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12:08 AM
@Semiclassical. You don't find convex more applied than real analysis (less proofs and less abstract is what I refer to by applied):0
12:19 AM
that nope was in reply to your earlier message
i.e., i havent' taken a course in convex optimization
12:30 AM
got it! thanks
if we have some drinks to sell. If it costs $1$ to buy coke (C) per pint that we can sell for $2$. And it costs $1.5$ dollar per pint for sprite (S). We also have twice as much of C pints for each 1 pint of S, meaning that $C\ge 2S$. Also, the company does only cell 3000 of pints from both C and S per week.

I am trying to build up constraints here please:
why the $Profit = C + 1.5S$ and not $profit = 2C + 3S$ based on how much we can sell them?
you probably don't want to use $'s for your currency here, b/c of mathjax
I forgot to use \
profit is revenue less expenses, so for every 1 coke you buy and sell you make 1 dollar
Oh!! Profit = value of cell - value of sell!
12:38 AM
I don't know why people are patient on me here
I should leave math site
you haven't said what sprite sells for, but i'm guessing 3?
I am sick of myself
yeah S buy for 1.5 and cell for 3
C buy for 1 and sell for 2
So it makes sense now for profit
Is my second constraint right throgh pls C≥2S
not sure. if it's twice as many units of C as for S, that'd be C=2S not an inequality
Because company sell any amount of C and S
but i'm guessing that's not quite what was assumed
12:41 AM
So I guess it's twice as much
well, you said "We also have twice as much of C pints for each 1 pint of S"
which sounds you'd be assuming that you sell exactly twice as many C as S, not at least twice
so may want to check the wording
You are right
your statement is a bit ambiguous to me. how much does Sprite sell for? also, i don't understand "only Sell 3000 of pints from both C and S per week". does that mean $C+S \le 3000$ or $C \le 3000, S \le 3000$?
It's written exactly as this: the company sells one pint of S for every 2 pints of C
So that means $2S = C$.
12:46 AM
Got it. Also, I see that C+S<= 3000 clear to me @copper.hat.
That is not clear to me from the wording.
@TedShifrin It's some kind of geometric description of cup product. It's written in Hatcher. I didn't understand it fully so doubt my argument. I thought you already know that.
No the constraint you wrote is right
Finally, if we are to draw the goemetric shape, what do you think about this pls:
I wrote two different sets of constraints, so I must also be wrong.
@copper.hat. Nope you are right.
We have 3 constraints
also both are positive
S>0 and C>0
12:49 AM
I think you did not read what I wrote.
I thought two $2$-cells in $CP^2\# CP^2$ are disjoint so they don't intersect
@Semiclassical. @copper.hat. This is the final geometric shape
Profit is found when we sell 2000 of C and 1000 of S
Now I have one question about graph though. X-axis represents amounts (pints) of C and Y-axis represents amoung (pints) of S. So, how is it possible that we draw profit to be $3500 on x-axis. Not sure if my question is clear pls?
12:53 AM
The wording is too ambiguous for me. I would read it as $2S=C$.
Also we are not allowed to have 3000 of S as the company only sell 2 of S for 1 C...So not sure about figure above :/
I don't want to spend time guessing what the question is.
I will write it again.
math.stackexchange.com/a/187417/668308 I have some question in this answer. It says $H_n(M\# N,S^{n-1})\to H_{n-1}(S^{n-1})$ is surjective because the boundary of $M$ minus a disk has boundary exactly a fundamental class of $S^{n-1}$. But why this is true? I mean the topological meaning of boundary same as image of cycle under boundary map?
Only do so if the wording is less ambiguous, there is no point in repeating.
12:57 AM
Company will only sell Moe one pint of Sprite (S) for every two pints or MORE of Coke (C) bought. Company does not sell more than 3,000 pints per week.
I forgot something :/ MOREEE
that'd do it
my mistake sorry
As you see profit is draw for each amount we found on both y and x axes although they should represent amount of pints'
Then I get $2S \le C, C+S \le 3000, 0 \le C, 0 \le S$.
It's just no clear to me how the figure is drawn
@copper.hat. Yep, my bad
If I understood correctly, the profit is given by the functional $C + {1 \over 2} S$.
Surely you can draw the area represented by each constraint separately?
Non negativity means it must be in the 1st quadrant. The $C+S \le 3000$ area is the area below the line $C+S=3000$, etc.
The picture above would seem to indicate that the profit of each $S$ is $1.5$. That differs from your statement above.
1:06 AM
@love_sodam I know the material very well, but your sentence was just plain wrong. Yes, you can choose disjoint $\Bbb CP^1$s in the two summands. When I inquire, you should reflect carefully on what you’ve written.
@copper.hat. I mean should not the x-axis represent pints of C and y-axis pints of S?
So, we are drawing the profts then and not pints above as it seems
For each pair of pints input we draw profit above
So what we have on both x-axis and y-axis is proft of S and C and not pints
To solve it based on constraints we have, we found the maximim profit is (2000, 1000)
@copper.hat. As S + C <= 3000, S+C <= S + 2S <=3000, so we have S <=1000, same for C<=2000. So we find the objective find substituion here pls?
@TedShifrin $X=CP^2\# CP^2$. As in Example 3.7 in Hatcher, I chose dual bases in
$Hom(H_1(X),\Bbb Z)$ correspond to two $2$-cells in $X$. Then since it's a connected sum, two duals don't intersect. So trivial cup product. Does it make sense?
Have you reread the sentence I complained about? You need to be precise with language. What you just wrote now is wrong, but it’s restating what I saud.
1:25 AM
@TedShifrin the word 'geometric description of cup product?' I wrote?
No, dammit. The original sentence to which I responded to which you responded. You can be so frustrating to deal with.
You mean 'they do not intersect transversely'
Guys, another question.

An industrial factory produces pieces of which 1/5 of them are defective. Two buyers A and B, rate these acquired pieces in two categories: I and II, paying $1.20 and $0.80 respectively as the following style:

Buyer A: Gets five pieces sample. If he finds more than a defective piece, he rate the batch as II.
Buyer B: Get ten pieces sample. If he finds more than two defective pieces, he rate the batch as II.

What is the probability of the buyers A and B found good pieces?
1:42 AM
@TedShifrin Oh once they don't intersect, they vacuously transverse...
We have $b \lor \lnot b \equiv T$
Also $b \lor \lnot b \equiv b$?
Not sure though why the first was written like that
Is the second statement true though?
@MatheusSousa i don't see the logic for your buyer A (or buyer B, but A is simpler). what ways are there to get less than two defects?
@Avra I'm not sure what your question was. There is a line indicating the points that generate equal profit, one of them is for \$3,000 profit. The line intersects the $S=0$ line at $C=3,000$ which shows that (as expected) there is a profit of 1 for each $C$ sold. The same line intersects the $C=0$ line at $S=2,000$ which shows that (contrary to the statement above) that the profit per unit $S$ is 1.5.
@Semiclassical suppose he gets a batch with pieces (of anything). If he finds more than one defective piece, he says this batch is bad batch quality. So, He will pay only $0.80 per piece.
@Avra The units on both axes are pints.
1:52 AM
@copper.hat. I was just asking the way graph is drawn. We should have S on x-axis and C y-axis, i.e., amount of pits for each drink
Though we can see profit lines intersecting both x and y axes
uh, okay? i'm not asking about the prices
@copper.hat. Okay, so why we have profit = $3500 intersecting x-axis if that represent pints pls?
Ah sorry, I understood now.
@Semiclassical Basically, don't have any form to pick up less than two defective pieces
Ah sorry
I made the wrong question
@Avra Each point on the plot corresponds to a particular $(C,S)$ pair, where each value is a number of pints. The profit in dollars for a given $(C,S)$ pair is $C+ {3 \over 2} S$. The lines shows points of equal profit. But the dimensions on the axes are pints.
@TedShifrin I think I should've said that not because they intersect transversely, but : Let $\alpha\in Hom(H_2(X),Z)$ be the dual of $a$ assigning the value $1$ to $a$ and $0$ to $b$ where $a,b$ are two $2$-cells in $X$ and similarly to $\beta$. Representing $\alpha,\beta$ by counting the number of intersections of each $2$-cells, $\alpha\cup\beta = 0$ since $2$-cells are disjoint.
1:58 AM
The right question is: What is the probability of the buyers A and B to find 1 good piece?
@copper.hat. tnx. Why 3/2 S?
that's what you started with
I see it's C+2S
are you trying to get to 0.737 for A, and 0.678 for B, or something else
2:00 AM
@Avra You need to do a little work yourself. Look at the line on the plot corresponding to a profit of \$3,000. Work out the equation of that line.
then my question still stands. for A, what does a I-rated sample look like?
for buyer A he only can get at most 1 defective piece
and 4 I-rated sample pieces
okay. so if we denote the pieces as G (good) or D (defective), one possibility is GGGGD.
what are the rest of A's options?
2:02 AM
the buyer B only can get 2 II-rated sample pieces.
@Avra Did you work out the equation of the profit=\$3,000 line?
@Semiclassical It's the permutation of the possibility you give now
yes, almost.
@copper.hat. I am thinking about it
and there's one more option
2:03 AM
It is a straight line, you just need the slope since you have the intercept.
so what's the probabilities of these cases?
for GGGGG, GGGGD, etc
5 * (4/5)^4 * 1/5
@Avra Quickly.
2:05 AM
why 4 * ?
is 5
okay. so crunching that, it's (4/5)^4+(4/5)^5. what does that come out to?
@copper.hat. I just used $2000S + 3000C = 3000 profit$
Not sure though this is correct to calcuate why 3/2 S, then I used fact that C=2S
This does not seem correct
@Avra The line of constant profit is not related to the sales constraints????
@copper.hat. As I told you the graph confuses me :/
2:08 AM
@Semiclassical $ \approx 0.737 $
the calculation for B is more of the same
I didn't think this. Thank you
@copper.hat. I will think about it more and see. Thanks.
2:10 AM
@Avra It just represents $(C,S)$ points. If there is some relation on this you can plot it on the graph.
One relation is the constraint $2S=C$.
So for each pair (C,S) we have profit
This is how we draw all lines of profits?
Another is $C+S \le 3000$. Another is $C \ge 0$ etc
@copper.hat. I mean pair (C,S) corresponds to profit and this is how we draw all profit lines above given the constraitss of C and S?
Then there is a formula for the profit $P(C,S) = C + 1.5S$. This is just a formula. If you pick a particular profit, say \$3,000, then $P(C,S) = 3,000$ will corresponds to some points on the graph.
@Semiclassical I didn't think this because I thought when he picks up the pieces, doesn't matter in what order he gets because the act of get bad or good pieces is independent.
2:13 AM
@Avra I need to go. The profit line is not related to the constraints.
@copper.hat. Thank you.
yeah, this is one where you gotta think through the cases
@copper.hat. Lastly, so the points we draw from p(s,c) have nothing to do with profit
@Avra You need to read what I write.
The formula $P(C,S)$ gives the profit corresponding to the particular $C,S$ value.
our NFT scam is never going to get off the ground at this rate
2:33 AM
i want to write an app for google called Ad nauseum.
2:59 AM
@copper you mean ad nauseam.
really great exhibit on the history of fake barf at the nauseum. don't miss the gift shop.
3:49 AM
@TedShifrin I was being proper :-)
the lawrence hall of science had many lowest common denominator displays, usually involving snot or barf.
i remember the terminal where you could play nim. and some polarized glass you could rotate to change your view of the bay.
i don't remember any snot stuff. i was a higher class of person i guess.
they had some cool stuff with real scientific principles involved, but they lost out to the beavis & butthead displays.
this must have been part of a general cheapening of the culture. i heard the exploratorium got dumbed down a bit when they moved it to the waterfront.
although some of the smart stuff is dumb too. those pendulums that knock over dominoes once an hour (or whatever) as they rotate are moronic and every science museum has one.
it's the perfect answer to, what makes the world's most visually boring exhibit.
or is it that we're rotating? whatever the point is, i missed it.
4:06 AM
i do like Foucault's pendulum, if for no reason than it shows that there is a greater force at play, and it keeps flat earthers at bay :-).
there's just a guy who comes in and nudges it a little bit every once in a while. or the chain it swings on is motor driven. the earth is still flat.
i went to the hyatt (?) 360 restaurant in SF a few times. it was much less interesting than i thought.
somehow i expect more action from a rotating restaurant
the fun of the hyatt is throwing paper airplanes from the upper floors of the indoor atrium until people from the hotel catch on and try to find who is doing it and eject them.
haha. I never tried throwing paper airlines :)
high anxiety.
4:10 AM
yes, the high anxiety hotel. that's what it will always be to me.
hmm, not sure michael o'leary would like me throwing him
i think i brought a few dates there for cocktails. its strange that my memories are more of places...
i was in the lobby of the hyatt the morning of december 31 1999 and saw more balloons being filled and loaded into nets than i have ever seen elsehwere.
sounds like fun :-)
lots of goofy memories of that time. the dot com bubble hadn't burst yet. some of those startups threw parties that would have been great, had they not been at the office.
i was so sad when the bubble burst.
4:22 AM
it's good that it did, it kept you closer to the people. otherwise you'd be shooting people into space and designing electric cars now.
i did save a few folks from going into dot com disasters.
i'm going to make a fortune by designing a foucault pendulum that doesn't rotate. i can sell it to flat earth museums.
you could call it the o'fouc pendulum.
if i work on code for a while and then return to maths it is like someone wiped my brain.
i get that feeling after talking to particularly difficult attorneys.
i think having the pendulum large enough that it could kill people would be an attraction.
4:36 AM
another good one would be a gimmicked pendulum that physics teachers use. you know, where they take this really heavy weight, hold it right up to their nose, then let it go. gimmick one of those so it can hit back every once in a while.
like a lottery that you don't want to win.
there is some gootube video with a physics prof (i think Dutch) who shows some conservation principle with a large & potentially dangerous pendulum. he is an antidote to flat earthers.
yes, i want a pendulum with some chaos built in. in fact, i think it would be a good addition to congress.
now is maybe the best time in human history to be a loud idiot in congress. it's always been a great time, but these days it's really something else.
you don't need to think, just be a rabble rouser.
it's literally everything we tell children not to do.
my daughter would be very evil in congress.
thankfully the swamp was drained recently.
5:15 AM
so that's sorted
5:33 AM
@Ted: we have to let our pup go tomorrow. She has two forms of cancer and a decubitous ulcer (like a bedsore on a joint) on her left front elbow from leaning on her left side (due to the edema in her right rear leg). The ulcer can't heal while chemo is being given and the cancer is growing too rapidly to wait for the ulcer to heal. We are bringing her home from the hospital tomorrow and sending her off from here.
5:43 AM
I think I finally cracked the case. I'll finalize things later today if not tomorrow assuming there aren't still any glaring mistakes, but I revisited the identity of $a\bmod bc$ and the results do appear more promising compared to before. Just take a look at this and it should give a hint of what I'm doing, but I'm too tired to make any meaningful progress right now. desmos.com/calculator/fs2sqj5bj0
rob, very sorry to hear this. my condolences.
@leslietownes thanks. I've been a real mess today.
at least we are bringing her home and a vet is coming here so that she is in comfortable surroundings.
she hates the vet.
oof. i told livvy, no getting sick until the daughter is out of the house.
@robjohn Very sad.
@robjohn I’m so sorry, but she has been a trooper and suffered enough. My fond condolences to your family.
5:50 AM
@TedShifrin Yes, she has, and she should not suffer any more.
She will still be under sedation from the vet's, but we're hoping she will know she is home at least.
I’m sure she will.
May your grievance be short and well-endured, Rob.
6:11 AM
Pardon my interruption. If $c$ is the modulus we want to compute, and $b$ is a power of two, then after computing $a \bmod bc$, we can compute a smaller modulus as $\left(a\bmod bc\right) \bmod b\prime c$. Simplest idea I have is to halve $a\bmod bc$ and find $b\prime$ such that $2^{\lfloor\log_2(b)\rfloor - \lfloor\log_2(b\prime)\rfloor} b\prime c = bc$. Is there any error in my reasoning?
I've been exposed to plenty of identities concerning a constant modulus, but not really any identities of the modulus itself and what holds for elementary arithmetic applied to it.
err the idea is to halve $a\bmod bc$ and find the largest multiple of $c$ less than $bc$ and for which $\lfloor\frac{a\bmod bc}{b\prime c}\rfloor = 1$ I believe is more descriptive. Something around that description is what I want.
I'm already aware that $\lfloor\frac{2^x}{2^{x-\lfloor\log_2(y)\rfloor - [\lfloor\log_2(y)\rfloor \neq \log_2(y)]}y}\rfloor = 1$. (Brackets are Iverson bracket notation)
6:35 AM
So how is everyone's morning at 2:35 AM?
1 hour later…
7:55 AM
im likely missing something obvious, im trying to show that for some fixed compact set $E \subset \mathbb{C}$, there is some monic polynomial of degree $n$ with a minimum maximum modulus on $E$
anyone have a hint?
min and max do not tend to succeed one another in the same clause. What does "minimum maximum modulus" mean?
oh, maybe I just identify the space of monic polynomials of degree $n$ with $\mathbb{C}^n$, and endow it with the norm that is the maximum modulus of any polynomial with associated coefficients, this norm is necessarily equivalent to the usual one, so bolzano-weierstrass still holds with respect to this norm, so we take a sequence of polynomials (i.e. their coefficients) each with norms approaching the minimum, so this is bounded in norm and has a convergent subsequence
the polynomial whose coefficients are the limit of this sequence is the minimizing one
or at least a minimizing one
oh wait, its not quite a norm though, because there can be nonzero polynomials with zero maximum modulus on $E$ as long as $E$ is finite
but if $E$ is finite the minimum maximum modulus is just zero so i don't think this is too much of an issue
uh, its only zero if the size of $E$ is less than or equal to $n$ I think..
8:35 AM
@leslietownes Thanks for the reply. The domain of $y(n)$ should be the natural numbers. The reason for the question was that, if $f_n(x)$ converges uniformly to $f(x)$, then one should be able to find an $N(\epsilon)$ given any $\epsilon$ such that $n>N(\epsilon)$ implies $|f_n(x) - f(x)|<\epsilon$ for all $x$ in the domain of $f_n(x)$ and $f(x)$.
By analogy with showing some function is continuous, I would try to obtain $n>N(\epsilon)$ from $|f_n(x) - f(x)|<\epsilon$; this only seems possible if $|f_n(x) - f(x)|$ is bounded above by some function only depending on $n$, so one can in turn bound it from above by $\epsilon$ and solve for $n$.
8:53 AM
Is someone around that can help me understand modular congruence better?
I'm trying to find a value that is modularly congruent to $38 \cdot (70 + 133)$ with modulus $n = 40$. As far as I understand, that means find a value $a$ such that $a - (38 \cdot (70 + 133)$ divides $40$, but is in also in the set {1, 2, ..., 40}, correct?
@UnderMathUate other way around. $40$ divides that difference
Oh, whoops ok. But otherwise, that's the idea?
@UnderMathUate if you want $38\cdot(70+133)\bmod40$, then you want a number in $\{0,1,\dots39\}$
Aaah, ok. I'll try to find one now. Thank you!
@robjohn Ok, it looks like the value $a = 34$ works.
9:09 AM
that looks right
Awesome, I get it now. Appreciate the help.
1 hour later…
10:15 AM
Dude how do I prepare for real analysis and calculus I,II,III exam?
I am preparing for exam to join top 10 university in my country ;_;
and the exam is so hard
the problems are extremely difficult
real analysis itself is difficult to understand how am I able to prove statements
2 hours later…
11:48 AM
I made a conjecture that's like Minkowski's lattice theorem, but instead of integer points, it's for prime points.
3 hours later…
2:41 PM
hi everyone, I am wondering if there's someone who's familiar with the min-cost max-flow problem, particularly in the context of integer programming and heuristics
basically I'd like to know the name of a similar problem: where instead of optimizing flow, one is given a flow and finds a minimum-cost capacity / topology that satisfies it
I am currently not sure whether the problems are identical, or whether what I'm saying even makes sense as a problem, but I'd like to know whether it's been considered and what is its name
3:27 PM
@robjohn that's very sad, so sorry to hear this professor Rob. My condolences.
2 hours later…
5:41 PM
hmm. i think i may have accidentally reinvented a very simple example of Van Kampen diagrams
so that's neat
hello please is this result exists ?
those are nice
vrouvrou, if it's theorem 1.2.4 of something, there's at least one data point suggesting that it's true. :) in english you might try searching for sobolev embedding theorems. they usually come out of inequalities involving integrals.
context: i've got a particular set $S$ of permutations $\in S_7$. i'm told (and can readily believe) that this set generates the entire group. given a particular permutation, how do we find the simplest representation as a product of the generators? @BalarkaSen
@leslietownes it is from a small exposition without proof and reference
5:49 PM
seems like a hard question.
you can solve it by a computer by first drawing the Cayley graph of $S_7$ wrt $S$ then taking the geodesic between $e$ and your permutation in the graph
was about to write that, yeah
finding geodesics on graphs has well known algos
the main difficulty being that said Cayley graph is big
5:51 PM
5040 vertices and (in my case) 45360 edges :S
it's actually not quite as bad as that: all of the permutations in my case are either transpositions or products of commuting transpositions
so a->b iff b->a, and thus we can just work with an undirected graph
but that's still 22680 edges :)
interesting question: randomly choose $S \subset S_n$ and $x \in S_n$. in expectation, how much of $Cayley(S_n, S)$ do you need to sample to compute a geodesic joining $e$ and $x$? most of the graph should be typically irrelevant given size of $S$
i expect persi diaconis has something to say
would presumably need to say something about how you're choosing S
uniformly randomly
so $2^{|S_n|}=2^{n!}$ options :)
another difficulty: my choice of $S$ has 9 elements
that's still 'small' relative to the 5040 permutations in S_7, but
still enough to make for a very painful graph
hmm. are van Kampen diagrams just Cayley graphs?
5:59 PM
no its subtly different but i dont remember what the difference is anymore
@TedShifrin please do you know this result ?
don't ping people randomly
@BalarkaSen i think the point is (mostly) this. suppose $a,b\in S$. then you can draw a van Kampen diagram to visualize some portion of the full Cayley graph
something like: vK's are to Cayley(S_n,S) as Cayley(S_n,S) is to Cayley(S_n)
i'm dubious that any of this can be made easy to do by hand, though
@BalarkaSen for specifics, my case is $S=\{(67), (57),(37),(46)(57),(23)(67),(45)(67),(26)(37),(13)(57),(15)(37)\}$
i can solve the word problem for my case of interest using SageMath: sagecell.sagemath.org/?q=ixlxcd
but while i can check the answer, my attempts to understand the algorithms involved have been unsuccessful
6:24 PM
wow @BalarkaSen I just found you in a stack post
the world is sm0ll
i think the difficulty is that most discussions of the word problem are in terms of Dehn's algorithm, which seems simple enough if you have a small presentation
but, uh, mine isn't small
is it sm0ll?
Hi @Sha
if i ask sagemath to come up with the presentation given these generators, i get this mess:
Finitely presented group < a, b, c, d, e, f, g, h, i | b^2, e^2, a^2, h^2, d^2, f^2, c^2, i^2, g^2, (c*f)^2, (f*a)^2, (h*b)^2, (c*a)^2, i*b*i*h*b, (a*g)^2*f, a*d*a*c*d, h*e*h*i*e, c*b*d*c*b, f*e*f*g*e, (h*c)^2*d, (b*h*a)^2, c*e*c*a*e*a, (a*f*b)^2, a*(b*f)^2*b, (e*a)^3, c*i*a*i*c*e, (c*i)^3, (e*b)^3, (a*e*b*e)^2, h*b*e*f*a*h*a*i*a*b*g*f >
some of that is easy, tbf
all of the elements are transpositions or products of commuting transpositions, so they all have order 2
some of the the products give transpositions, and those have order 2 as well. also products of adjacent transpositions always have order 3
but...that still leaves a lot of crap in there
@Ted olas!
7:06 PM
$$\Prod_{A \ne \varnothing} A = \varnothing$$
and 10 other reasons why you dont care about logic
7:19 PM
Hello guys,
For contradiction compound statement that is always false. Suppose we have contradiction statement $c$ and another statement $p$, then if $c=F \land p=F$, then why $c \land p = F$ and not $T$?
i.e, we know that $F\land F = T$, right?
$F$ means false.
7:38 PM
The statement $2$ is odd and $1$ is even all of a sudden becomes a true statement?
Go back to the beginning.
For me Prof?
why would the conjunction of two false statements be true?
@Semiclassical. I am confused here b/c of F and T usually I follow truth table
So not sure here in logical statements world if $F$ for statement stands for another thing!
okay, what's your truth table for $\wedge$?
Okay so it's false
7:46 PM
I see :/ my bad
OR and AND agree that two falses make false and two trues make true
@Semiclassical. Last question pls, for $F$ when it comes in conjuntion with another statment, does it mean a stement that is always false pls?
but OR is happy with at least one true, while AND requires both
Does $F$ statement in $P \lor F$ means:
7:47 PM
\watch out, $\vee $ is OR not AND
no no. I don't mean thier compound truth. I mean only for $F$ that appears in $P \lor F$. Does it stands for statment always false:
what's the truth table for $\vee$? you gave it for $\wedge$ before
@Semiclassical. My question it seems is not clear :/
you wrote $P\vee F$
that doesn't match the order you gave for $\wedge$ before
can you give the first two columns of your truth tables as well?
Got you. So how about $F$ statement that stands for $F$? Does it mean only FALSE, so we have for $P \lor F$ the following:
P -- F
T -- F
F -- F
F -- F
Is this how the truth table looks like for $P -- F$ before taking $P\lor F$?
7:53 PM
i don't understand what you just wrote
I know, one second pls
a truth table for $P\wedge Q$ would be
P&Q&P\wedge Q\\
let's see if i can make what i wrote nicer
Can you see bold T and bold F pls above?
If yes, I am just confused about what they mean! That's all
I got you for truth table.
7:56 PM
okay. note that the order is FTTT given my P,Q columns
oh drat
\begin{array}{cccc} P&Q&P\vee Q & P\wedge Q \\ \hline F&F&F&F \\ F&T&T&F \\ T&F&T&F\\ T&T&T&T \end{array}
there we go
Great. How about $P \lor F$:

\begin{array}{ccc} P&F&P\vee F\\ \hline F&F\\ F&F\\ T&F\\ T&F \end{array}
latex on the fly
sure. you could drop two of the rows here
b/c they're identical
Sure, so $F$ in bold means always false?
\begin{array}{ccc} P&T&P\vee T\\ \hline F&T\\ F&T\\ T&T\\ T&T \end{array}
Same for $T$ statmenet meaning always true
7:59 PM
but you can also just do this from looking at the truth table i wrote. if $Q$ is false, then you can restrict to the first and third rows
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