Basic Mathematics - 2020-01-12

user21820

02:17

@user8718165 Not every teacher is a "sir". The distance between two sets A,B of points is defined as inf( { dist(p,q) : p∈A ∧ q∈B } ). "inf" is the infimum. In some cases, such as when A,B are lines, this infimum is in fact the minimum. So you want to find points p∈A and q∈B to minimize the distance between p,q. One general method is to simply write down the general formula for p,q and minimize dist(p,q)^2 by any means such as differentiation.

An ad-hoc method for lines is to observe that the line through p,q must be perpendicular to A, otherwise you can move p along A to decrease dist(p,q). Similarly that line must be perpendicular to B. Thus it suffices to find a vector v that is perpendicular to both A,B, and then solve for p−q = kv for some real k. Since p,q each have one real parameter, you are effectively solving a vector equation for 3 real parameters, and there is a unique solution if the lines are not parallel.

user8718165

03:20

@user21820 Hello. Sorry, I was just a bit unsure so went with "sir". No worries I'll refer to you with the username :-)

user21820

@user8718165: Hello. That's great. Do you get what I said above?

user8718165

@user21820 I got the second part of the message. I'm just a beginner so please bear with me :)

user21820

@user8718165 If you're in high-school, you can ignore the first part, as it is the more general definition of "distance between point-sets".

user8718165

@user21820 I'm confused about the equation in Cartesian form. Where are the determinants coming from?

@user21820 yes. In high school. Thanks for telling me though. Still, I read the 1st part a few times. :)

user21820

@user8718165 If you have a determinant, it is most obvious via the ad-hoc approach, because a 3d vector equation with 3 real parameters is nothing more than 3 simultaneous linear equations in those parameters.

user8718165

03:28

@user21820 okay got it. Can I show you once?

user21820

Definitely please do.

user8718165

let's say we have two lines given by $$l_1:\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}=k_1$$
$$l_2:\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_1}=k_2$$

so the perpendicular distance between them is given by $$d_\perp=\dfrac{\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1\\
a_1 & b_1 & c_1\\
a_2 & b_2 & c_2\end{vmatrix}}
{\sqrt{\left(\begin{vmatrix}b_1 & b_2\\c_1 & c_2\end{vmatrix}\right)^2+\left(\begin{vmatrix}c_1 & c_2\\a_1 & a_2\end{vmatrix}\right)^2+\left(\begin{vmatrix}a_1 & a_2\\b_1 & b_2\end{vmatrix}\right)^2}}$$

user21820

@user8718165 I don't like this formulation. It unnecessarily excludes lines parallel to an axis, so the formula fails to be meaningful in those cases.

The general equation of a line should always be in vector form, namely { a+bq : q∈R } for some vectors a,b.

user8718165

@user21820 yes, in the book the case of parallel lines is given separately. This is from my book :)

user21820

a,b can of course be represented by coordinates a[1..3] and b[1..3].

It's better to use the proper formula from the beginning.

user8718165

03:39

@user21820 yeah. I found it like $\vec r=\vec a+\lambda\vec b$

@user21820 okay. I got it. So the equation I posted before is a bit incomplete... Is it?

@user21820 hello :) Are you there?

user21820

@user8718165 I'm thinking what is the best way to do this.

Normally, if you just want to compute the answer, you just follow the adhoc method, since it is trivial to find a vector perpendicular to both lines.

But if you just want the distance...

user8718165

03:57

@user21820 hello. Thank you so much. Sad, but I know of no other way of doing this. That's the only way stated in my textbook.

user21820

Why don't you try the way I gave you? Name me a vector perpendicular to b,d.

user8718165

@user21820 okay, let's say it's p

user21820

What I mean is can you tell me immediately a vector perpendicular to b,d?

Have you learned about cross products?

user8718165

@user21820 yes. I know a bit about them Is this what we're talking about $$d=\left |\dfrac{(\vec {b_1}\times\vec {b_2})\cdot(\vec{a_2}-\vec{a_1})}{|\vec {b_1}\times\vec {b_2}|}\right|$$

@user21820 hello. Did I say something wrong? Sorry.

user21820

04:15

@user8718165 No I was away from my computer for a while.

user8718165

@user21820 okay no worries. :-) We're talking about cross products.

user21820

@user8718165 No. I asked for a vector perpendicular to b and d.

The two lines we are interested in can be expressed as a+bq and c+dr, where a,b,c,d are vectors.

We want a vector v such that v ⊥ b and v ⊥ d.

There are many, but name me one.

You said you learned about cross products, so it's odd that you can't answer my question. b×d is automatically perpendicular to both b,d.

user8718165

@user21820 Sorry. I didn't think about it. Got it.

user21820

(You can take a look at this intuition concerning that fact.)

So immediately we have a vector equation. We want to find q,r,k such that (a+bq)−(c+dr) = vk.

And that's it; we would have found the exact points on the lines that have the minimum distance between them.

user8718165

@user21820 Yeah . Got it.Thanks for the link. I saw it.

@user21820 can I ask you another question?

user21820

04:29

Sure.

user8718165

@user21820 I got this approach. I'll use it for sure :) Thanks a lot.

user21820

(Generally I dislike formulae involving determinants, as they look nice but are very inefficient in practice. I'm also unable to find a simple proof of the formula you asked about. Google brings up this and this, both of which involve geometric arguments.)

user8718165

@user21820 I don't understand this. "there are many, but name me one." are there multiple lines which can be perpendicular to both b and d?

user21820

If v ⊥ b,d, then obviously vk ⊥ b,d as well for any real k.

user8718165

@user21820 thanks, I saw the first one. But the second one is new to me :-)

@user21820 okay got it... Thank you so much.

user21820

04:35

The point is that we know that the minimum distance is along a line segment which is perpendicular, and it happens to be the case that every vector perpendicular to b,d is in fact a multiple of b×d (if b,d are not parallel).

user8718165

@user21820 Is it that they'll lie on the same line b×d. Just that they'll have different lengths because of k?

user21820

A vector is not a line.

1 hour ago, by user21820

The general equation of a line should always be in vector form, namely { a+bq : q∈R } for some vectors a,b.

All lines parallel to v will be of the form { a+vq : q∈R }. But vectors parallel to v must be vk for some real k.

And yes, different (length,direction) because of k.

user8718165

@user21820 okay. I got it. Thanks a lot.

user21820

You're welcome!

adesh mishra

05:29

@user21820 I have started reading the book which you have linked in Logic , I have got my first doubt

@user21820 it writes An equivalence structure is a pair (A,$\approx$) where A is a set, $A \neq \emptyset \\ \approx \subseteq A \times A$

such that the following axioms are satisfied : (A1) For all x , $ x \approx x$

My doubt is what does $\approx$ represents? Is it a set (beacuse only a set can be a subset of another set, it is written that $\approx ~ \subseteq A\times A$ ) or is it a relation (because it is written $x \approx x$)

Physics freak

05:54

@user21820 I am not able to fix broken indentation(kindly excuse that). For the first blank I realised the error so here it goes. Given any real x:
(2x^2−1)^2 ≥ 0.
... [4x^4+1-4x²≥0 » 4x²(1-x²)≤1]
4x^2·(1−x^2) ≤ 1. Please help me with the second blank

user21820

09:12

@adeshmishra It is both. When studying logic, it is important from the start to know that you are working in a meta-system (MS) that can talk about objects, strings and sets of them, sometimes sets of subsets. The part you quoted says that an equivalence structure is a pair (A,≈) where A is a set (of objects) and ≈ is a relation on A, namely a function from A^2 to bool, where bool is the two boolean values true and false.

In some kinds of MS, such a concept is native. In the modern de facto MS, which is typically a kind of set theory (due to social and historical factors), there is no such thing as a set of truth values. But one can encode such a concept as a set. In this case, as a subset of A×A.

That is why that textbook says it is a subset S of A×A that satisfies certain properties. Under this encoding, ( x ≈ y ) is encoded as ( ⟨x,y⟩ ∈ S ).

From the perspective of the axioms, you should not care about the encoding at all, because it is irrelevant to the structure itself. But from the perspective of MS, encoding is necessary otherwise we cannot even define what a structure is.

@Physicsfreak As I said, you must press "fixed font" before "send". Then whatever you paste (including spaces and tabs) will be preserved. Your first blank is now correct. Note that there are 3 statements and every one of them is true under the context "x is a real".

To fill the second blank, use the following fact:

> For any reals a,b, if a ≥ b ≥ 0 then √a ≥ √b.

Let me repeat the outline I gave you, with your filled in first blank.

Given any real x:
  (2x^2−1)^2 ≥ 0.
  4x^4+1−4x^2 ≥ 0.
  4x^2·(1−x^2) ≤ 1.
  If |x|≤1:
    1−x^2 ≥ 0.
    √(4x^2·(1−x^2)) ≤ √1.
    ... [fill this in] ...

Maybe I should have given you a bit more detail.

Given any real x:
  (2x^2−1)^2 ≥ 0.
  4x^4+1−4x^2 ≥ 0.
  4x^2·(1−x^2) ≤ 1.
  If |x| ≤ 1:
    x^2 ≤ 1.
    1−x^2 ≥ 0.
    4x^2·(1−x^2) ≥ 0.
    √(4x^2·(1−x^2)) ≤ √1.  // by the fact I just stated
    ... [fill this in] ...

The last line is just a trivial algebraic simplification for you to do, so I won't give it. Fill it in and paste-fixed-font-send and I'll check.

adesh mishra

11:55

@user21820 Ma’am I have read your reply many many times but I think it is so well written that I’m unable to understand it. I concluded from your reply that $\approx $ is a relation and $\approx~ \subseteq ~ A \times A$ means that it takes an input from A xA and gives output a bool I.e. either true or false value, am I right?

adesh mishra

12:07

@user21820 what does $x \approx y$ means? Does it mean that relation $\approx$ is true between x and y ?

user21820

16:31

@adeshmishra I need to know your background. Which year undergraduate are you in? Pure or applied mathematics? Can you code in a programming language?

As I said in the other room, I believe you have completely no familiarity with basic logic. But mathematics relies completely on such familiarity. If one cannot do basic FOL reasoning, one cannot have full understanding of any branch of mathematics. Even more in the branch of mathematics that studies logic itself, because you must both work inside an FOL system (called the meta-system) as well as reason about FOL systems, without getting confused about what you are doing.

Based on our interaction so far, I actually recommend you hold off on a logic text (which studies FOL), and make sure you know how to simply use FOL first, in the following steps:

(1) Propositional logic.

(2) First-order logic.

(3) First-order Peano Arithmetic.

As I suggested earlier, look at the Examples linked from my profile under "Natural deduction", and go through the Rules up to "Boolean operations". If they seem to make sense for you, try the following exercises and post your attempts here (paste, then click "fixed font", then click "send"), and I will check them.

Dec 9 '18 at 7:19, by user21820

(1) A and ( B or C ) implies ( A and B ) or ( A and C ).
(2) A or ( B and C ) implies ( A or B ) and ( A or C ).
(3) ( A or B ) and ( A or C ) implies A or ( B and C ).
(4) ( A or B ) and ( B or C ) and ( C or A ) implies ( A and B ) or ( B and C ) or ( C and A ).
(5) ( A implies B or C ) implies ( A implies B ) or ( A implies C ).

Once you get them all correct, we will move on to FOL.

adesh mishra

17:42

@user21820 okay I will do it.

@user21820 where is the exercise?

Transcript for

♦ $Mathematics$ Basic Mathematics

Transcript for

♦ Basic Mathematics

♦ $Mathematics$ Basic Mathematics