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Thank you for your videos even after 6 years. I have a question, please how do you negate “For every strictly positive epsilon there exists a strictly positive delta such that |f(x)| < e (epsilon) whenever |x| < delta ?
Work with turning it into symbols first. Three things you’re talking about. Any integer x and any integer y. Then there’s a z in between For all x, for all y, there exists a z such that x
Correct. Remember that a statement can be true or false. In this episode, I think Trev was just trying to show how to translate from words to notation, not necessarily giving a true theorem. You'd need absolute values around x and y (or some other modification) for the theorem to be true.
the statement is: if x^2 > y^2, then x > y. let p = x^2 > y^2 let q = x > y the statement is p➡q Truth Table : T for True F for False p q p➡q T T T T F F F T T F F T The statement p➡q would be false if p is True and q is False. That's the second row. So for your example, if x^2 > y^2 is True then x > y is False. Input values for p (hypothesis) before you input values for q (conclusion) x = -5 (but it could also be 5 since we input value on x^2 first) x^2 = 25 y = -3 (but it could also be 3 since we input value on y^2 first) y^2 = 9 Substitute (on the second row): If 25 > 9 is True, then -5 > -3 is False. This is true. But not for all x and y, because if x=5 then x^2 could still be 25 and if y=3 then y^2 could still be 9. If 25 > 9 is True, then 5 > 3 is False. This is false because 5 > 3 should be true right? Therefore, the second case in the video is correct. You just got confused with the correct statement "if x^2 > y^2, then x > y" because you thought it was "if x > y, then x^2 > y^2". Thanks for reading.
Consider the compound proposition (Allm∃n [P(m,n)]implies( ∃nAllm[P(m,n)]) where both m and n are integers. Determine the truth value of the proposition if (a) P(m,n) is the statement “m < n”. (b) P(m,n) is the statement “m | n”. [By “m | n”, which we say as “m divides n”, we mean that n = km for some integer k.] how to do these? And it will be a great help if you make a video with similar problems like these.
Hi, I hate taking notes on paper, what note taking software are you using to make this video? I'm currently using my stylus/wacom tablet but the drawing software that comes with it is wayyyy more than I need to just take notes.
@@mohamedasim776 The point is that you didn't declare what is x. If you put S(x) then it would be recognizable as a student. If you say for all x (m(x) or c(x)) then you're including everything (or everyone) 'in the entire universe' to be taking major in Math or CompSci even though they are not students. This is what I understood, sorry if I interpreted it wrong.
A little confused here. In the previous video we learnt that ∃xPx ~Vx[~P(x)] So why is the first negation example in this video showing that ∃x (Px ^ Qx) Vx (~P(x) v ~Q(x)? Shouldn't the V quantifier have a ~ in front of it? I also tried applying the trick you showed in the previous video to negate quantifiers, and it gave me ~Vx (~Px v ~Qx ) If anybody could clear this up it would be greatly appreciated
~Vx[~P(x)] is equal to ∃xPx, its not the negation of it. Since the question wants the negation of ∃x (Px ^ Qx) its going to be Vx (~P(x) v ~Q(x)). its like saying A = ~~A , but the negation of a is not ~~A its just ~A. i hope this cleared it up.
@5:09, you meant "There is some number that is odd AND even", right?? since that's what [∃x(Px＾Qx)] translates to, no? Then you negate it to "There is not a single number that is odd or even", or ¬[∃x(Px＾Qx)]
I think if you write ¬[∃x(Px＾Qx)] and translate it, it would become ∃x(¬Px v ¬Qx)]. The AND would change into OR if the negation is distributed inside the parenthesis. And as far as I have learned, it's a rule in propositional logic. Please correct me if I'm wrong, I'm just a beginner on this topic, sorry.
Express this statement using quantifiers: “Every student in this class has taken some course in every department in the school of mathematical sciences.”
My attempt: ∀ x, ∀ D [S(x) -> ∃C [D(C) ∧ C(x)]]. For all people x, for all departments D, if x is a student in this class: S(x) then there exists a course C such that a) this course was taken by the student: C(x) and b) the course belongs to a department: D(C).
Hi mate , can u help me on this? 1. a) Consider the statement: For every integer x and every integer y there is an integer n such that if then . i. Translate the statement into symbols. Clearly state which statement is the proposition p and which is the proposition q, the quantifiers and the connectives.
![[Discrete Mathematics] Unique Quantifier Examples](http://i.ytimg.com/vi/HIaNF4O3i7U/mqdefault.jpg)
![[Discrete Mathematics] Unique Quantifier Examples](/img/tr.png)
Check out my new course in Predicate Logic: trevtutor.com/p/master-discrete-mathematics-predicate-logic
It comes with video lectures, text lectures, practice problems, solutions, and a practice final exam!
Thank you for your very clear examples!
Thank you for your videos even after 6 years. I have a question, please how do you negate “For every strictly positive epsilon there exists a strictly positive delta such that |f(x)| < e (epsilon) whenever |x| < delta ?
∃ε > 0, ∀ δ > 0 [ |x| < δ ∧ |f(x)| > ε ]
I am not entirely certain though, and even if 1 year has passed I hope it is helpful.
@@CodeKali The correct one would be : ∀ε > 0, ∃δ > 0 [ ( |x| < δ ) --> ( |f(x)| < ε ) ]
In the last exercise, how did you go from Ex [ - (Px -> Qx) ] to Ex [ -(-Px v Qx) ] "-" denotes "not" and "E" denotes "there exists"?
Because it's equivalent to the original value and is stated as the "law of implications" go browse this k 👍
5:16
Q:- Write a nontrivial negation:
- Between any integer and any larger integer, there is a real number.
What the solution???
Work with turning it into symbols first.
Three things you’re talking about. Any integer x and any integer y. Then there’s a z in between
For all x, for all y, there exists a z such that x
nice examples, thank you
Isn't the second case at 2:06 incorrect? if x were to be -5 and y were to be -3/ 25 is >9 but -5 is not > -3.
Correct. Remember that a statement can be true or false. In this episode, I think Trev was just trying to show how to translate from words to notation, not necessarily giving a true theorem. You'd need absolute values around x and y (or some other modification) for the theorem to be true.
the statement is:
if x^2 > y^2, then x > y.
let p = x^2 > y^2
let q = x > y
the statement is p➡q
Truth Table :
T for True
F for False
p q p➡q
T T T
T F F
F T T
F F T
The statement p➡q would be false if p is True and q is False. That's the second row. So for your example, if x^2 > y^2 is True then x > y is False.
Input values for p (hypothesis) before you input values for q (conclusion)
x = -5 (but it could also be 5 since we input value on x^2 first)
x^2 = 25
y = -3 (but it could also be 3 since we input value on y^2 first)
y^2 = 9
Substitute (on the second row):
If 25 > 9 is True, then -5 > -3 is False.
This is true. But not for all x and y, because if x=5 then x^2 could still be 25 and if y=3 then y^2 could still be 9.
If 25 > 9 is True, then 5 > 3 is False.
This is false because 5 > 3 should be true right?
Therefore, the second case in the video is correct. You just got confused with the correct statement "if x^2 > y^2, then x > y" because you thought it was "if x > y, then x^2 > y^2".
Thanks for reading.
Consider the compound proposition (Allm∃n [P(m,n)]implies( ∃nAllm[P(m,n)]) where both m and n are integers. Determine
the truth value of the proposition if
(a) P(m,n) is the statement “m < n”.
(b) P(m,n) is the statement “m | n”. [By “m | n”,
which we say as “m divides n”, we mean that n = km for some integer k.]
how to do these? And it will be a great help if you make a video with similar problems like these.
Hi, I hate taking notes on paper, what note taking software are you using to make this video? I'm currently using my stylus/wacom tablet but the drawing software that comes with it is wayyyy more than I need to just take notes.
windows journal
Very helpful video.....thx man
Thanks for nice and clear video
Why is it for all x (s(x) arrow m(x) or c(x))?
Is it supposed to be for all x (m(x) or c(x))?
My question is on the first sentence of translation
If you say "all x (mx or cs)" then you're saying everything IN THE ENTIRE UNIVERSE is either m(x) or c(x).
@@Trevtutor So What's the point in that?
@@mohamedasim776 The point is that you didn't declare what is x. If you put S(x) then it would be recognizable as a student. If you say for all x (m(x) or c(x)) then you're including everything (or everyone) 'in the entire universe' to be taking major in Math or CompSci even though they are not students. This is what I understood, sorry if I interpreted it wrong.
thank you
Hi, firstly thanks for this beneficial video :) My question is where did we know that (Px -> Qx) = ~Px v Qx, Thanks beforehand :)
Definition of the conditional. P -> Q is equivalent to ~P v Q.
Thanks so much)
You can use that equivalence to use "OR" instead of "Implication" for some equations
~(Px -> Qx)=Px and ~Qx
@@Trevtutor or law of implications, that is, right?
A little confused here. In the previous video we learnt that ∃xPx ~Vx[~P(x)]
So why is the first negation example in this video showing that ∃x (Px ^ Qx) Vx (~P(x) v ~Q(x)?
Shouldn't the V quantifier have a ~ in front of it? I also tried applying the trick you showed in the previous video to negate quantifiers, and it gave me ~Vx (~Px v ~Qx )
If anybody could clear this up it would be greatly appreciated
~Vx[~P(x)] is equal to ∃xPx, its not the negation of it. Since the question wants the negation of ∃x (Px ^ Qx) its going to be Vx (~P(x) v ~Q(x)). its like saying A = ~~A , but the negation of a is not ~~A its just ~A. i hope this cleared it up.
The reason why V needn't to have - is cuz it's already the negated form is E
If i have some predicate so that it's statement is considered to be false, when i negate it does it become true? Is that how it works?
Yes.
@5:09, you meant "There is some number that is odd AND even", right?? since that's what [∃x(Px＾Qx)] translates to, no? Then you negate it to "There is not a single number that is odd or even", or ¬[∃x(Px＾Qx)]
I think if you write ¬[∃x(Px＾Qx)] and translate it, it would become ∃x(¬Px v ¬Qx)]. The AND would change into OR if the negation is distributed inside the parenthesis. And as far as I have learned, it's a rule in propositional logic. Please correct me if I'm wrong, I'm just a beginner on this topic, sorry.
i like the 3:05 part
4:16 that looks like a C function pointer a little
what aspect, bro ???
What if there are two variables like x,y
does any one know why he did add "-" at 3:51
I LOVE YOU
Express this statement using quantifiers:
“Every student in this class has taken some course in every department in the school of mathematical sciences.”
My attempt: ∀ x, ∀ D [S(x) -> ∃C [D(C) ∧ C(x)]]. For all people x, for all departments D, if x is a student in this class: S(x) then there exists a course C such that a) this course was taken by the student: C(x) and b) the course belongs to a department: D(C).
Why didn't you negate all your quantifiers as example? It would be so helpful
4:00
ty
Hi mate ,
can u help me on this?
1. a) Consider the statement:
For every integer x and every integer y there is an integer n such that
if then .
i. Translate the statement into symbols. Clearly state which statement is the proposition p and which is the proposition q, the quantifiers and the connectives.
What does this mean...i mean it's kinda incomplete I guess
Im watching that in 2024 💀💀💀
@@abderraouf2m me too have exam tomorrow
my brain is not braining atp
YAWA KJA
What does any of this even mean?
Were you even speaking English?