[Discrete Mathematics] Negating Quantifiers and Translation Examples
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In this video, we translate English sentences into quantificational logic and then negate some quantificational statements.
Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.
![[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!
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 👍
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.
nice examples, thank you
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.
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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
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
Very helpful video.....thx man
thank you
Thanks for nice and clear video
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.
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.
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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.
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.
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?
Why didn't you negate all your quantifiers as example? It would be so helpful
ty
i like the 3:05 part
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).
does any one know why he did add "-" at 3:51
4:16 that looks like a C function pointer a little
what aspect, bro ???
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
4:00
@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.
my brain is not braining atp
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What does any of this even mean?
Were you even speaking English?