something that would follow from genericity.

I thought I could show that by contradiction, what do you think of this? That is, assume that $m_G$ is finite and then show that $G$ can't be a filter.

It is apparent in your other forcing questions too. You are uncomfortable with genericity arguments, but those are the main tools of forcing.

You simply add a brand new subset to the universe which has a very specific property. So avoiding genericity is just... not the right thing to do.

What do you mean I add a new suset to the universe?

This means that you want to show that for every $k\in\omega$ and $(s,x)$ there is some $(t,y)$ which extends it, and $t$ has at least $k$ elements. This means there are dense sets whose finite sets (left coordinate) get unboundedly high; the generic filter meets all those dense sets and thus the union is infinite.

@Matt The notation $M[G]$ is not the same as $\mathbb Q[\sqrt 2]$ for no reason. You add a new element to the universe.

Where did the "how many digits of 3^100" question go?

To sleep.

Hmm.

How would one go about figuring out that question ?

I'd say 9^50 has slightly fewer than 50 digits

Literally, I would say that, in an answer, if the question were not "asleep"

@AsafKaragila So for every $k \in \omega$ I construct a dense set $D_k$?

@Matt Quite.

And I would say that the value of $$ \large \int_{-\infty}^{\infty} \frac{\sin x}{x} \, dx $$ is slightly less than $4$. But to a mathematician this doesent cut it.

We want exact numbers ^^

@AsafKaragila Nice, thank you. I'll try that.

No, the question was about estimating the number of digits...

then

$$ \log\left( 3^{100} \right) = 100 \log(3) \approx 100 \cdot 1.1 = 110$$

But you can calculate it.

@N3buchadnezzar I'd be satisfied with the fact that it is bounded 8-). In analysis we rarely need exact numbers.

Great. 110 digits, base e.

I feel like I'm talking to Skullpatrol 2.0!

smacks my own head

(thank you!)

Ofcourse we need log base 10

Mmm, I do not know the value of log_10(3) in my head... Screw it puts forth calculator

Forbidden by the OP...

Challenge accepted

I mean, we could approximate (pen and paper) the mantissa of 9, and that should do

But there was all this talk about ln(3) ~ 1.1

And change of base. And now the question is gone.

$$ \log_{10}(3) = \frac{\ln(3)}{\ln(10)} \approx \frac{1.1}{2.3} \approx .48$$

Memorizing the approximate value of $\ln(10)$ is good for converting logarithms rather quickly.

Indeed.

Well that was fun. If anyone (10k+) has an intelligent answer for why the question is gone, ping me.

all you basically need to know is $\sqrt{2},\sqrt{3},\sqrt{5}$ and the same logarithms. We also could use the babylonian method for calculating the roots, but I often find this method slower.

Bonus question:

What are the two last digits in $3^{100}$?

Enjoying a nice piece of cheese with a fine wine. Ahh.

Enough with your cheese and wine.

Drink arak cut with coke or something.

I'll go for whisky after this.

Or captain morgan and coke

Just stop with the wine and cheese.

This is not a grad student's drink. This is a fat-cat grad student's drink. It is not like you.

Even wine is fun, if you drink enough

@Matt These sort of arguments are the main arguments with forcing. The sooner you get used to them, the better.

@N3buchadnezzar Yes, but after one big nice glass of arak I have taken more alcohol per volume than I would in a bottle of wine.

@AsafKaragila I know. I'm not trying to avoid it, I'm just really behind with things.

Ouzo !

@N3buchadnezzar Too sweet.

@AsafKaragila Really?

@N3buchadnezzar Quite.

@AsafKaragila I thought I could define $D_k = \{ (k,x) \mid x \in \mathcal{U} \}$ but that's not dense.

@AsafKaragila Asaf I guess you could try mixing Vinegar, lemons and Everclear. Or something from the physiclab.

@Matt Why not?

@N3buchadnezzar I guess you could try mixing arak with cyanide and semen.

I find Ouzo quite sour enough, thank you very much =)

@N3buchadnezzar Well, you are clearly not used to drink arak. When the pub is out of arak I get to drink Ouzo and it's just not quite the same. Arak is bitter, much like life itself.

@AsafKaragila Because dense would mean that I can pick any $(s,x) \in M_\mathcal{U}$ and then find a $(k,y)$ in $D_k$ such that $(s,x) \leq (k,y)$. But that would mean $s \subset k$ which is not the case for arbitrary $s$.

Yeah, arak tastes like some cleansing agent.

@JonasTeuwen Probably, seeing as it tastes of aniseed.

@Matt Then show that there is a definable dense set which does that, for example: $$D = \{(k,\omega\setminus k)\mid k\in\omega\}$$

@Matt Arak is much much much bitter than Sambuca, and much bitter than Ouzo.

@JonasTeuwen Your sense of taste is clouded by too much mayonnaise, Dutch-boy.

Ouzo is extremely sweet.

Where I live we use something called Karsk...

Basically it`s strong coffee mixed with 60 or 80

@AsafKaragila No, I want $D$ to depend on $k$.

@Matt Why?

@AsafKaragila I need $G$ to meet each $D_k$ otherwise I don't get the union to be unbounded.

@Matt But if $D$ is already dense then $G$ meets it, and it would do so unboundedly often (why?)

@AsafKaragila No, it meets it in at least one element. Now let me think about unboundedly often.

You can take, instead: $$D_k = \{(n,\omega\setminus n)\mid n>k\}$$

This is perverse.

@JonasTeuwen What is? That we talk about forcing in a set theory chatroom?

Yes.

@AsafKaragila This looks good.

@Matt Of course it does :-)

In general, proofs by forcing should go like this: "Density argument. No? Oh, right... a different density argument!"

Density arguments? :)))?

Yeah.

Generic filters meet all the dense sets.

Stuff like $C_c$ is dense in $L^p$?

I should probably not drink more tonight.

And then take limits?

@JonasTeuwen Only the topology is the order topology generate by cones.

@AsafKaragila Thank you. Just starting to grasp it.

@Matt This is the important part. After you understand that 93.6357% of forcing is about the right poset and the right dense sets, you're fine.

@AsafKaragila If I'm drinking more, sleep is a good idea.

@AsafKaragila This message is a self reference, but it's not a cliche! This is also a proof that the chat is an incomplete system.

@AsafKaragila Yes it is.

You are one big cliche.

@JonasTeuwen What about yo momma?

(is joke 8-))

So is ma momma, yes.