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4:00 PM
I know it's called blackboard bold, I'm just horrible at explaining properties of fonts...
 
I'm pretty sure I've seen fonts that had proper blackboard bold for lowercase too, but MathJax doesn't seem to have it. :(
 
2
Q: Fourier transform of $f[n]$ and $f[-n]$

freak_warriorHi I am just wondering, If I have a signal $f[n]\in \mathbb{C}^L$, i.e. $f$ is $L$-periodic, i can also define $h[n]=f[-n]$. Is it true that the Fourier transform of $f$, say $\hat{F}$, and the Fourier tranform of $h$, say $\hat{H}$ is related by the equation below? $$|\hat{F}[k]|=|\hat{H}[k]|...

 
I wonder if a bigger bounty would have made any difference to math.stackexchange.com/questions/1059379/…
 
I love how the number of people who have earned the "Self-Learner" badge has pretty much doubled since the start of Winter Bash 2014...
Nvm, I was thinking of a different badge
 
@Chris'ssis did you get this interview question correct at ur interview ? math.stackexchange.com/questions/866921/interview-riddle
 
4:08 PM
is there a symbol $A \subset B$ where $A$ and $B$ are multisets?
 
@TheArtist Wait ( to read it again) - Yes, I think so.
 
Wait... If $\frac{0}{0}\not=1$, does that mean $\frac{x}{x}\not=1$?
Did I really just ask that? I need more sleep..........
 
Get some rest (or coffee), @teadawg1337. It happens to the best of us when we're tired.
 
(^(^_(^_^)_^)^)
 
4:11 PM
@teadawg @Khallil: There is no blackboard bold for lower-case letters, only for upper-case, i.e., capitals.
 
@Chris'ssis wow, and how long did the interviewer give you to find the answer?
 
Hey, @TedShifrin! How's it going?
(Also, I didn't know that. What are the lower-case blackboard boldy looking letters called?)
 
@skullpatrol Conservation of mass implies that there aren't enough particles in the entire universe to write a comprehensive guide for understanding women
 
Sorry, @teadawg. I would never read $\epsilon$ of that, let alone upvote it.
There aren't any, @Khallil, except a few custom ones in the AMSfonts grabbag, I believe.
 
@TedShifrin 𝕆𝕙 𝕪𝕖𝕒𝕙?
 
4:14 PM
@teadawg: And women would say the same of men.
 
Or did you just mean in LaTex / MathJax?
 
Oh, interesting, @Ilmari. What font is that? I meant in standard LaTeX, yeah.
 
@TheArtist I don't think they were interested in a certain answer, but they expected to see the way I approach it.
 
@TedShifrin It's just Unicode, whatever font your browser is using.
 
@TedShifrin cries silently in a corner
 
4:15 PM
I'm sorry, @teadawg ... I've been nice to you other than that :P
I'm grumpy because I'm grading absolutely horrendous final exams ... which I guess I should go back to doing. :(
 
@TedShifrin I don't feel insulted at all, it took me five days to notice that I made two tiny mistakes in that behemoth of an answer xD
 
@Chris'ssis okay :)
 
Well, @teadawg, it was only after the publication of one of my best papers in one of the best journals that I found a very basic (multivariable chain rule) error in it. :(
Fortunately, it is just embarrassing; it didn't affect the validity of the result.
Since when did @skull turn into a cute rat? :P
 
He's become a karma chameleon
 
Oh, it's the multifarious hatted one.
 
4:19 PM
@TedShifrin the hat transformed me :-)
 
@Mike: You should thank me gratefully for doing as you requested and "answering" that question.
 
Thank you, @Ted :)
I actually think that's a really good answer!
No scare quotes.
 
@Mike That wasn't grateful enough, do it again >:(
 
That's all he gets @teadawg1337
 
I think you got some good answers, actually.
 
4:20 PM
@TheArtist In my country, the smartests persons I've ever met it happened to be during the interviews. I'd like to work with some of them and present them the real face of the reality. :-) (well, you only remain with personal satisfaction, nothing more)
 
I agree, @Ted. A bit too much to present in one course... but I think a few of them can be good motivating examples.
And now the next time someone asks me 'what's this good for?', I can say actual things instead of muttering "Uh... uh... it's used pretty much everywhere"
 
I was very pleased when I figured out (writing the linear algebra book) a cool argument that the vandermonde matrix is nonsingular by applying the change of basis formula.
Well, you should actually learn some of what you plan to mumble.
 
Yo @Mike
 
OK, back to the horrors of grading.
 
@Chris'ssis :)
 
4:22 PM
@TedShifrin Which is why I posted the question... of course I'm not going to find it sufficient to just glance at the answers if I want to talk about them. :)
(One of the answerers, the one who answered about computer graphics and rotation matrices, is a UCLA first year... that's 4 of us that can be considered vaguely active users now. Almost 20%!)
 
@Chris'ssis These questions are super super hard, especially when one is stressed during an interview
 
That's actually one of the fun things about teaching: You get to actually learn new, cool stuff. I only wish my probability students had learned some. And, it turns out, many have no clue how to set up the easiest double-integrals, despite their having had homework on this.
 
@TheArtist by the way, do you know to answer it?
 
Eric seems to have disappeared, @Mike ... or I haven't seen him 'cuz I'm disappearing.
 
I'll find out how the calc students do this afternoon, @Ted. Test is at 1130, grading's 'til I'm done.
 
4:24 PM
multivariable? @Mike Huh?
 
Eric uses less actively but still posts questions semi-frequently.
Thinko @Ted
 
Ah, in his old days, I answered a lot of his questions.
Nice morphing there, @Mike ...
 
@MikeMiller Did you see what I did yesterday?
 
Now Kevin Carlson, one of ours, is answering a lot of his questions.
No, @BalarkaSen.
 
That's best, @Mike. I've seen Kevin around for ages.
 
4:25 PM
Please Professor @TedShifrin unignore BalarkaSen :-)
 
@Chris'ssis Yes I've read the answers in the post, $\frac{ab^2}{6}$ is the method I like, however I don't know how one would come up with that ....I don't understand at all Omrans answer, and it seems to be an interesting answer
 
When I visit LA, we'll have to have a MSE reunion :P
@skull: Is there a good reason?
 
Kevin's a good friend. Funny how it all works out.
 
@TedShifrin 'tis the season
 
@TheArtist I see. By the way, I also consider myself an artist. :-)
 
4:26 PM
I'm a long-time grinch, @skull.
 
Thanks @skull but I don't think he is unignoring me again, ever.
 
@TedShifrin he has learned his lesson and will try much harder
 
I'm not sure what lesson he learned, @skullpatrol
 
I am even doing algebraic topology without complaining.
 
LOL, you planning to be one of those shyster lawyers who defend drunk drivers, @skull?
 
4:27 PM
:D
 
@Chris'ssis Omrans answer is mathematics I've never seen. You know, before I joined mathematic stack exchange I never knew such brilliant smart people like you existed.
 
That's a firm yes.
 
What, @Mike? @skull's answer to my question?
 
Yep
 
You should've expected that there were unbelievable people out there, @TheArtist!
 
4:28 PM
@TheArtist I'm just a retarded ... :-)
bbl
 
What do I have to do to be unignored?
 
@Chris'ssis :)
 
@MikeMiller a smile is just a smile
1 min ago, by skullpatrol
:D
 
That's a lawyer answer, @skullpatrol
 
yes sir
 
4:29 PM
sigh So much for saying algebra > geometry
 
OK, back to grading.
 
See ya
 
@Chris'ssis: lab is really a good teacher Very refreshing.
 
@KhallilBenyattou I was aware about the discoveries and inventions, but I've never seen their workings,etc
 
@MikeMiller $\Bbb Z_2$ acts on $\Sigma_3$ by rotating the three-squashed-torii around the middle axis. Quotienting gives $\Sigma_2$.
 
4:30 PM
Right.
 
In fact the action is properly disc.
So $\Sigma_3 \to \Sigma_2$ is galois.
Similarly, $\Sigma_{11} \to \Sigma_3$ is galois.
 
@KhallilBenyattou "never seen their working" it's kinda hard to describe wot I totallly mean over here
 
So now we have a bunch of Galois covers $\Sigma_{11} \to \Sigma_3 \to \Sigma_2$
 
Mhm
 
@Chris'ssis hehe
 
4:32 PM
It's cool!
 
And we want a SES $1 \to Aut(\Sigma_{11}, \Sigma_3) \to Aut(\Sigma_{11}, \Sigma_2) \to Aut(\Sigma_3, \Sigma_2) \to 1$
 
Mhm.
 
This boils down to $1 \to \Bbb Z_5 \to \Bbb Z_{10} \to \Bbb Z_2 \to 1$
Which exists.
 
I'm not convinced that $\text{Aut}(\Sigma_{11},\Sigma_2) = \Bbb Z_{10}$.
 
@MikeMiller An object with 10 fingers, with a single hole in each finger and one hole in the middle.
 
4:34 PM
(But whatever it is, we automatically have your SES.)
@BalarkaSen But that's not the original covering map.
 
No it is.
 
Why do you say that?
 
Oh wait the composition of the two covering map is what we want.
 
Right
I have no idea what that one Aut is. I have the vague but unjustifiable feeling that it's probably $D_5$.
 
Wait I think I can do this.
@MikeMiller I'd be surprised if it was so.
I don't expect it to split.
 
4:39 PM
Things can split even if they don't canonically split.
@BalarkaSen Suggested approach: show that the composite covering map is the same as the composite $\Sigma_{11} \to \Sigma_6 \to \Sigma_2$; consider the degree two deck transformation from the first map as an element of $\text{Aut}(\Sigma_{11},\Sigma_2)$; check how it commutes with the other guy.
(Of course, this requires you write down a homeomorphism between the two ways of looking at $\Sigma_{11}$ we're using.)
 
Actually I was thinking something along the lines of this : $p_1$ and $p_2$ be two different covers of $\Sigma_{11} \to \Sigma_2$, where $p_2$ is the composite and $p_2$ is the $\Sigma_{11} \to \Sigma_{11}/\Bbb {Z_{10}}$. This means that there is a self-homeo $h$ of $\Sigma_2$ with itself such that $p_1 = h \circ p_2$
Dunno, might or might not work.
 
@BalarkaSen Let $q:E\to X$ be a covering map, $A\subset X$ locally path-connected. Then the restriction of $q$ to each component of $q^{-1}(A)$ is a covering map onto its image.
Can u share a hint?
 
$A$ is evenly covered by $q$?
 
@Nick Yeah, a bit long, but it's good.
 
@BalarkaSen No
 
4:45 PM
What do you mean by "component" then?
 
Is this an ok proof of $A \backslash (B \cap C) = (A \backslash B) \cup (A \backslash C)$?
$$ \begin{aligned} \text{Suppose that } x \in A \backslash (B \cap C) & \iff x \in A \land \lnot ( x \in B \cap C ) \\ & \iff x \in A \land \lnot ( x \in B \land x \in \cap C ) \\ & \iff x \in A \land (x \not\in B \lor x \not\in \cap C ) \\ & \iff ( x \in A \land x \not\in B) \lor (x \in A \land x \not\in \cap C ) \\ & \iff (x \in A \backslash B) \lor (x \in A \backslash C) \\ & \iff x \in (A \backslash B) \cup (A \backslash C) \end{aligned} $$
 
@BalarkaSen I was hoping you knew...
 
I dunno.
$A$ is evenly covered I presume.
 
The only thing said about $A$ is that it's locally path-connected
 
pass @Mike
 
4:49 PM
@Venus @Chris'ssis Is there any way to ignore all non-integral users in one click?
 
Nope @Integrator
You have to click one-by-one
 
Yes there is, @BalarkaSen.
 
LEL
 
You should totally do that @Integrator
 
4:51 PM
If you can explicitly write down the order 2 deck transformation that'd do it, @BalarkaSen
I'm having trouble visualizing it
Which is why I suggested the other approach
 
@TheArtist have you seen this one I created yesterday? $$\int_0^{\infty}\left( \frac{\log ^2(\gamma x+1)}{(\gamma x+1)^{\large \sqrt[3]{\log(2)}}}-\frac{\log ^2(\pi x+1)}{(\pi x+1)^{1+\sqrt[3]{\log(2)}}}\right) \frac{1}{x} \, dx=\frac{2}{\log(2)}$$
 
@MikeMiller I mean I pass you @iwriteonbananas's question.
 
I know, I don't want to think about it.
 
OK
Me neither.
 
No offense to him. :)
 
4:52 PM
I'm hurt and heartbroken
 
Will you feel better if you know that I love your gimmick, @iwriteonbananas?
 
@Chris'ssis omg , no I havent
 
@TheArtist Skip the appearance. It can be done without pen and paper.
 
4:53 PM
@Chris'ssis how do you create these?
@Chris'ssis for real?
 
@TheArtist Yeah, for real.
 
I considered responding to your torus question with a banana picture, but I'm out of bananas @iwriteonbananas :(
 
@TheArtist I think I'm just a lucky person when talking about integrals, series and limits.
 
@MikeMiller how dare you be out of bananas
go to the grocery store right now!
 
@Chris'ssis you sure is....also gifted :) how do you do this ? Even with a pen? :p it looks impossible to me
 
4:57 PM
@TheArtist I let you think of it for a while.
 
What is the flattest fruit?
 
@Chris'ssis okay :) let me think
 
Flatapples, @user130018
 
No, @MikeMiller. It's a flat mangodule.
 
Hello @user130018 how many more papers to go?
 
5:01 PM
@JasperLoy I finished all my papers
@JasperLoy I only have final exams left
 
@user130018 Ah, that's what I meant.
 
@JasperLoy Two more
 
@Chris'ssis hmmmm no idea
 
@MikeMiller Exactness of $1 \to Aut(X, Y) \to Aut(X, Z)$ is obvious.
 
@Chris'ssis I've been thinking without a pen or paper too :) coz I don't have access to that now :/ (in the car)
 
5:10 PM
@TheArtist Use my approach here (take the file, after that I delete it)
 
Actually @Mike. The covering maps $X \to Y \to Z$ must be all finite-sheeted.
 
Nah.
 
Consider $\Bbb R \to S^1 \stackrel{(-)^n}{\to} S^1$
 
@TheArtist did you take the file?
 
Oh, I see your complaint.
 
5:12 PM
@Chris'ssis I have posted my answer
1
A: Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$

VenusHere is an elementary way to evaluate the integral without involving any special functions or advance formulas. Notice that $$\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}$$ Hence we have \begin{align} \int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\in...

 
Actually, no.
 
Huy
Sup, everyone.
 
Your sequence is $0 \to \Bbb Z \xrightarrow{\times n} \Bbb Z \to \Bbb Z/n\Bbb Z \to 0$.
 
ohhh right
my bad
 
5:13 PM
@Venus Nice (+1).
 
hi @Huy
 
@Chris'ssis Thanks. It takes lots of efforts for me
 
@Chris'ssis Yes I did :) I'm reading it :)
 
@TheArtist OK
 
@Chris'ssis jus finished reading it :) gonna apply this, now this looks promising :) let me see if I can write my solution. Il show u
 
5:17 PM
@TheArtist OK :-)
@Venus I see. Well, I only wrote down the main idea.
 
@Chris'ssis Thank you :) , brb (back to work on that)
 
@Chris'ssis I don't know about Binet 2nd formula, but thanks for giving it :)
 
@Venus Ups, I forgot to upvote M.N.C.E.'s answer .... Done.
@Venus It's nice this last integral from his answer $$\int^1_0\frac{1}{\ln{x}}+\frac{\sqrt{x}}{1-x}\ {\rm d}x$$
I think there is a very fast way of finishing this one ...
 
@Chris'ssis I don't know the fastest one
I'm studying his answer
 
5:28 PM
@Venus Do you want me to show you a brilliant way? (one line proof)
 
@Chris'ssis Yes, please.
 
$$\int^1_0\frac{1}{\ln{x}}+\frac{1}{1-x} {\rm d}x-\int_0^1\frac{1-x^{3/2-1}}{1-x}\ {\rm d}x$$
Done.
In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: It is the first of the polygamma functions. == Relation to harmonic numbers == The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as == Integral representations == If the real part of x is positive then the digamma function has the following integral representation...
 
@Chris'ssis Clever!
 
@Venus Thanks. :-)
 
Random Set Theory Question : What do you mean by an equinumerosity among sets? Could someone give me an example of $\Bbb A \approx \Bbb B$. From what I understand, if there exists some bijection $f : \Bbb A \to \Bbb B$, then the sets are equinumerous.
 
5:39 PM
What you understand is correct, @Nick. That's the defiinition.
 
So, if I understand correctly, then equinumerosity is the relation between the domain and codomain of a bijective function.
 
If I understand you correctly, yes
 
@MikeMiller That's pretty vague. There could be an infinite number of bijections between any two sets. Does this mean that all sets are equinumerous to one another or is this property only referenced in light of some specific bijective function.
 
@Venus When will I get to see you? ;)
 
I just answered a lhf. I now need 30 points to reach 3,000.
 
5:42 PM
The definition of equinumerous doesn't involve a specific function. It's just the existence of one.
You can pick two sets $A$ and $B$ for which there is no bijective function $f: A \to B$, @Nick.
Say, $A = \{0\}$, $B = \{1, 2\}$.
 
@JasperLoy lhf ?
 
@Integrator Low hanging fruit.
 
@JasperLoy Ah!
 
That's a big hat @Mike
 
I answered another lhf.
 
5:45 PM
@MikeMiller That's neat. That brings to my mind a more twisted question. In the sets you've given what would you call $f: B \to A$ where $f(x) = 0$ . Sure, it isn't a bijection but what is the relationship between its domain and codomain called?
 
i was joking
 
I could hear myself in a DeLorean from 2050 laughing his ass off.
 
What's $0$, @Nick? Is $x$ just all elements? So you're saying $f$ is a map that sends every element to $0$ (which is some element of $A$)?
I doubt that, @Nick, it wasn't very funny.
 
@Integrator What did you mean?
 
@Venus Oh, You've forgotten! Nothing! (>‿◠)✌
 
5:49 PM
$f = \{(1,0),(2,0)\}$ ... A many-to-one function.
 
@Integrator Nice answer anyway.
Forgot of what?
 
@MikeMiller Hey, he was 2050, he could have lots of reasons to laugh.
 
@Nick I don't think that's an interesting relation (so there's no name for it). Of course, you can do this for any set.
any pair of sets, rather.
 
@Venus Your picture!
 
@Integrator I said if Jack gets elected as a mod
 
5:51 PM
$Aut(X, Z) \to Aut(Y, Z)$ looks nontrivial
We want to "restrict" the equivalences of $X$ to $Y$ somehow.
 
I have gotten 3,000. It's time to retire.
 
Bye @Jasper and see you in a week.
 
@Venus Does that mean Forget it? ^^
 
@BalarkaSen I am still coming to chat.
 
OK.
 
5:53 PM
Out for joggggiinggggg!!! Yeah!!!:-)
@JasperLoy ^^^
 
@MikeMiller So, the term Equinumerous function. Does that make sense? Shouldn't it be 'Bijection or Bijective Function' instead?
 
I've a question. $(a,b)$ where $a,b\in \Bbb Q$ form a basis for $\Bbb R$, right?
 
@Venus And which answer were you talking about?
 
Yeah @Studentmath. You just proved that $\Bbb R$ is separable.
 
@Integrator Do you hope Jack will not be elected?
 
5:54 PM
@Chris'ssis Yeah, I've been constipated while jogging too. I hear castor oil to be a good remedy.
 
I've not heard that term, @Nick. Usually one says bijection or bijective function, like you thought.
 
@Mike and also that it is second countable
 
@Integrator The latter one. Beta
 
@Studentmath Oops, that's the one I meant
 
Separable is due to $Q$ being dense
Right?
 
5:54 PM
@Venus I wish he should, at-least for a day! ;)
 
Mhm
 
@Venus Oh, I see!
 
@Studentmath The same argument, by the way, shows all separable metric spaces are second countable.
 
@Chris'ssis was busy around many things, coz I'm travelling right now....finished doing it :) but I'm getting a different result :/
 
OK I guess there is a map @Mike
 
5:57 PM
@Mike oh that's cool. So $R_s$ (which is separable yet not second countable) in't a metrizeable?
 
Consider the map $Aut(X, Z) \to Aut(Y, Z)$ by taking an element $h : X \to X$ of the former and mapping it to $p \circ h$ where $p$ is the covering map $X \to Y$
 
(The one with the basis $[a,b)$)
 
Yup @Studentmath
@BalarkaSen $p\circ h$ is a map $X \to Y$...
 
maybe then $p \circ h \circ p$?
 
Look at the domains and codomains of those functions. That doesn't make sense.
 
5:59 PM
OK, coffee is what I need.
 
@Chris'ssis I'm getting $\frac{2}{\sqrt[3]{\log 2}}$
 
Do you see how the argument generalizes, @Studentmath?
 
@Mike Not really, not sure how to make use of the metrization of the space (barely touched it yet)
 
problem is that $p^{-1}$ is not a function
grumph
 
@BalarkaSen said the grinch to his pup.
 
6:09 PM
i dunno @Mike i'm confus. you have a map in mind?
 
@BalarkaSen See this please.
I'm in a dilemma. I am confused!! Arghh..
 
Well $\Bbb Q(\pi)$ is a field ext over $\Bbb Q$ and there is $\pi \in \Bbb Q(\pi)$ such that $\Bbb Q(\pi) = \Bbb Q(\pi)$
Yet $\Bbb Q(\pi)/\Bbb Q$ is not algebraic.
You meant $\Bbb Q(\pi)$ instead of $\Bbb Q[\pi]$
 
I got carried away with the "[" sign in the question!
 
Eh, the question is garbled.
$F[\alpha_i]$ is a field iff $\alpha_i$s are $F$-algebraic.
 
"garbled"means wrong?
 
6:15 PM
Alright
 
Sorry for my vocab.
Oh. Thanks @MikeMiller
 
@SwapnilTripathi The statement in the question is true, and the proof is tautology/
This fellow is simply connected.
 
I just noticed that most proofs end with either QED, ■ or both. The Latin, I get. The quadrilateral, not so much.
 
@Nick It's just a symbol, that is all. Nothing to get.
 
@Mike That space is homotopy equivalent to the boquet of 6 spheres, right?
 
6:21 PM
@JasperLoy That's like saying the Eye of Horus is just a symbol. There is no way a common full stop like that has no meaning or history.
$\square$
 
$\square$ Sounds like a big conspiracy $\square$
 
$\blacksquare$
 
@TheArtist You did some mistakes I think.
 
6:30 PM
@skullpatrol The Tombstone is $\blacksquare$ and the rudin is $////$. These trends are trendy in this sphere.
 
@Nick yes, it is a matter of style
 
@TheArtist check again the third line from below.
 
I miss wordplay and playful banter. I feel so dull. I should rest my head. Put myself into bed. Toodles, with that said :D
 
@BalarkaSen Yes. Thank you. I feel so dumb. :/
 
later pal
:D
 
6:35 PM
@skullpatrol $\mathfrak{LATER}$
$$\mathcal{\Large G \large ood\ \Large N \large ight,\ \Large C \large omrades}$$
 
@Mike any hint? (for your generalization) got me curious
 
@Balarka Ask me later
@Studentmath Enumerate the dense countable subset $x_n$; pick your basis to be $B(x_n,r)$ where $r$ is rational
 
And that gives you a countable basis, awesome
 
Do a similar idea to show metric spaces are first countable :)
Yes it is @Balarka. Wedge of spaces is \vee, not \wedge. The latter denotes smash product in the context of spaces.
 
Precisely the same idea, just that for each $x\in X$ we have the basis to be $\{B(x,r): r\in Q \}$
 
6:46 PM
Aye
 
2 hours ago, by Khallil Benyattou
Is this an ok proof of $A \backslash (B \cap C) = (A \backslash B) \cup (A \backslash C)$?
$$ \begin{aligned} \text{Suppose that } x \in A \backslash (B \cap C) & \iff x \in A \land \lnot ( x \in B \cap C ) \\ & \iff x \in A \land \lnot ( x \in B \land x \in \cap C ) \\ & \iff x \in A \land (x \not\in B \lor x \not\in \cap C ) \\ & \iff ( x \in A \land x \not\in B) \lor (x \in A \land x \not\in \cap C ) \\ & \iff (x \in A \backslash B) \lor (x \in A \backslash C) \\ & \iff x \in (A \backslash B) \cup (A \backslash C) \end{aligned} $$
Does this line of reasoning validate $A \backslash (B \cap C) \subseteq (A \backslash B) \cup (A \backslash C)$ and $A \backslash (B \cap C) \supseteq (A \backslash B) \cup (A \backslash C)$ thereby meaning that the two sets are equal?
 
Hello!!!

I have a question about a program in C. Can someone help me??
 
@Khallil seems right to me
 
Thank you, @Studentmath! I made an embarrassing blunder in an earlier question, so I'm running low on confidence. ^_^"
 
Happens :) You grow it back up slowly
 
6:59 PM
Hello @DanielFischer !!! Are you familiar with the programming language C ??
 

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