@10:13 since outer loop has y and inner loop has x so in each ordered pair first element is y and second is x but by mistake i have considered first element as x and second as y...... @10:13 i said 1 is greater than 1 , 1 is greater than 2 and so on but i should have said 1 is lesser than 1 , 1 is lesser than 2 and so on.... because first element is y and second is x and question says y
Jo jo log sir ke kiye gaye mistake ko pakad le rahe ho congratulations..tum log sir se sahi gyan le pa rahe ho..... actually DMS is not easy subject but sir made this easy for us...
Sir I have watched all graph theory videos then I understand that I have to watch logic first then I started logic. Really impressive contents. Sir my humble request is please make content on engineering maths. Please🙏
@39:30, In current example of 3 varible nested quantifier, i.e. x + y + z = 10, we can say that, from all the options, the option that has ∃z at last postions will be True, b/c we can choose any one value for z from domain and get 10.
Sir aapki short trick se + jo aapne indepth understanding di usse sare ques apki video pause krke kree sare ek dum correct aur speed se hogye .really sir apki short trick aur smjhane ka tarekaa superr se bhi uparrr
Really Really Thank you sir this is an nightmare topic for me. I tried it uncountable time but every time , unable to get proper understanding but now this is crystal clear thank you so much sir.
Yeh phela comments section jaha bahuto ko doubt h even mujhe bhi kal fir video repeat krke dekhna padega But thnx sir apne apna best diya merko nhi lagta isse best pure yt mai koi explain kr sakta h....i guess yeh topic he bahut tough tha
Thanks for the detailed explanation sir, pehele ye sab dekh kar he darr lagta thaa, iis video ke bad dekh kar he pata chal zata hai kya horaha hai🙂🙂🙂!!!!!
Sir ji at 09:18 when you changed the inner loop to x and outer loop to y , p(x,y) will make pattern like ((1,1)^(2,1)^(3,1))v((1,2)^(2,2)^(3,2))v ((1,3)^(2,3)^(3,3) ) , Please correct me sir
@@pavansingh8089 sir interchanged the position of x and y should this be done ??? means loop me to y bahar and x andar hoga but kya oredered pair me bhi (y,x) ho jayega ??? sir plse reply
sir interchanged the position of x and y should this be done ??? means loop me to y bahar and x andar hoga but kya oredered pair me bhi (y,x) ho jayega ??? sir plse reply
@@cosmophile8218 As per the Q , y is the first coordinate in the ordered pair i.e., (y,x) since the outer loop represented by Y runs first. Since the general form of an ordered pair is (x,y) hence the positions interchange making it (1,1) , (2,1) and so on.
Sir ek doubt hai.... 53:40 par ∀x ∃y ∀z mein ham x ki sabhi values k liyea check krege ki koi ek y aisa hona chahiye jiske lea sabhi z dalne par true aye Yani ki sabhi x ki liyea check krna hai to ham x1 k liye krege fir x2 fir x3....... Xn To maan lete hai ki x1 k liye check krte samay hame y ki ek value mil jati hai (y1) jiske liye sabhi z dalne par true ata hai..... (Maan lete hai ki aise values mil gy..... Actually nahi hongi, par man lo hai) Ab ham x2 ke liye bhi same krege.... To is baar kya ham y ki koi alag value le skte hai.... Ya y1 he lena pde ga.... Means y ke sath there exist hai....to kya x ki sabhi values k sath y ki same value leke prove krna pde ga... Ya alag alag le skte hain jise x1 k liye y1 aour x2 k liye y2........ And so on??
@53:00 Sir, on question no. (g) while solving by short cut you said that y must be independent from z that's why it is false. but on question no. (d) also we can use short cut and declair that x must be independent from y that's why it is false but you said that is true. Please, sir can you clear my doubt. Thank you
In option g) by using shortcut method y must be independent on z but in equation (y=10-x-z ) , y is dependent on z that's why it's false. Now, in option d) x must be independent on y but in equation (x+z=10-y) , z create problem in making dependency of x on y. Hence, it's true.
@@AmitKhuranaSir sir pair ka order change kyu karenge ?? wo to sirf loop ki bat hai na ki y ki value fix rake aur x change karte jaye jaise loop chalta hai .. and it will be (1,1) (2,1)(3,1) (1,2)(2,2)(3,2) (1,3)(2,3)(3,3)
sir pair ka order change kyu karenge ?? wo to sirf loop ki bat hai na ki y ki value fix rake aur x change karte jaye jaise loop chalta hai .. and it will be (1,1) (2,1)(3,1) (1,2)(2,2)(3,2) (1,3)(2,3)(3,3)
@@it7264 sir pair ka order change kyu karenge ?? wo to sirf loop ki bat hai na ki y ki value fix rake aur x change karte jaye jaise loop chalta hai .. and it will be (1,1) (2,1)(3,1) (1,2)(2,2)(3,2) (1,3)(2,3)(3,3)
Hi... Watch video again you will definitely get it.... I have explained everything in easiest possible way.... Thoda pain to aega samajhne me because main deep me ja ra hu topics ke......
yes ..m also having same doubt becoz we need just one value then ..in this way all are true exept first right? but sir did it in another way ,please clear my doubt if u got it .
If we fix x=0, then we check for, 0+(for all y)+(for all z)=10 Now, for make condition true equation must be 0+1+(for all z) =10.but it's not possible Hence, it's false.
Part 6 is saying For every y their exist at least one x such that x>y that is y is less than x Since domain has only 3 values so one by one check all values When y is 1 then x can be 2 or 3 When y is 2 then x can be 3 But when y is 3 then we can't find x So this is false..... Hope it clears your doubt.....
@10:13 since outer loop has y and inner loop has x so in each ordered pair first element is y and second is x but by mistake i have considered first element as x and second as y......
@10:13 i said 1 is greater than 1 , 1 is greater than 2 and so on
but i should have said 1 is lesser than 1 , 1 is lesser than 2 and so on....
because first element is y and second is x and question says y
sir same thing you have said @13:20 i think it should be Yx
Jo jo log sir ke kiye gaye mistake ko pakad le rahe ho congratulations..tum log sir se sahi gyan le pa rahe ho..... actually DMS is not easy subject but sir made this easy for us...
27:19 for short cut of nested quantifiers
28:00,30:20 imp statement
Thank you so much sir....we can never be enough grateful for your knowledge and delivery of complex concepts in such a lucid manner...🙏🙏🙏
Thanks sir for explaining in such simple ways. Your patience and humbleness is incredible.
sir ap itna cool hoke kese padate ho bahut hard hota hai board ke samne without student samjana
Iske liye ghor tapasya karni hoti hai..
Sir you converted a nightmare topic into an easy one 🙏🏻✨
sir one by one video i have gain alot of knowledge thnkyu sir its very helpful
Best ever explanation for this topic on internet. Sir, you make this hard topic like a cakewalk.
Sir I have watched all graph theory videos then I understand that I have to watch logic first then I started logic. Really impressive contents. Sir my humble request is please make content on engineering maths. Please🙏
@39:30, In current example of 3 varible nested quantifier, i.e. x + y + z = 10, we can say that, from all the options, the option that has ∃z at last postions will be True, b/c we can choose any one value for z from domain and get 10.
Sir aapki short trick se + jo aapne indepth understanding di usse sare ques apki video pause krke kree sare ek dum correct aur speed se hogye .really sir apki short trick aur smjhane ka tarekaa superr se bhi uparrr
Really Really Thank you sir this is an nightmare topic for me. I tried it uncountable time but every time , unable to get proper understanding but now this is crystal clear thank you so much sir.
Best video sir!!!!🤩🤩🤩🤩 I never understood this before watching this video... Thank you so much...
sir , app maths ke jaadugar ho.... ek dum se hi subject ko interesting bna diya + baccha maths kar diya ......:))))))))))
Yeh phela comments section jaha bahuto ko doubt h even mujhe bhi kal fir video repeat krke dekhna padega
But thnx sir apne apna best diya merko nhi lagta isse best pure yt mai koi explain kr sakta h....i guess yeh topic he bahut tough tha
Thanks for the detailed explanation sir, pehele ye sab dekh kar he darr lagta thaa, iis video ke bad dekh kar he pata chal zata hai kya horaha hai🙂🙂🙂!!!!!
Math never looked so easy before
Sir ji at 09:18 when you changed the inner loop to x and outer loop to y , p(x,y) will make pattern like ((1,1)^(2,1)^(3,1))v((1,2)^(2,2)^(3,2))v ((1,3)^(2,3)^(3,3) ) , Please correct me sir
Yes you are right. Sir made a mistake. he has already mentioned in his pinned comment.
@@pavansingh8089 sir interchanged the position of x and y should this be done ??? means loop me to y bahar and x andar hoga but kya oredered pair me bhi (y,x) ho jayega ??? sir plse reply
sir interchanged the position of x and y should this be done ??? means loop me to y bahar and x andar hoga but kya oredered pair me bhi (y,x) ho jayega ??? sir plse reply
@@cosmophile8218 As per the Q , y is the first coordinate in the ordered pair i.e., (y,x) since the outer loop represented by Y runs first.
Since the general form of an ordered pair is (x,y) hence the positions interchange making it (1,1) , (2,1) and so on.
21:34 ans is option a and c
i wish i found this channel earlier before wasting my time on half explained time constraints channels
Amazing Sir
13:34 (5), (6) is also satisfied.
5th is false, 6th is true
@@kuldeepkumarprajapati4596 nah i think 5th is true because for pairs ((3,1),(3,2),(3,3)) here all these are satisfying x>=y for some x and all y.
THANKS A LOT SIR
love you total concept perfectly understood at 52:00 minutes time
sir app bhaut aacha padate hain jinhe basic nahi aata hain to thoda basic bhi clear kra dijiyr please bhaut log h basic ke wait mein 👍🙏🏼
Sir ek doubt hai....
53:40 par
∀x ∃y ∀z mein ham x ki sabhi values k liyea check krege ki koi ek y aisa hona chahiye jiske lea sabhi z dalne par true aye
Yani ki sabhi x ki liyea check krna hai to ham x1 k liye krege fir x2 fir x3....... Xn
To maan lete hai ki x1 k liye check krte samay hame y ki ek value mil jati hai (y1) jiske liye sabhi z dalne par true ata hai..... (Maan lete hai ki aise values mil gy..... Actually nahi hongi, par man lo hai)
Ab ham x2 ke liye bhi same krege.... To is baar kya ham y ki koi alag value le skte hai.... Ya y1 he lena pde ga.... Means y ke sath there exist hai....to kya x ki sabhi values k sath y ki same value leke prove krna pde ga... Ya alag alag le skte hain jise x1 k liye y1 aour x2 k liye y2........ And so on??
We can take different y also
Yes because y will change with x and z
@53:00 Sir, on question no. (g) while solving by short cut you said that y must be independent from z that's why it is false. but on question no. (d) also we can use short cut and declair that x must be independent from y that's why it is false but you said that is true. Please, sir can you clear my doubt. Thank you
In option g) by using shortcut method y must be independent on z but in equation (y=10-x-z ) , y is dependent on z that's why it's false.
Now, in option d) x must be independent on y but in equation (x+z=10-y) , z create problem in making dependency of x on y. Hence, it's true.
the 3 var one was a bit confusing at the start, but later on i felt better about it.
This session was1 hour but I completed in 2 hours and finally I feel confidence this topic.
Maza agaya sir 😊
Kaise mujhe aap mil gaya, kismat pe aayi na yakeen.
13:21 sir tuple m phla element to y h or dusra x so we should get T v F v F = T instead of F v F v T
yes i have done a small mistake here thanks for identifying that.....
THEN ALSO It will be F^F^F
@@AmitKhuranaSir sir pair ka order change kyu karenge ??
wo to sirf loop ki bat hai na ki y ki value fix rake aur x change karte jaye
jaise loop chalta hai ..
and it will be
(1,1) (2,1)(3,1)
(1,2)(2,2)(3,2)
(1,3)(2,3)(3,3)
sir pair ka order change kyu karenge ??
wo to sirf loop ki bat hai na ki y ki value fix rake aur x change karte jaye
jaise loop chalta hai ..
and it will be
(1,1) (2,1)(3,1)
(1,2)(2,2)(3,2)
(1,3)(2,3)(3,3)
@@it7264 sir pair ka order change kyu karenge ??
wo to sirf loop ki bat hai na ki y ki value fix rake aur x change karte jaye
jaise loop chalta hai ..
and it will be
(1,1) (2,1)(3,1)
(1,2)(2,2)(3,2)
(1,3)(2,3)(3,3)
Can anyone elaborate @21.54 problem? Would be a great help.
sir apne jo short trick bataya hai quantifiers ke wo mujhe samajh nhi a rhe hai.....
Hi...
Watch video again you will definitely get it....
I have explained everything in easiest possible way....
Thoda pain to aega samajhne me because main deep me ja ra hu topics ke......
at 55:55 super sir i cant show my exictness
sir samajh mein aagya
thanks^-^
Irrational numbers are also real number
Here is the outstanding session ❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️❣️
Thank you sir🙏
thank u very much sir!
iss topic pe sirf number based questions bante ??
free mei koi nahi pada skta itna sab kuch impossible!!!!
mere paas words hi nahi hain kuch kahne ko ..............
@25:10 If we have x=5, then to obtain the sum=10, y should also be equal to 5. Is it okay if x and y are same integers?
Ha because answer same. Ayega
Yes
Big supporter 😁
sir at 53:04 in (f) if we fix x to 0 then it is possible right? then why it is not true as they are asking for just one x pls clear this.
yes ..m also having same doubt becoz we need just one value then ..in this way all are true exept first right? but sir did it in another way ,please clear my doubt if u got it .
If we fix x=0, then we check for, 0+(for all y)+(for all z)=10 Now, for make condition true equation must be 0+1+(for all z) =10.but it's not possible Hence, it's false.
@@enjoylearning210 watch video again from starting.
Thanks Sir.....
You are god for me
Will anyone suggest any more detail, to understand further about 4th Proposition @44.00 i.e, Ex Ay Ez(P(x,y,z)). E- There exists and A- For all.
Ping me on telegram and send your doubt.....
44:19
Sir I didn't understand the difference between 3 and 6 part at 11:08
Part 6 is saying
For every y their exist at least one x such that x>y that is y is less than x
Since domain has only 3 values so one by one check all values
When y is 1 then x can be 2 or 3
When y is 2 then x can be 3
But when y is 3 then we can't find x
So this is false.....
Hope it clears your doubt.....
30:05
👍
Mja aagya sir
2/7/23
SIR AAP TOH TOPIC SIKHANE KI BAJAYE GADHO KI TARAH RATTWA RHE HO SIRF
Root 2 is irrational number not a real number.
Its still under real number