answer: 1.False , because when x= -2 . (x (x^2-1) became (true) -> (false) = false 2.A.True , because when x= -2 (x !=2) ^ (x^2=4) became (true) ^ (true) = true 2.B.True , because when x=-2 (x (x^2-2>1) became (true) ^ (true) = true Correct me if im wrong :)
brother with due respect, i think 1. is TRUE , 2(a) is FALSE(as it is existensial quantifier except 2 all integer should be equal to 4 but only -2 satisfy the function) and 2(b) is also FALSE as no value less than 0 satisfy the function.
@@mdmuktadirmazumder284 but in 2nd a and b part that sign is of "there is" which means if it is true for any single value of x in domain so it is truth value will be true.
We can use implications only with "for all" as you said in the video then why are you using it with "there exists"?? Can anyone help me in clearing this???
It may at first seem that "Some x satisfying P(x) satisfies Q(x) '' should be translated as ∃x(P(x)⇒Q(x)), like the universal quantifier. To see why this does not work, suppose P(x)="x is an apple'' and Q(x)="x is an orange.'' The sentence "some apples are oranges'' is certainly false, but ∃x(P(x)⇒Q(x)) is true. To see this suppose x0 is some particular orange. Then P(x0)⇒Q(x0) evaluates to F⇒T , which is T, and the existential quantifier is satisfied.
1. False for all x less than 0 (-2^2 =1 ) 4 =1 which is false 2.a) True since we have -2 in which gives -2^2 =4 b) True the integers more from -3 to -inf obeys the rule
Your way of teaching is fantastic,does'nt feel like boring lectures,👏👏😎
Thank you sir. As always very clear in your explanation.
Which defines why the use of -> for all x0) and why use of ^ in the case of their exixts
answer:
1.False , because when x= -2 .
(x (x^2-1)
became (true) -> (false) = false
2.A.True , because when x= -2
(x !=2) ^ (x^2=4)
became (true) ^ (true) = true
2.B.True , because when x=-2
(x (x^2-2>1)
became (true) ^ (true) = true
Correct me if im wrong :)
brother with due respect, i think 1. is TRUE , 2(a) is FALSE(as it is existensial quantifier except 2 all integer should be equal to 4 but only -2 satisfy the function) and 2(b) is also FALSE as no value less than 0 satisfy the function.
@@mdmuktadirmazumder284 you are completely wrong
@@mdmuktadirmazumder284 but in 2nd a and b part that sign is of "there is" which means if it is true for any single value of x in domain so it is truth value will be true.
@@mdmuktadirmazumder284 you are completely wrong
1st is true because -1 satisfy
-1 < 0 ^ (-1)^ 2 = 1
Means true ^ true is always true
Answers of HW:
1. False
2. (a) True (b) True
Thankyou Sir you are really helping a lot...very awesome explanation...ek ek topic dhang se clear ho raha hai...Thankyou so much Sir 🙏
Q2:a:T when x=-2
b:T when x=-2
How is b true for x=-2? Wouldn't that be 0>=1
@@niccolopaganini1782 when x=-2,
(-2)^2 - 2>=1
4-2>=1
2>=1
hence, True.
Waiting for the next lectures...pls upload as much as possible...
Please post all videos of Discrete Math as soon as possible.
We can use implications only with "for all" as you said in the video then why are you using it with "there exists"??
Can anyone help me in clearing this???
same Q.if you have found the answer then let me know
@@maliaqakhan7502 no I haven't
Binod 😁
ikr
im also confused with the same that is Q2 (b) part
I am too😂
que 2 part (b) u said that implies in for all and AND for exist.
same doubt
Even now we still visit your lectures
Sir aap bhout achcha padate ho please please sir poora discrete mathematics upload karo
Sir plz my doubt clear about it
¥x is ->
and there is an = ^
But you use -> in at least one how?
In homework question 2. B
Any one explained why sir wright implications in (b) second home work problem there has to be ^
Was wondering the same thing
1)F
2
a-T
b-T
But the symbol (implication) shd not be there in the question 2.b) sir
It may at first seem that "Some x
satisfying P(x)
satisfies Q(x)
'' should be translated as
∃x(P(x)⇒Q(x)),
like the universal quantifier. To see why this does not work, suppose P(x)="x is an apple''
and Q(x)="x is an orange.''
The sentence "some apples are oranges'' is certainly false, but
∃x(P(x)⇒Q(x))
is true. To see this suppose x0
is some particular orange. Then P(x0)⇒Q(x0)
evaluates to F⇒T
, which is T, and the existential quantifier is satisfied.
1. F
2a. T
2b. T
1--- F
2 a---- T
2 b ----- T
Please post all videos of discrete mathematics sir...
Please upload all videos of discrete mathematics
1. False for all x less than 0 (-2^2 =1 ) 4 =1 which is false
2.a)
True since we have -2 in which gives -2^2 =4
b)
True the integers more from -3 to -inf obeys the rule
But for 1 we can use -1 , -1 square is 1 rite.. then it becomes TRUE..
@@KandhaMaaran-10 it is exist for only -1 but in question there is for all so the statement for all x is false
but in que(2(b)) we can't use implication in Existential Quantifier
1) False
2) a. True, b. True
Observe 2 b there exists and for all both occurs so it is a contradiction
Thanks you sir
a) true because we use x=-2
b)true because we use x=-3
But it says not equal to two and if you use 2 you will get two but the question says not equal to two. Please explain further to me
Separate fan base for neso academy....
I need of next lecture link
Thanks for this sir
Sir first ka answer false
Second ka answer true
Third ka answer true hoga
Notes available hai kahi ??
1. F
2.(a) . T
2.(b) .T
Sir please send the link of next lecture......i want to clear the question which you hiden the Ans in next lecture
Discrete Maths: goo.gl/7VUE7z
Quantifier over domain for finite video send
1. false
2. true
3. true
1) False
2) a - True
b - False
Sir next lecture link...
Answers False, True, True
Sir please make the next lecture
Sir network theory complete kr dijiye please 🙏
Y playlist m khan h .?
Where are Java lectures?
Ok thanku
i think he is following kenneth h rosan discrete mathematics.....
I dont get that last question...how?
Solve the equation , for x less than negative root 3 there exists a solution hence true
home work
q1) false
q2) i) false
ii)true
How did you get 2.i) as False? X=-2 is the single case to prove it as True.
(false, true, true)
Second view
Homework problem
F T T
Answers of HW:
True
False
False
false
true
true
false
true
trye
F,T,T
Q1:T
False. You have to prove all values in domain with condition for universal quantifier.
First view
F, T, T