Basic Mathematics - 2021-09-28

Pax Daga

04:04

If Neg B and neg A:
                           If A or B:
                                        If A:
                                              A
                                              Neg A
                                               Contradiction
                                         If B:
                                               Contradiction
                          Contradiction
                          Neg(A or B)

@user21820

Am I wrong

@Prithubiswas Is my argument fine

user21820

06:14

@PaxDaga It's wrong. Nearly everything you post is sloppily done and full of errors. Put in more effort, and try again.

Pax Daga

06:52

If Neg B and neg A:[Assumption]
                           If A or B:[Assumption]
                                        If A:[Assumption]
                                              A
                                              Neg A
                                              Contradiction
                                         If B:
                                               Contradiction

                          A impies Contradiction
                          B implies contradiction

@user21820

Its the same thing but know I tell you the stpes

Also my earlier proof was not wrong just some steps were missing if I'm not mistaken

user21820

07:07

It was wrong. Don't post here again until you have found at least one error.

cOnnectOrTR 12

07:53

Hi, There is always a tangent at a point parallel to secant adjacent to it. So geometrically it is correct that the limit of the slope of secant is the slope of tangent and so the approximation is converted into instantaneous dy/dx ?

user21820

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@cOnnectOrTR12 Don't attempt to use geometric intuition when doing real analysis. The only correct approach is via the rigorous definition of the derivative.

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cOnnectOrTR 12

@user21820 you mean while calculation. Ok. But is it right what I said if we have to understand what derivative is intuitively?

user21820

08:10

@cOnnectOrTR12 Intuitively, yes, the derivative of f at c is the limit of the gradient of the secant to f with one endpoint at (c,f(c)), as the other point goes toward the first point.

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cOnnectOrTR 12

Thanks! @user21820

Prithu biswas

@user21820 I am running out of variables while doing long proofs using the "fresh variable" requirement of "∃elim rule". Is there a way to solve this issue?

user21820

@Prithubiswas There are 3 ways to solve this problem: (1) As in programming, we can allow multi-character variable names, such as "a1" and "a2" and so on. (2) We can factor out lemmas. (Though in some sense this doesn't solve the problem because we can't copy-paste the lemmas' proofs due to the same restriction.) (3) We can relax the ∃elim rule a bit. I'm seriously considering this last option.

The relaxation would be something like this:

Jun 27 at 18:26, by user21820

@F.Zer Well in the system I gave, I actually defined "fresh variable" to mean "variable that does not appear in any previous statement", not just in surrounding contexts. In practice, this only poses a problem if in your proof you do many ∃elim steps, which is not typical unless you are writing a gigantic proof without lemmas. I think we can relax that a bit to "variable that does not appear in any statement in the surrounding contexts or in subcontexts of the current context".

This would mean that variables declared via ∃elim in a subcontext would be fresh again once you jump out at least two levels.

Actually there is another possible solution: (4) Make the ∀intro rule stricter so that we can relax the ∃elim rule even more. To understand the issue, consider the following:

Jun 27 at 18:40, by user21820

Given x,y∈S:
	∃z∈S ( z∈S ).
	If x≠y:
		Given z∈S:
			x≠y.
			∃t∈S ( x≠t ).
		Let t∈S such that t∈S.
		∀z∈S ∃t∈S ( z≠t ).  [∀intro!]
		∃t∈S ( t≠t ).  [∀elim??]

Jun 27 at 18:42, by user21820

Of course, there are many possible ways to evade this problem. I allowed ∀intro to work multiple times after the same "Given ..." subcontext, so that it is user-friendly when getting multiple ∀conclusions from a subcontext. You can ban this just to avoid the above problem, but then the system suffers in usability in other respects.

(The "Let t ..." line is invalid with both the (current) strict ∃elim rule as well as the relaxed rule I cited earlier.)

@cOnnectOrTR12 I should say that the geometric intuition is not really the best way to understand derivatives intuitively. The best way is via the asymptotic behaviour: A function f is differentiable at c iff there is some r such that f(c+t) ∈ f(c) + r·t + o(t) as t → 0. This "o" is little-o notation (see the table of formal definitions of Landau notation at wikipedia).

Pax Daga

08:35

If (A implies neg B) and B:
 B
 (A implies neg B)
 If A:
  Neg B
  B
  Contradiction
 A implies contradiction
 Neg A

I don't have programmers notepad I use pages which is not bad I was not indenting properly I reaslised

For indent I use 1 space is that Ok?

user21820

Fine, that's correct. But next time use either tab or at least 2 spaces for each indent. Also, I asked you to post (P1) and (P2) first. I don't want to see anything else until you post them.

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Pax Daga

I did post it and you said its correct....

user21820

@PaxDaga I want to see the whole proof. You never posted an entire proof of (P1).

@PaxDaga I don't care what you use as long as you produce good work. Your work isn't good. If you refuse to use PN, at least use a normal text editor (even Windows' Notepad will do) and use tab for indentation.

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One more stupid spam and I will permanently ban you.

hyper-neutrino

Hey, everything alright here?

user21820

@hyper-neutrino This room is recently getting overrun by people whose behaviour borders on trolling. I may have been too lenient. If this goes on, I will inform the Math SE moderators myself.

hyper-neutrino

08:48

@user21820 Ah, I see. Alright. I just came here because of the automatic notification for too many repeated kicks; I'll see myself out then and hopefully the room situation improves.

user21820

@hyper-neutrino I also hope so. Thanks!

user21820

09:26

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Pax Daga

09:40

 A∨B∧C ⇔ (A or B) and (A or C)

Proof the of the first direction

If A or B and C:
                        If A:
                              A
                              A or B
                              A or C
                              (A or B) and  (A or C)
                        A implies (A or B) and (A or C)
                        If B and C:
                                         B and C
                                         B
                                         C
                                         A or C

now I replaced it @user21820 please now help me find my mistake in P4

user21820

@PaxDaga Wrong. There are 3 errors (1 completely bogus line and 2 missing lines), and your indentation is horrible. Fix them all before posting again. And don't ask me to do anything if you refuse to do (P2) and (P3).

I even told you one of the missing lines before, but you stubbornly ignored it:

Sep 24 at 16:08, by user21820

And one more missing line was "(A or B) and (A or C)" at the end of the subcontext under "If A or B and C:". Well I guess that counts as a really missing line, though I know that was your goal that so I didn't notice it was missing earlier.

Pax Daga

I did p3 and you said its correct?

user21820

@PaxDaga Yes, but you didn't do (P2) yet.

Get on with your work and stop wasting my time with your inane objections.

Mark

10:01

Hi, I am facing with a problem. If I have a set S, how can I "easily" find a vector space V that contains the elements of S that are closed under the XOR operation? With closed I means that all the result of the XOR operation are in V.

user21820

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Mark

I would like to find the biggest vector space V. For example let S = {0000,0011, 1001}, V can be <0000,0011> since 0000 \oplus 0011 = 0011 \in V or <0000,1001> since 0000 \oplus 1001 = 1001 \in V. But V CAN'T BE <0011, 1001> because 0011 \oplus 1001 = 1010 \notin V. Does anyone has any ideas? Thanks in advance

user21820

← 2 messages moved from Logic

@Mark This doesn't make sense, because you ask for a V that contains S, so why are you trying Vs that do not? Also, why are you looking for the "biggest"?

Are you misinterpreting "V that contains the elements of S"?

user21820

10:28

Spam removed.

Mark

10:45

@user21820 I ask for a V that contains the element of S that are closed under the XOR operation. Another example in {0,1}^5 could be this: S = {00000,00001,00100,01000,01100,10000,10100,11000,11100} and V (what I want) = {00000,00100,01000,01100,10000,10100,11000,11100} and in other words V is the biggest vector space that is contained in S. This clarifies?

user21820

@Mark No. That's wrong. You're not a native English speaker, correct?

Mark

@user21820 No I am not

user21820

@Mark Ok. So I'll explain why you're wrong. The phrase "vector space V that contains the elements of S" means "vector space V such that the elements of V contains the elements of S". And "contains" and "is contained in" are completely the opposite directions.

Mark

@user21820 I have a set S and I want to extract from the elements of S a vector space V, that is a set of points that is closed with respect to the XOR

user21820

@Mark No! I keep telling you that that is not what the question you cited wants!

Mark

10:57

Which question?

user21820

55 mins ago, by Mark

Hi, I am facing with a problem. If I have a set S, how can I "easily" find a vector space V that contains the elements of S that are closed under the XOR operation? With closed I means that all the result of the XOR operation are in V.

Mark

sorry for the misunderstanding. What I want is this: I have a set S and I want to extract from the elements of S a vector space V, that is a set of points that is closed with respect to the XOR. For example let S = {0000,0011, 1001}, V can be <0000,0011> since 0000 \oplus 0011 = 0011 \in V or <0000,1001> since 0000 \oplus 1001 = 1001 \in V

Pax Daga

@user2180 Im really sorry for angering you I guess I am bad in logic as I'm a weak student ;is logic really needed for higher level math? I guess ill stop the excersices and focus on other math since I am not good at writing formally.

thank you for your help (earlier and now)

user21820

@PaxDaga Of course basic logic is needed for all mathematics. Otherwise I wouldn't be spending hours trying to convince you to learn it. Do you think I enjoy spending my time like this? Whether you want to learn basic logic or not is your free choice. Whether I want to teach you is also my free choice. I will not teach you if you refuse to learn what is crucial for mathematical understanding, and that's final.

@Mark Then your "vector space V that contains the elements of S" in your original question is wrong. But in your new question, there may not be a unique answer because what you are looking for is a maximal subset of S that is closed under ⊕. However, it is possible that there are distinct maximal subsets.

Mark

11:17

@user21820 Yes there may be distinct maximal subsets and the important thing is to extract one of them, it doesn't matter which one

user21820

12:13

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@Mark Then just add one by one if possible. Mathematically, this is trivial. If you're asking for an algorithm, then you need to say so.

Mark

@user21820 Sorry, yes I am looking for an algorithm, or, if exists, some theorem proven in the literature.

user21820

@Mark But as I said it is a trivial theorem, especially if your S is finite. Just start with T = ∅, and go through the members of S one by one and add it to T iff the ⊕-closure of the result is still a subset of S. At the end you will have a ⊕-maximal subset of S. I leave it as an exercise for you to prove.

There's no need to cite trivial theorems like this one, nor any need to even find it in the literature.

What I just said still works if your S is countably infinite. But if you want to prove it for potentially uncountable S, then you need to tell me how much mathematics and set theory you know.

Mark

12:31

@user21820 S is finite. The problem (for me) is that this algorithm depends on the order in which the members of S are considered. For example S = {00000,00001,00100,01000,01100,10000,10100,11000,11100}. If I bring into T 00000 and 00001, as consequence, T cannot be extended further, so |T| = 2, while I know that the $\oplus$-maximal subset of S is |T| = 8

user21820

@Mark Maximal subsets need not have the same size. From that last sentence, it seems you want a subset with maximum size, not a maximal subset. But it's still a trivial theorem, because you said S is finite, so there are finitely many ⊕-closed subsets of S and obviously their sizes has a maximum.

So what are you really looking for?

Mark

@user21820 I am looking for the subset with maximum size (if there are distinct maximal subset with the same maximum size it doesn't matter which one will be extracted)

user21820

@Mark I already told you its existence is trivial.

Mark

S is finite but if I try all the ⊕-closed subsets of S and take the maximum size this is exponential

user21820

@Mark So why when I ask you whether you want an algorithm, you say "algorithm or theorem"?

Mark

12:44

I need to implement it, so an algorithm. I say also a theorem if it exists in the literature (if this has been already studied)

user21820

Anyway it's still easy. You can use the method for maximal subset, but with backtracking. If you think carefully, you should be able to figure out how to get a quadratic-time algorithm, because the depth of the recursion is just log(n) and each node takes O(n) time to expand the current set to a ⊕-closed set including any added element.

Hold on. Maybe not quadratic-time. I'll need to think more. But it's certainly bounded by O( n^log(n) ).

Mark

@user21820 this approach is very similar to an exhaustive one, and it is too expensive. For this reason I was referring to some theorem / proof existing in the literature but I couldn't find anything

user21820

13:02

@Mark But it's not anywhere near exponential, unlike the naive brute-force you suggested in your 3rd-last comment.

But there probably is a better algorithm.

Mark

this characterization of V / T is correct? V = {s \in S | \forall s1 \in S, s \oplus s1 \in S}

Pax Daga

13:52

@user21820 Thankyou for your comment there now I am even more confused two high rep users have condradicted each other can you please exlapin why its wrong

Pax Daga

14:05

Not even one step that I did not explain

please help me know I am sorry for the past

If neg A and neg B:
  neg A and neg B[Restate]
  neg A [and elim]
  neg B.[and elim]
  If A:
    A [restate]
    neg A [restate]
    Contradiction [neg elim]
  If B:
    B [restate]
    neg B [restate]
    Contradiction [neg elim]
  A implies Contradiction
  B implies Contradiction
  (A or B) implies contradiction[or elim]
  neg(A or B) [neg intro]

user21820

14:22

@Mark I do not know what you mean by V / T. Although I'll think about a better algorithm, you should first of all figure out the simpler algorithm.

Mark

@user21820 ok thanks, I understood the simpler algorithm

Pax Daga

16:28

I figured out my thanks to G kemp mistake one of my reasonsigs was wrong @user21820

namely the or elim one'

Ill do p6 now

should I also redo P1?

The mistakes there were just silly mistakes

user21820

@PaxDaga Yes that is the first error. The next line is also wrong, for a related reason. But...

Ok so you've finished (P1) and (P3) and half of (P4). You haven't done (P2) and the other half of (P4).

Pax Daga

If neg(A or B):
  If A:
    A or B
    Contradiction
  Neg A
   If B:
     A or B
     Contradiction
  Neg B
  Neg A and Neg B

@user21820

Why is this wrong?

Should I redo?

I think that its correct @user21820

Will this also be thrown in trash?

If neg(A or B):
  If A:
    A or B
    Contradiction
  Neg A
  If B:
    A or B
    Contradiction
  Neg B
  Neg A and Neg B

Transcript for

♦ $Mathematics$ Basic Mathematics

Transcript for

♦ Basic Mathematics

♦ $Mathematics$ Basic Mathematics