Part-01: Can You Solve This Lost IIT Problem ??? | Aman Sir
Вставка
- Опубліковано 15 жов 2023
- Discussion on an important topic, that's forgotten from all the textbooks of IIT.
Subscribe to @BHANNATMATHS for brain-twisting videos around maths.
Join our Telegram : t.me/bhannatmathsofficial
----------------------------------------------------------------------------------------
#bhannatmaths #amansirmaths #jee #jee2024
For any query/doubt mail us at: bhannatmaths@gmail.com
For All Notifications Join Our Telegram Group: @bhannatmaths
Join our Telegram : t.me/bhannatmathsofficial
Lets assume given statement is correct then iota = 1/ iota
Sending iota to LHS then (iota ) square =1
-1 = 1 which is not possible our assumption is wrong and hence iota ≠ 1 / iota
False hai sir by 1=-1(never possible)
Sir apne Denominator jo ki √1 hai apne ise √[(-1)×(-1)]=√1 =1 kar Diya hai jo ki galt hai
√[(-1)×(-1)]=√(-1)²=-1 hota hai
Sir yaha par mistake hai see this
√(-1/1)≠√{(-1)/[(-1)×(-1)]}≠√[1/(-1)]
Name Asfar
From Pakistan
Reason ye ha keh √-1 complex field ha or √1 real field ha dono ko combine nahi kia ja skta kabi
The answer if False
Because LHS = i, while RHS = 1/i
IIT to itna seedha sawal nhi dega: That's why FALSE 😂😂
The rule is you cannot multiply inside root if the numbers are not positive real numbers
Right.
Because at the last when you will cross multiply then you will get a false equality as {-1 = 1} so this is false and i is not equals to 1/i .
Here is division, not multiplication
😅 division and multiplication are same bro a/b. means
a * 1/b
No you can if both are negative or both are positive
I think there is an issue in the third step where you multiplied -1 on top and bottom, so what you essentially did was multiply √-1 to both numerator and denominator. It was not a problem with the denominator but in case of numerator you multiplied √-1 with √-1 and wrote 1 (the operation you did was √-1.√-1= √ (-1).(-1)=√1, but we cant really do that here as both the numbers are -ve (√-a)×(√-b)≠√ab if both a and b are negative, one of them has to be positive)so √-1.√-1 should be equal to i.i= i²= -1 and not 1 ...that was the issue with this problem.
Definitely this must be the correct explanation 💯
Bro lekin exactly multiply thodi kiya bas sign ko upar se neeche shift kiya
If there is error then it will be in 3 to 4 step
@@praveertechnogaming9584
Bhai to usse aise likhne ke liye technically maths main upar neeche multiply karna padhta hain
Definitely it must be the correct
Explanation 💯
Last step is incorrect:
√(a/b) = √a/√b has condition
it only holds when -π < Arg(a)-Arg(b) ≤ π
but in √(1/-1), Arg(1) = 0 and Arg(-1) = π
so the difference will be -π, and equality will not hold
basically, denominator cannot be taken as negative
Sir from converting to second step to third step we multiplied numerator and denominator by root -one but there is a concept that 2 roots can't be multiplied if both of them have negative sign inside.
I think the second step is wrong because in the backend of this step we would have to multiply by iota in the numerator and denominator and then it would be correct.
Sir First of all thank you for giving us that kind of rare problems that we can't get in today's books. Now I am explaining it :- First of all 1/iota = iota^-1 and iota^-1 not equal to iota. Second thing is iota is a part of complex number and the part which is purely added with iota that is call pure complex part. So we can't calculate the value of pure complex part by using simple algebraic process. e.g :- √-6 × √ -6 we can't write it √36 = 6 . May be I am wrong with my answers but I as a student I should attempt it. And at last I am saying honestly that sir , you are doing very good job.❤❤❤🎉🎉🎉
That's one of first question of complex no in cengage
So I don't think it's that rare,
I love your teaching skills
Sir I think in order to convert -a/b to a/-b we have to multiply both numerator and denominator with -1,but in this case doing that causes an imaginary number to change from imaginary to a real number and that causes the whole number to change. Therefore it is not possible.
Logic is not correct
because in rational domain -1/1 is always equal to 1/-1
so you don't need to multiply numerator and denominator, you can just substitute
only last step is wrong
Kyuki ham square root ko split nhi kr sakte hain..... square root ka distribution aur recombination sirf positive real numbers ke liye hi defined hota hai.
To equate 1st and last step it says that
(iota ÷√1)=(√1÷iota)
Collect iota at 1st side:
(iota)^2={√1×√1}
As number in both root is +ve we can multiply them
So,(iota)^2=√1
But √1=(-1),(1)
So if we take √1 = -1 so,
{iota}^2=(-1) which is right statement
So if we take √1=-1 then
Answer can be true
❤❤❤❤ wow amazing question
Agar koi simple number fractional form me hai that is -a/b agar ham issme numerator and denominator Dono ko '-ve ' number se multiply kre to hame a/-b milta hai pr root ke ander esa krna possible nhi hai
Problem is in third step
As /-1×/-1≠/1 but = -1 it is a rule that we have follow for complex no multiplication
in 2 second step we only multiply iota but we don't divide it with iota hence the 3 step is not equal to 1st step
I'm thinking about two kinds of things:
This can be true because, √-1/√1 = √1/√-1 , since √1/√-1 can be written as, √(-1)(-1)/√-1, here √-1 gets cancelled, so we have √-1 that is equal to √-1/√1
(If we're just talking about if the intial and final quantities are equal)
Or this can be false if we look at the steps because -p/q = p/-q isn't valid for imaginary numbers
I'm really curious about the actual solution
First and last quantities can also be proved unequal :-
First quantity is iota (i)
Last quantity is 1/iota (1/i)
If we multiply and divide last quantity by iota (i)
1/i × i/i
= i/i²
= i/-1 (• i²= -1)
= -i
Hence first quantity is iota (i) and last quantity is -iota (-i). Hence they are unequal.
But, 1/i = √1/i = √(-1)(-1)/i = i × i / i = i
I know you're right as well, but how would you explain this? Let me know if I'm wrong somewhere
1/i = √1/i = √(-1)(-1)/i
Till this it is true but after this you cannot write it as √(-1) × √(-1) and then i × i
Because the following property √a×√b = √ab satisfies only when a and b are positive number.
For negative number the property becomes √-a × √-b = -√ab
The proof of this property is mentioned below :-
√-a can be written as √a × i
√-b can be written as √b × i
So now √-a × √-b = √a × i × √b × i
= √ab × i²
= -√ab
So √-a × √-b = -√ab
Hence in you proof you cannot write it as √(-1)(-1)/i = √(-1) × √(-1)/i = i × i/i
The proof I mentioned here, you can verify it on Google.
The property I mentioned also has a famous Olympiad question
√-4 × √-9 = ?
The answer most people will give is 6 because they will just multiply things under the square root but you have to follow property :- √-a × √-b = -√ab. And hence answer would be -6 to question I mentioned.
You can search this question and its solution on UA-cam.
@@shauryakansagra8922 okay thanks a lot for this great explanation
Very nice problem ❤
If iota = 1/iota . then, iota square = 1. Thus this equation gives two possible values of iota that are 1 and -1. But iota = root -1.
I have a book named ML Khanna which was based on IIT paper back in 1980s, it was bought in 1987 and it has this same question
Which page no bro ?
And in which part algebra 1st or algebra 2nd ?
Namaste Ma'am, Do you have the book - "Calculus on several variables" by ML Khanna?
The one on extreme left is simply iota by but the one on extreme right is (iota)^raised to power -1 which is in form of 4n+3 form and i gives -(iota) 👍
what 4n+3 form?
Hi, sir as i know the second step you did on taking square root on Nr(-1) and Dr(1) was not valid because this shifting of whole root is allowed if the number in individual root is positive . I think this may be a possible reason that the overall ans is FALSE.
And also sir interchanging sign is giving you conjugate of the complex given.
And we know conjugate and complex is not equal
Sir is this reason is also correct
yes it can be
yes it can be
Sir maan lete hai lhs aur rhs barabar hai toh inka square bhi same hoga chahiye aap iota ka square kijiye ya 1/iota ka answer -1 hi aayega so by contradiction ye statement true hai
CONCEPT- √a . √b not equals to √ab when both a , b are negative . According to this concept,
we can't write √(-1/1) = √(1/-1) because when we cross multiply each other LHS not equals to RHS.
You have caught it ❤❤❤❤🎉🎉🎉🎉 congratulations your answer is true ... brother ❤❤❤
The concept you told is different, and the thing which you are writing is different, you told that if a and b, both are negative, but here, -1 is negative while 1 is positive.
@@AdityaKumar-gv4djhe's confused but he's got the spirits!
You can multiply numerator and denominator by a complex number (sqrt(-1)) only when none of them are complex.
In step 3 when you shifted -1 in denominator by multiplying and dividing by √-1.
And √-1x√-1 is not defined which is a blunder. If you solve it furthur, then you will get 2 different answers
√(-1)x(-1) = √1=1 and we will do it different way
√-1x√-1= (√-1)^2= -1.
Yes it can be possible as it is the property of under roota and basic algebras....
root
Not sure, but one way to prove this could be by method of contradiction.
Let's assume that the end result is True, which will then mean that: i = 1/i
which then means i^2 = 1
But, this is a contradiction because, the proven fact is that i^2 = -1. Therefore, this can't be true.
firstly, the operation we did gives us conflicting results BECAUSE by definition, the thing you're taking the square root of cannot be negative, unless we're dealing with imaginary numbers. and we are, so we cannot just apply real number techniques to an imaginary number and expect it to work.
RHS = i, while LHS = 1/i = i^4/i = i³ = -i, which ofcourse is predictable as we performed the operation a/-b = -a/b, an operation defined only for real numbers, for a complex number.
we violated the law and thus we get a conflicting result. that's about it
Logic is not correct
because in rational domain -1/1 is always equal to 1/-1
so you don't need to multiply numerator and denominator, you can just substitute
only last step is wrong
@@theseusswore-a/b = a/-b is true for all complex numbers.
the law which got violated is √(a/b) = √a/√b which doesn't always apply to complex numbers
Sir pls bring more such videos ❤❤❤❤❤❤❤
Sir, apki hi to videos dekhkar maine Jee ki preparation start ki thi
I think Its false,
👉 First of all it is the basic rule that when you merge or split the roots both numbers should be positive whether it might be product or quotient , like, let a and be are 2 positive natural numbers √(-a/-b) is defined though √-a/√-b individually not defined. Also if we consider a,b as two functions then their domain will change after splitting and merging which no longer be equal to original function.
👉Here in statement indirectly i=1/i is given which I think is false statement. Cause in this cause basically function is changing though it may be constant. Many people give false proof of 1=-1 by this method.
Sir please let me know if I'm correct or not ?
-a/b and a/-b ka value same hee. But matlab alag alag hee.
And we can't multiple (-) to a number when it is under root.
I try it with equatting
After rationalization of 1st and last terms
I think it is true because sir in 1st one if we see root-1/root1 let it ans be i and when we see the last step root1/root-1 it's ans had to be -i but we can't forget that root 1 can be +-1 so the ans can be be i too of last step.
Sir we can't take combined square root on step 2 and not also reverse in step 4 because for this both the numbers in the numerator and denominator should be positive
sir in step of -a/b is equal to a/-b if we multiply both numerator and denominator by (-1) or by taking minus common. But in (-1)^½ me ham minus common nikal he nhi sakte sakte hai....
That's why it is false
First step is wrong, although the equality remains after the first step..it is mathematically incorrect. When we have sqrt(ab), we can write it equal to sqrt(a)sqrt(b) only when a>0 and b>0, else we can not split the square root like that. For example if we write sqrt(-4)*sqrt(-9) (which is equal to -6) as sqrt(-4*-9) then it will sqrt(36) which is 6 and is absolutely incorrect. As a sidenote a comprehension was asked in jee advanced in physics in a topic which involved square roots. We all know that n(refractive index) = sqrt(relative permeability*relative permittivity). But there are meta materials which have negative refractive index and for them both relative permeability and relative permittivity are negative so the standard formula fails and we have to write n(refractive index) = 1/(sqrt(relative permittivity*sqrt(relative permeability)
I think first thing came in mind first expression was i and second was 1/i so definitely they are not equal however question is where is the mistake see we know that -a/b and a/-b are equal but we cannot apply this concept in roots like example √-2*√-3 ≠√6. Same way in division we cannot write √1/√-1 as whole √1/-1 or √-1/√1
exactly!
sir i think its true as we not left iota in denominator so when we rationalize lhs=rhs
(-a)/b= a/(-b) means multiply by -1 on both denominator and numerator. If square root is applyed on (-a)/b, in order to obtain (-b) you should multiple by sqrt(-1).which gives minus square root of "a" on numerator.
Transformation from 2nd to third is incorrect . As minus is missing before square root in 3rd step.
Logic is not correct
because in rational domain -1/1 is always equal to 1/-1
so you don't need to multiply numerator and denominator, you can just substitute
only last step is wrong
@@harsh_will_iit how do we can prove that , in rational domain -1/1 is always equal to 1/-1 ?
The second step is wrong as that operation in sqrt is only valid for positive no.s. hence the given result is false. The way we cannot equate sqrt(a)sqrt(b) = sqrt(ab) where either of a and b is negative....similarly we cannot equate the above.😊
❤I like your teaching sir. ❤
While dealing with negative numbers inside a square root, the basic rules can't be applied.
Here, clearly √-1/√1=i whereas √1/√-1=-i. Surely i≠-i.
Hence it is false.
Here 1 is equal to 2 is equal to 3 but the fourth part is not equal to the rest because the first three symbolise iota while the last one symbolises 1 by iota
It is based on minus into minus equals to Plus
Sir 1 baar tetration kai uper bhi video bnna do uski history btta denna mzza aaiga❤❤
Sir that particular property of splitting is only valid for non negative real numbers. And -1 is negative
The second step is false because when there are positive integers inside the root, then only we can separate the roots of numerator and denominator..
It is false because lhs is i and rhs is -i,we cannot multiply two numbers under root,if they are not positive,it is just like the fact that 3i multiplied by 3i is actually -3 but when we multiply by taking them together under one root ,it gives 3
I think the last step is wrong. Not sure, but my expln is: We can pull out or push in positive no.s inside or outside the root. The last step pulled out -1 from the root, which caused the contradiction, otherwise we would have had i=sqrt(1/-1) which is correct.
The simple answer is No as we cannot Assign same root if Negative and Positive Integers occour simultaneously
This is 100% true, because if take the very first term and the very last term we get √-1/√1= √1/√-1 which can also be written as i/√1 = √1/i, if we square botu sides we get i^2/1=1/i^2 therefore by cross multiplication we get i^4 = 1 which is infact true therefore the equation is correct.
Sir third step is wrong as when we go from -a/b to a/-b we hate multiply and divide by -1
Since there is a root over it so we have multiply divite by iota to make it possible.
Also when it will be multiplied by √(-1/1) × √(-1/-1) the property says if √a×√b = - √ab if both a & b are negative hence √(-1/1) × √(-1/-1)= - √(1/-1)
Hence going to third step was wrong as it does not follow fundamental rules of maths.
sir i think its wrong because if we have a given function as y = ((f(x))^0.5)/((g(x))^0.5) we cant write this as y = (f(x)/g(x))^0.5 because in first case both f and g functions cant be negative but in second case both f and g can be negative simultaneously...as it change the range of function this step is wrong
Flase h sir kyuki aapne jo first step kiya पूरे uper of niche par ek saath root laga diya wo nhi lagega
Sir, according to me the problem is in whiles proceeding from 2 to the third step
because my ma'am said -1 is not sustained in the denominator that's why we put "minus " with both the denominator and nominator
so according to me, the problem is in the third step.
Ans- False
Name- Ritanjay kumar mishra
class ( xii )
It is wrong because negative should always be in numerator , keeping negative sign denominator is wrong ( I think this ,it may be wrong )
The rule (-a/b=a/-b ) fisible where a and b are real numbers.but this is a complex number
It can't be bcs... at first its i. but the result its 1/i if we simplify we get i/i² which means -i
The answer is false because i = -1/i ( by values we know )
but in iota we can't apply the 2 nd step which sir have done
If i would have been equal to 1/i it should have been correct but that is not true if it was the case then we would get -1=1 more precisely -a/b=a/-b works only if we apply it outside the root not inside which you did
pehla wala √-1/√1 = √-1/1 = √-1 = iota
and last wala √1/√-1 = 1/√-1 = 1/iota
as we can se that the last one is the reciprocated form of the first one, also, iota is not equal to 1/iota, hence it's false and they are unequal.
Now, we know the answer is false (100% sure), the reason must be that we cannot apply the exponential formula of √(a/b) = √a/√b , if either of them is a negative number or something like that.
Maybe am wrong lol since haven't studied much about complex numbers yet.
False hai kyuki second step me numerator me ek imaginary quantity hai aur uska ham division kar sakte iska karan ye hai ki imaginary quantity cannot hold inequality
Thank you sir for this good IIT Ques. ❤️😊
I think it is true
FIRST TERM = ±iota
Last term = ± 1/iota aur yisko rationalize krne pr bhi ± iota hi aa rha hai
Means both are same
√a.√b=√(ab) ; for real nos
But if both a and b are imaginary nos
Then,
√a.√b=-√ab
Sir easy hai
first step ka solution iota milta hai
and last step ka solution -1 milega its false
last step ko samjo
1 upon iota
upper and niche iota se multiply kijiye to iota upon iota square milega jiska answer -1 hoga
pure false by this explanation if i m wrong then pls reply right and correct it
If we compare the ratios root -1/ root 1 and root 1/root -1 and cross multiply then we would get 1i^2= 1 which is not satisfying the equality
the first one reults to be equal to iota whereas the other one results to be the mulplicative inverse of iota.
iota is an imagenary number and 1 is a real number so taking comon root is wrong because after involment of an imagenery number we must use complex number rules.
Sir if we solve some part we get i^2=1 if this is true we can solve any quadratic equation and get real roots if this happens maths will break down at this point
Root 1 = +1,-1
False hoga kyuki 2 cases banenge jab hum dono side ko compare karenge ki root 1=-1 ho ya root 1=+1 ho or dono cases ko solve karne par (iota)² = +1 aa raha h jo ki galat h.
since sir wrote -a/b=a/-b which is true because after cross multiplication its ab=ab but now in the given question i/1=1/i and after cross multiplication its iota squared = 1 which is not true since iota squared is -1 therefore the ans is false
Sir ap logo ne energy 2024 cancel kr diya kya?I was waiting for it eagerly ,came from very far but found it was cancelled.Please reply...
अगर हम दोनों को एक साथ बराबर लें और √-1 की जगह आयोटा रखें तो i^2=1 आएगा जो कि असंभव है क्योंकि i^2= -1 होता है।❤
We dont write 6i*6i as +6..SO i guess we cant expand and write it as whole root,So aswer is iota not 1/iota
sir false. reason being root-1/root1 ko whole root -1/1 nhi likh sakte because whole root -1/1 ya phir plus ayega ya minus toh dono same sign ke saath open honge root mei se but when we talk of root -1/root 1 toh kya pata denominator wala neg open ho ya positive, neg hua toh ans 1/i hoga lekin positive hoga toh ans i hoga therefore it depends and matters ki whole root likha hai ya seperately root likha hai.
We apply -a/b=(-1)(-a)/(-1)(b)=a/-b but since we are dealing inside a sqareroot where multiplication through negative numbers is not allowed this statement is false
sir false hoga kynki 1st mea iota aayega
aaur 2nd mea (1/iota= -iota) aayega
and underoot negative value par root minus a x root minus b is not egual to root ab isleye steps false hai
what i find the error is that (√-a)×(√-b)≠√ab if both a and b are negative, one of them has to be positive)so √-1.√-1 should be equal to i*i= i²= -1 and not 1
Sir plzzz iski explaination dijiyega. Love u sir from Pakistan.. you are the best maths teacher i have seen in my lyf❤❤
It is not possible because if we write √-1/1 as √1/-1 then we are multiplying the numerator and denominator by -1 and then two -1s come under root √-1×-1 /-1 but it is not appropriate to write in such a way that's why it is not possible
I think sir there is mistake in 3rd step
As if we are multiplying √-1 with numerator & denominator we'll get (√-1)(√-1)/(√1)(√-1) this is the only mistake according to the rules of complex no. We cannot multiply 2 roots of negative no. Therefore the answer is false.
If equal honge then cross multiply krke 1=-1 hojayega which is not equal so dono equal nhi honge
Sir second step likhane ke liye donon number positive real number hone chahie isliye is question ka answer
In carrect hoga
False because given condition tab hi use kar sakte hai jab root me dono nos positive ho
If we cross multiply the constants i/√1 and √1/i , then we see that i^2 = +-1 whereas √1.√1 = +1
Hence, false
There is false step while going 1st step to 2nd step because Ratio of complex no. to real no. isn't applicable in the laws of exponent as square root law. ...Please justify whether true or false. ❤❤❤❤❤❤❤❤❤
Sir i think there is a mistake in the last step that (a/b)½≠ a½/b½ when a is positive and b is negative
Wrong because
If √1/√-1=√-1/√1,then
√1=1,put in
1/√-1=√-1/1
1/√-1=√-1
So put,√-1=i
Then,1/i=i
Multiply both sides by 1/i
1/i×i=i×1/i
1/i²=1
We know i²=-1
Then,1/-1=1
Multiply both sides by -1
1/-1 ×-1=1×-1
1=-1
Fault in 3rd stap because you can't give negative sign in divisor when it have a square root
Third step is wrong. To get the third step we need to multiply numerator and denominator by √(-1). But to get the third step it is assumed that √-1 * √-1 =√1 which is wrong.
Amazing 😍
Sir it is false because original was iota and final it is 1/iota which is -iota and the problem was in between 2nd and 3rd step where we wrote √-1/1 as √1/-1 which was done by multipling it with √-1/-1 and we know we cant multiply a root with negative quantity inside with another root with negative quantity inside as it gives incorrect result. Example if we say √-3×√-3=√(-3)×(-3) which will be √9 which will be 3 but it actually is -3.
This is false because 1st vla iota ke equal hoga and last vala -iota ke to ye equal nhi hoga !!👍 Because 1/iota= -iota 👍
1/iota=iota^-1 🧐
1/i = -i kab se hone laga😢😢
@@souryajitghosh9341 bhai denominator rationalize kar
Bhai 1/i =-i kr barabar he hota h
1/i ×i/i =i/i² =(-i)
1/iota should be equal to iota^-1?
Sir 2 step me galti h kyuki - ka sign upar se niche lee jaate time aapne upar niche √-1 se multiply kiya to numerator me √-1×√-1 ho gya jo directly √((-1)(-1)) ke equal nhi hota that's why this step is wrong
🙏 according to me it is false because jo aapne identity Lagayi hai √a/√b= √(a/b) ye only real numbers ke liye valid hai but √-1 is considered as imaginary....
If i/√1 = √1/i
which by cross multiplying gives
i^2 = 1 that is false...
as we know i^2= -1..
So, should be false..
in the second step you did an error .
inside a square root we cannot change sign from numerator and denomentor .
AND if one no. is negative and other positive then we cant seprate the root
Similarly using this same error , any no. can be prooved to be its negative
Logic is not correct
because in rational domain -1/1 is always equal to 1/-1
so you don't need to multiply numerator and denominator, you can just substitute
only last step is wrong