6. practice questions on minimization
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- Опубліковано 9 вер 2024
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I have done a small mistake in Question 10 b part.....
intersection is distributive over set difference.....
Sir ,
1.union is distributive over set difference True or false ??
2.intersection is distributive over set difference true or false ??
sir what about third question when you are converting qand p' bracket then dot sign is come inside bracket also so this will be like P.(Q.P') then we will proceed sorry sir if I am wrong Sir please clear this doubt also because you write P.(Q+P') Sir kindly look into it. Thanks
@@Singhhvs conclusion from Q10.
Set Union is not distributive over Set Difference but Set Intersection is distributive over Set Difference
Set Difference is right distributive over Set Union and Set Intersection and when Set Difference is applied at left of Set union , it changes to set intersection following the principle of distributive law and vice versa
you have done many mistakes not just one. In some places you are writing intersection as +
@@badsanta7356 yes bro
i also found mistakes but was confused earlier that am i wrong
Timestamps :
Question 1 : 00:14
Question 2 : 01:00
* Question 3 : 02:06
Question 4 : 03:01
Question 5 : 04:05
* Question 6 : 04:29
Question 7 : 05:55
Question 8 : 06:55
Question 9 : 07:45
Question 10.a : 09:42
* Question 10.b : 12:14
Question 11 : 15:08
Question 12 : 15:30
Question 13 : 15:51
Question 14 : 17:14
Question 15 : 19:01
* Question 16 : 19:35
Question 17 : 23:01
Question 18 : 24:06
* Points and Corrections :
* Question 3 : 02:13 After converting to boolean algebra the expression will be *P.(QP')*
* Question 6 : 05:21 NOTE : (PQ)' is not the same as P'Q'
* Question 10b : 12:46 After converting to boolean algebra the expression will be *A(BC') = AB(AC)'*
* Question 16 : 22:16 *AB + A'C +BC = AB + A'C* is called the Consensus theorem
Thankyou bro ❤️
I'm following this playlist 🙏
@@Enigma-fk7mh Me too :) keep grinding 💪🏻
Thankyou for the rectification 🌟
Pin this 📌
thanks bro for the correction
Question 10 b: After converting to boolean algebra the expression will be A(BC') = AB(AC)'
Sir ne q10 ka second galat solve kiye kya?
RHS:
(AB) (AC)' 𝓬𝓸𝓷𝓿𝓮𝓻𝓼𝓲𝓸𝓷
(AB) (Ā+C') 𝓭𝓮𝓶𝓸𝓻𝓰𝓪𝓷𝓼
ABĀ + ABC' 𝓭𝓲𝓼𝓽𝓻𝓲𝓫𝓾𝓽𝓲𝓿𝓮
AĀB + ABC' 𝓬𝓸𝓶𝓶𝓾𝓽𝓪𝓽𝓲𝓿𝓮
0.B + ABC' 𝓬𝓸𝓶𝓹𝓵𝓲𝓶𝓮𝓷𝓽
0 + ABC' 𝓭𝓸𝓶𝓲𝓷𝓪𝓽𝓲𝓸𝓷
ABC' 𝓲𝓭𝓮𝓷𝓽𝓲𝓽𝔂
LHS:
A.(B.C')
ABC'
HENCE LHS = RHS
yeah i got the same thing was confused ki mera wrong hai ya nhi
Thx bro
Thx
Thnks vro 👍
Q3 )) small mistake
you have taken OR instead of AND
yes answer is 0
It's just error in quick solving if you can spot the error made by sir then you already knowing the concept on a good level, congrats!!
in que 10 part b for intersection..... dot will be use n.... aap ne + use kiya hai
Yes aur wo prove ho Raha hai agar sahi sign use kare to
@@thesumitkumar0 yes , sir ne galat kiya na?
Q3 it is both "and" inside and outside but you have taken "and" outside and "or" inside in boolean algebra expression
Hi
Yes I have done a small mistake thanks for identifying that....
In actual question "or" is inside "and' is outside....
@@AmitKhuranaSir I am getting 0 as the answer for the given question in the video. is it correct?
@@mr.prashant9139 yeah.
@@mr.prashant9139 me too
Yes
Hello sir...I am your new subscriber (So called your new student) ... Your way of teaching and the approach you provide us to solve logic questions are just awesome.... Thank you so much sir ... For all this quality content for us and for free... !! I promise you sir your channel will soon grow as top quality content in UA-cam for Gate exam...!!
conclusion from Q10.
Set Union is not distributive over Set Difference but Set Intersection is distributive over Set Difference
Set Difference is right distributive over Set Union and Set Intersection and when Set Difference is applied at left of Set union , it changes to set intersection following the principle of distributive law and vice versa
dusra statement thoda samajh nahi aya pls thoda samjhadenge ?
@@subhrakantsahu8857 yes second statement thora confusing hai
can u explain by writing expressions for right and left distribution?
can u expain me q 10 and 17 pls
@@subhrakantsahu8857 are kuch he is saying that question 10 mai do problems the na upar wala galat hai necha wala sahi hai sir ne galti kardi hai neche wale mai bas
10B is wrong, Set intersection is distributive over set difference. RHS was not solved correctly.
the left part also
hello sir. in Q3 how did you put + instead of and . It should be . right?
ya seems like sir made a mistake
@@aditithakur5349 tq
In 10th ques also he did the same thing
Yess same issue 🥲
p•(q•p') commutative
(p.p)'.q compliment
0.q domination
0
2:21 m question 3rd m And + thodi aayega vo to × aayega
wahi
thank you sir, i am starting your discrete math series for gate 2023
can u explain me q 10 and 17
A(BC') = AB(AC)'
The equation is not equal.
Proof:
Let's expand both sides using Boolean algebra properties:
Left-hand side (LHS):
A(BC') = ABC'
Right-hand side (RHS):
AB(AC)' = AB(A' + C') // De Morgan's Law
AB(A' + C') = ABA' + ABC' // Distributive Law
ABA' = 0 // A and A' are complements
So, RHS = 0 + ABC' = ABC'
sir i am a second year student and want to crack gate with good rank in the first attempt so sir i need your guidance
12:58 the expression should be (A.B).(A.C)' and the final result should be equal can anyone confirm it?
Yes you are correct I have made a small calculation mistake here....
intersection is distributive over set difference.....
@vidhayak proof it ?
I'm stuck
yes I got same in 10th second part
@@anmolthewonderboy
A INTERSECTION (B - C) = (A INTERSECTION B) - (A INTERSECTION C)
=> A.B.C' = A.B.(A.C)'
= A.B.(A' + C')
=A.B.A' + A.B.C'
=> 0 + A.B.C'
= LHS =RHS
in ques 10 :-
sir did mistake at b part ,,, a part is correct
conclusion of that ques is :-
union is not distributive over set difference.
intersection is distributive over set difference.
great video
Thanks sir 😊😊😊
in question number 8 in step 2 while converting from set to booolean you take and as + so whole problem after that cant understood
by the way awesome teaching , love your dedication and passion towards the series
hoping to enormous growth of your channel
no he didn't. there was only one OR operator so he put one + and rest are . as well . please check again
Q)13 i am getting AB as an answer i.e AnB
Thank you Sir🙏
this is best playlist for discrete mathematics
Thank You So Much for this Amazing Lecture 😊
thnks for this
so, sir in 10.b intersection is distributive over set differnce that is true
right ??
Yes
Good explanation
14:38
sir in the question number (3), The problem is : P ^ (Q ^ P' ) , But in the solution part when you converting inside the braces ( Q ^ P') HERE IT IS INTERSECTION , SO ho you converted intersection as + , while breaking. Please elaborate sir. Thank you.
Yes I have done a small mistake here....
Please read comments....
@@AmitKhuranaSir Okay sir, thank you for guide.
sir, maths was never interesting before watching this video !
Really sir
🔥🔥🔥
Happy teachers day sir
Day 2
Is the answer for 3 is 0 , anybody please reply
yes
Thank you
Sir there is a mistake in quetion no. 3, while converting into Boolean Expression instead of + ,there will be .
Pls REPLY
yes
q13 me .A+^AB +A+^BA
IS ME ANS A HAI
Q10 b is lhs = rhs
Q10a is lhs not equal to rhs
vahi, sir ne + likh diya hai
16:25 (Q13) ANS will be A BCOZ B+B`=1
Can you show your solution?
actually maine yeh vps dekha na maine mistake ki h lkin yeh sir n krvya h iska sol toh isme A+AB AYGA TOH A(1+B) hoga toh hume pta h 1+B =1 HOTA H TOH ANS "A" hoga isme ( B+B`) AAYGA NHI !
@@diyagupta9572still I get A+B as my answer.
Let me show you.
A +A'B +AB'
= A+B+AB' (By absorption)
= A+AB'+B
= A(1+B')+B
= A+B
Sir set walle question nahi samjh main aa rahe
Ab lagrha hain GATE ki preparation start kardi mene.
how's it going then
Q-3 me logic and hai to usko dot me change kar denge to fir apne beech me plus kui laga die hai
It's a Mistake.
Hi, Sir. If there is any correction somewhere then could you please pin your comment so that it can be always on the top and you don't need to reply again and again. Thank you.
thanks sir
Very nyc
Have some doubt on distributive law application in q.12
Please sol in detail
Q.8 was easy one step solution using absorption law
can u explain me q 10 and 17
there r mistakes sir. u have written and symbol as plus in 10 question
Jaldise 50 k karwa do sir ko sab, please
Q.3 ) p^(q^p')
= p.(q.p')
=p.(p'.q) commutative law
=(p.p').q associative law
=0.q
=0
Is this correct????? (2:43) ?????
bro i was thinking the same
@@outsmart2042 have you find the sol bro?
can u pls explain me q 10 and 17 pls
yes it is....
QUES 3 MISTAKE :-
ANS =0
Sir boolean ,set theory,yeh sab kya hai mujhe kuch samaj mai nahi aa raha aur apne kaha ki discrete mathematics se start karo
For question no. 13,the ans is (A+B)
i think its A+B
@@matrix6794 yeah, my bad😬
Sir how we will know that it is our last answer like
P.Q.P.R.S'
Ans-PQRS'
It is one step question
Cz u can't simplify further
Can anyone solve question 16 please
Fun fact sir is doing mistake Intentionally becauz they want student to current his mistake so after reading your comment he understood that everyone jad understood the concept warna tumlog je socho sir ek he video mein itni sari galtiya kasie kar sakte hai
😊😊😊😊
You have also solved Q.3 incorrectly, i guess. The conversion of P ^ (Q ^ P') will be P.(Q.P')
Q 3
P . (Q . P' )
P . Q . P'
P . P' Q
0 . Q
=0
Is it right sir please correct me if it is wrong ????
yo did you check if this is right or wrong
Q.14 my answer is a+b.
there is a mistake in 3rd solution sir u have taken "+ " in place of "." and solution is "FALSE "
Q -6)
PQR + P'Q'R +PQR(commutative law)
PQR + PQR + P'Q'R
A+A =A (idempotent law )
PQR +P'Q'R
taking R common
R(PQ + P'Q')
(compliment law)
R(1)
(identity law)
R.1 =R
ans => R
is this correct ?
No , coz here p' and q' have separate compliment p'q'=(p+q)'
mistake in 3rd Q. 0 will be the answer
❤❤❤
Sir aapne question 3 mein mistake kar di hai
q10 b part in conversion some error is there
Thanks for pointing it out yes I have done a small mistake in q10 b part..... intersection is distributive over set difference.....
@@AmitKhuranaSir sir ,plz check it once. Intersection is distributive over set difference.
3 is wrong
did lecture 6: 16 feb 2023
Sir, at 2:26 I think it should be P and Q and P', and the answer should be 0.
ya right i was thinking the same
Yes
My answer is right or wrong for question 10 (b)
A.B.C'=(A.B).(A'.C')
=A.B.A'.C'
=A.B.C'
sir already said about 10 b, there is small error. Check the comment
@@krishan4292 Yes , but i want to check my answer by the way thankyou
@@krishan4292 Are you preparing for gate 2023
You can join sir telegram channel, students help there
Wrong A. A'=0 A+A'=1
sir q3 me dono me dot hoga .
Aaj ka pura din ye video samazne me gaya
Weak hu na
#done(30.6.2024)
Sir, mujhe y samajh nhi aa rha hai... Property use krke aap kaise solve kr rhe hai..., , Mere mathematics kafi poor hai.. kaise kru mai..., Aur aapne bola hai ki logic pahle kro fir Krna TOC, uske chalte mai TOC bhi start nhi kr paya hu..,
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@@AmitKhuranaSir Ohk, Sir Thanks 🙏
@@AmitKhuranaSir sir 17th or 10th question samajh nh aya
Q. 17) kuch samja hi naho😢😢
Sir in Qno3 there is a mistake
sir third question you made wrong
sir mujhe to set me convert krke easy lag rah h boolean algebra me to smjh hi nhi aa rah kaise kre
q 10 a explain krdo sir ne kese distributive law laga diya wahan pe toh + and . h
Sir 15:43 pe smjh hi laga aapne shortcut laga diya video me kayi ques smjh nhi Lage kese hue direct shortcut laga diya
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If in q8th 1+RST=1(According dominative law)
Then in q6th
=Pqr+ p'q'r+pqr
=r(pq+p'q'). "According complement law a+a'=1 == pq+p'q'=1"
=r(1)
=r
p’q’ != (pq)’
Noo .
PQ + P'Q' = 1 is not always true statement
its actually true when we have,
= r(pq + (pq)')
= r(1)
= r
But here we have compliment for induvisual terms
ie, p'q' hence the law will not be applicable
it's p' q' not (pq)', so law is not applicable
@@jakoick1298 but cant we say both are equal
q3 getting 0
Please elaborate your doubt and mention timestamp of video....
@@AmitKhuranaSir all are AND gates so p and p complement will give 0 and 0 will dominate the result
@@sunny90908 thanks
sir is it compulsory to covert in boolean algebra
Converting to boolean algebra will make things easy and saves lot of time....
@@AmitKhuranaSir ohk sir
@@AmitKhuranaSir can u explain me q 10 and 17 again pls
Can anyone plz explain ques 4
Apply distributive law,
Sir in question no 6 why can't we apply complement law , then the answer will be PQ +P'Q' = 1
Did anyone know the answer of my question, then please clear it. Thank you
If there is whole bar on PQ then it's applicable.
10 ka 2nd problem LHS=RHS h , I respectfully say you did wrong.
Yes
lec 5 private kyu hai
Now please check...
@@AmitKhuranaSir
Thank you sir
Q.10) kuch samja nahi soln likega kohi comment mai
so many mistakes in one video. Confuse kr diya pura
true in this video theres many mistakes but sir is good
Thank you sir
Thank you sir
Thank you sir