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
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)]
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.
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
2:26 this step is wrong Thus the statement is false Because √(-a) × √(-b) ≠ √((-a)(-b)) Thus √(-1/1) = √(-1/1) × √-1/√-1 ≠ √((-1)×(-1)/(1×(-1))) => √(-1/1) ≠ √(1/(-1)) Hence the statement is false
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.
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 ?
Sir i think that in the step where we convert root(-1/1) to root(1/-1) is wrong. Because when we croos multiply in that step. We get -1 = 1. So the answer is false. Thank you sir for giving such good problems.
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)
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.
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.
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
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.
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.
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.
√-1/√1 = sqrt 1/ sqrt -1 As we know i ^ 2 = - 1 then i ^ 4 = 1 so , √- 1/1 =1/√-1 i / (i ^ 4) = i ^ 4 /i 1 / (i ^ 3) = i ^3 i^4/i^3 = i^3 i =i^3 i =-i So it is false
Sir i am sayan.... I'm in class 10th i think its true beacuse if √a² = | a | so. If √-1/1 = 1/√-1 So 1 = √ (-1)² according to the rule its true 1 = | -1| =1 If its not. .true then sir i want to know why √-4 . √-9 = 2 (iota) . 3 (iota) = 6 i² = 6 . (-1) = -6. Then why i² = -1. ?
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.
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.
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.
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
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
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
Answer to this question is false. Because iota=-1/iota We can prove this as follows: i= √-1/1 i= -√-1/(-1) -1= (√-1) ^2 So i= -√-1/(√-1) ^2) i= -1/√-1 Therefore, i= -1/i
Mistake here is that √a.√b is onlu possible if both a and b are positive 2:07 but here one of them is negative to we cannot take the root for the whole
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.
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.😊
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
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 iska answer dono h i is equal to -i (false ) for true i=-i (a) i+1=-i+1 2i=0 i=0 So , put i value in a 0=-0 : (0 is neither positive nor negative) 0=0 Hence proved true statement
Sir this should be false because in 1st step it is :-> (root(-1)/root(1)) which means i In next step it is also i but in next step it is 1/i. And they have given second step equal to third step, they mean -> i = 1/i When we transfer i from RHS (that is in divide) to LHS it will multiply with i on LHS and make it i². Now there statement says i² = 1, that is not true. Because all of us know i² = -1. Hence proved. Imagine if jee advance had these type of easy questions in exam now too.😅
-a/b = a/-b is obtained by multiplying -1 in Nr and Dr and this is true . But if you done this operation in square root, then this becomes √{(-1)*(-1)}/{-1} =. -√1/√-1 because √a×√b=. -√ab if both a and b are negative . So our expression becomes -1/i which is equal to i
It is false because sqrt(-1)/sqrt(1) = i but sqrt(1)/sqrt(-1) = -i. In the given equation the mistake is in 3rd step. sqrt(-1/1)is not equal to sqrt(1/-1) because while shifting the negative symbol from numenator and denominator we are not actually shifting it but multiplying both numenator and denominator by -1. In this case the correct steps are sqrt(-1/1)=sqrt(-1×-1/1×-1)=-1/sqrt(-1)=-1/i because sqrt(-1×-1) is not equal to sqrt(1) but = -1. This is the mistake in the given equation. And after that -1/sqrt(-1) = -1/i = -1×-i = i. This one is correct
According to me Its True .... 👇 Because lets √-1 / √1 => √-1 = iota (i) => i² = -1 ------ equation no. 1 Similarly... , √1/√-1 = 1/i =>squaring both sides... => 1/-1 = 1/i² By Cross Multiplication:- => i² = -1 ----- equation no. 2 Hence from eq. 1 and 2 We conclude that √-1/√1 = √1/√-1... Its my Opinion if any mistake explain in your next video
Sir I am a 7th class student but still...... 👇🏻 √-1 / √1 = √1 / √-1 ➡️ i / 1 = 1 / i ➡️ i×i = 1×1 [Cross Multiplication] ➡️ i^2 = 1 ➡️ i = √1 Because (i = √1) is False.... The given equation (√-1 / √1 = √1 / √-1) is also False.... 🤔 [I don't know if I'm correct]
sir answer false hai kyuki jin char chijo ko equate kiya hai , according to that i = 1/i it means 1 = -1 and that is not ture so that's why answer is false
Absolutely...this is false because 1st term has the value equal to iota(i) and the last term gives us -iota if we rationalise the denominator then 1/iota = -iota then we can clearly say that both are not equal...means false..lovely question
(-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.
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.
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.
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
Sir as we know that product of two complex number is always complex number . Show if we open root one we get i×1/i =1 show this statement is false i think that i was right 👍
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 this is false because in complex numbers we have studied that √(a•b)=√a•√b iff a,b>0 and same goes to the vice versa of this statement and while writing the solution, you violated this rule in 2nd step.
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. ❤❤❤❤❤❤❤❤❤
Agar aisa hai to i^2 = i * i= sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = sqrt (1) = 1 Lekin i^2 = -1 (we all know it.) So, my opinion is that the given statement is false because square roots cannot be taken common or be distributed (though it is done generally and we come out with the correct answer because, over there, we deal with positive quantities, so that action does not matter).
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
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.
Answer is false When u write √(-a/b) = √(a/-b) for this we have to multiply by √(-1/-1) And if multiplied √(-1/1) by √(-1/-1) it would actually be i²/√-1. and final output would be -1/i U can't write i²=√-1 * √-1 = √1=1
The fault in this method is that √(-1) / √1 = √(-1)/√(-1 x -1)= √1/√-1 Here we cant write √1 as √(-1 x -1) that is the major blunder here. In general square root of a positive number is always a positive no. But if we write √1 as √{(-1)²} gives us -1 that is universaly wrong.
I think the first inequality is false because √ab=√a√b only if voth a and b are positive Same goes with √a/b=√a/√b only if both a and b are positive I think so
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 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 √(-1/1) cannot be written as √(1/-1). This is because when we write -4/6 as -6/4 we actually multiply and divide by -1. Same is the case here. For making √(-1/1) as √(1/-1) we will have to multiply it by √(-1/-1). After this step, we cannot multiply √(-1/1) and √(-1/-1) directly to give √(1/-1) because there is a rule that √-a * √-b = -√ab. So √(-1/1) * √(-1/-1) will actually be -√(1/-1) which is equal to -√1/√-1 = -1/i = (-1/i)*(i/i) = -(-i) = i {not 1/i}.
sir statement is false because 2nd step is wrong you can't write -a/b=a/-b for any random number logic behind it was multipling -1 at numrator and denomirator in 2nd step multiple by i
Sir this question is wrong because the property of a / - b = -a/b Is only valid for real n.os not for iota. As the √-1 is not a real n.o the formula can't be applied. Thus the step of √(1/-1) is not equal to √(-1/1) as √-1 is not considered as a real n.o
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.❤❤❤🎉🎉🎉
√(-1)/√1 = √(-1)/√1*√(-1)/√(-1) = √[(-1)^2]/√(-1) Now the question arises if √[(-1)^2]= +-(-1) or not..... Here since it is positive square root, it should be +(-1) So finally the expression becomes -1/√(-1) What the question does is that it takes the negative square root [-(-1)] which is incorrect, therefore it is a false statement....
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.
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
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.
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.
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
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
sqrt(-1)/sqrt(1): This step is incorrect. You can't simply take the square root of -1 and the square root of 1 and equate them in this manner. The square root of 1 is 1, but the square root of -1 is "i," not 1. So, this step should be i/sqrt(1), which is still i.
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
It should be false because, if we assume 1st and the last expression to be equal i.e √-1 / √1 = √1 / √ -1 we get i² = 1 that gives -1 = 1 , which contradict our facts, hence our assumption was wrong, and the answer to the question is. "False" Considering for intermediate steps , we are not allowed to make operations under square root, if expression within are not all positive real.
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
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.
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
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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
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.
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 💯
Absolutely correct broooooo 🙌🙌🙌🙌🙌
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
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
√a.√b=√(ab) ; for real nos
But if both a and b are imaginary nos
Then,
√a.√b=-√ab
Problem is in third step
As /-1×/-1≠/1 but = -1 it is a rule that we have follow for complex no multiplication
2:26 this step is wrong
Thus the statement is false
Because
√(-a) × √(-b) ≠ √((-a)(-b))
Thus
√(-1/1) = √(-1/1) × √-1/√-1 ≠ √((-1)×(-1)/(1×(-1)))
=> √(-1/1) ≠ √(1/(-1))
Hence the statement 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
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
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 ?
Sir i think that in the step where we convert root(-1/1) to root(1/-1) is wrong. Because when we croos multiply in that step.
We get -1 = 1. So the answer is false. Thank you sir for giving such good problems.
Bhai -1×-1=1×1
1=1 hi to hota hai
@@NishikantYadav-ig3yiroot ke andar - hai , iota square ki value-1 hoti hai....
@@Xyztaken50 aree bhai -1=1 aagya to ek baar or dono taraf whole square kar de....😅
@@NishikantYadav-ig3yi lekin-1=1 possible hi kaise hogaa??
Ek. Cheez aayi dimag me 0multipy 1=0 alright then
1=0/0 butt 1 is not equal to infinite? 😂😂
@@Xyztaken50bhai 0 ki to baat hi nhi chal rhi thu fiir ye 0×1 wala concept kaha se litaya..😅
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)
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.
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?
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.
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
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
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.
good way to think
I think this is the correct explanation❤@@amangamerzpro595
√-1/√1 = sqrt 1/ sqrt -1
As we know
i ^ 2 = - 1
then
i ^ 4 = 1 so ,
√- 1/1 =1/√-1
i / (i ^ 4) = i ^ 4 /i
1 / (i ^ 3) = i ^3
i^4/i^3 = i^3
i =i^3
i =-i
So it is false
Sir i am sayan.... I'm in class 10th i think its true beacuse if √a² = | a | so. If √-1/1 = 1/√-1
So 1 = √ (-1)² according to the rule its true 1 = | -1| =1
If its not. .true then sir i want to know why √-4 . √-9 = 2 (iota) . 3 (iota) = 6 i² = 6 . (-1) = -6. Then why i² = -1. ?
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..
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!
It is wrong because negative should always be in numerator , keeping negative sign denominator is wrong ( I think this ,it may be wrong )
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.
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
Sir, it is wrong
Because the property of exponent is not suitable here that is √-1/√1 is not equal to whole root under -1/1
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
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
But, bro √1=1 not -1or1
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
Answer to this question is false.
Because iota=-1/iota
We can prove this as follows:
i= √-1/1
i= -√-1/(-1)
-1= (√-1) ^2
So i= -√-1/(√-1) ^2)
i= -1/√-1
Therefore,
i= -1/i
Mistake here is that √a.√b is onlu possible if both a and b are positive
2:07 but here one of them is negative to we cannot take the root for the whole
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.
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.😊
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
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 iska answer dono h
i is equal to -i (false )
for true
i=-i (a)
i+1=-i+1
2i=0
i=0
So , put i value in a
0=-0 : (0 is neither positive nor negative)
0=0
Hence proved true statement
If i=0 then 1/0=0 what are you saying bro
I think NO because -a/b=a/-b but √(-a/b)=/ √(a/-b)
Sir this should be false because in 1st step it is :-> (root(-1)/root(1)) which means i
In next step it is also i but in next step it is 1/i.
And they have given second step equal to third step, they mean -> i = 1/i
When we transfer i from RHS (that is in divide) to LHS it will multiply with i on LHS and make it i².
Now there statement says i² = 1, that is not true. Because all of us know i² = -1.
Hence proved.
Imagine if jee advance had these type of easy questions in exam now too.😅
-a/b = a/-b is obtained by multiplying -1 in Nr and Dr and this is true . But if you done this operation in square root, then this becomes
√{(-1)*(-1)}/{-1} =. -√1/√-1 because √a×√b=. -√ab if both a and b are negative . So our expression becomes -1/i which is equal to i
It is false because sqrt(-1)/sqrt(1) = i but sqrt(1)/sqrt(-1) = -i. In the given equation the mistake is in 3rd step. sqrt(-1/1)is not equal to sqrt(1/-1) because while shifting the negative symbol from numenator and denominator we are not actually shifting it but multiplying both numenator and denominator by -1. In this case the correct steps are sqrt(-1/1)=sqrt(-1×-1/1×-1)=-1/sqrt(-1)=-1/i because sqrt(-1×-1) is not equal to sqrt(1) but = -1. This is the mistake in the given equation. And after that -1/sqrt(-1) = -1/i = -1×-i = i. This one is correct
You can multiply numerator and denominator by a complex number (sqrt(-1)) only when none of them are complex.
According to me Its True .... 👇
Because lets √-1 / √1
=> √-1 = iota (i)
=> i² = -1 ------ equation no. 1
Similarly... , √1/√-1 = 1/i
=>squaring both sides...
=> 1/-1 = 1/i²
By Cross Multiplication:-
=> i² = -1 ----- equation no. 2
Hence from eq. 1 and 2
We conclude that √-1/√1 = √1/√-1...
Its my Opinion if any mistake explain in your next video
The simple answer is No as we cannot Assign same root if Negative and Positive Integers occour simultaneously
Sir I am a 7th class student but still...... 👇🏻
√-1 / √1 = √1 / √-1
➡️ i / 1 = 1 / i
➡️ i×i = 1×1 [Cross Multiplication]
➡️ i^2 = 1
➡️ i = √1
Because (i = √1) is False.... The given equation (√-1 / √1 = √1 / √-1) is also False.... 🤔
[I don't know if I'm correct]
sir answer false hai kyuki jin char chijo ko equate kiya hai , according to that i = 1/i it means 1 = -1 and that is not ture so that's why answer is false
Absolutely...this is false because 1st term has the value equal to iota(i) and the last term gives us -iota if we rationalise the denominator then 1/iota = -iota then we can clearly say that both are not equal...means false..lovely question
(-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.
@harsh_will_iit how do we can prove that , in rational domain -1/1 is always equal to 1/-1 ?
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.
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.
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
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
Sir as we know that product of two complex number is always complex number . Show if we open root one we get i×1/i =1 show this statement is false i think that i was right 👍
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, apki hi to videos dekhkar maine Jee ki preparation start ki thi
If we cross multiply the constants i/√1 and √1/i , then we see that i^2 = +-1 whereas √1.√1 = +1
Hence, false
Sir this is false because in complex numbers we have studied that √(a•b)=√a•√b iff a,b>0 and same goes to the vice versa of this statement and while writing the solution, you violated this rule in 2nd step.
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. ❤❤❤❤❤❤❤❤❤
Agar aisa hai to i^2 = i * i= sqrt(-1) * sqrt(-1) = sqrt(-1 * -1) = sqrt (1) = 1
Lekin i^2 = -1 (we all know it.)
So, my opinion is that the given statement is false because square roots cannot be taken common or be distributed (though it is done generally and we come out with the correct answer because, over there, we deal with positive quantities, so that action does not matter).
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
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.
Answer is false
When u write √(-a/b) = √(a/-b) for this we have to multiply by √(-1/-1)
And if multiplied √(-1/1) by √(-1/-1) it would actually be i²/√-1. and final output would be -1/i
U can't write i²=√-1 * √-1 = √1=1
Please like if u agree
The fault in this method is that
√(-1) / √1 = √(-1)/√(-1 x -1)= √1/√-1
Here we cant write √1 as √(-1 x -1) that is the major blunder here. In general square root of a positive number is always a positive no. But if we write √1 as √{(-1)²} gives us -1 that is universaly wrong.
It has two sol
1. i = 1/ i
By squaring both side
-1=-1
But in 2. Method
i =1/i ~~ i² =1 ~~ but -1 ≠ 1
Hello sir can you please make a new series
Kon kon agree hai mere baat par
I think the first inequality is false because
√ab=√a√b only if voth a and b are positive
Same goes with √a/b=√a/√b only if both a and b are positive
I think so
Yes it can be possible as it is the property of under roota and basic algebras....
root
Flash statement
root-1= i
(-1)/1=i/1×i/i =i²/i=-1/i= -1/root-1
hona chahiye
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 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.
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...
this can be proved by contradiction . let's assume that i=1/i
squaring both sides, we get
i^2= 1
but i^2=-1. therefore, i is not equal to 1/i
This is false because √(-1/1) cannot be written as √(1/-1). This is because when we write -4/6 as -6/4 we actually multiply and divide by -1. Same is the case here. For making √(-1/1) as √(1/-1) we will have to multiply it by √(-1/-1). After this step, we cannot multiply √(-1/1) and √(-1/-1) directly to give √(1/-1) because there is a rule that √-a * √-b = -√ab. So √(-1/1) * √(-1/-1) will actually be -√(1/-1) which is equal to -√1/√-1 = -1/i = (-1/i)*(i/i) = -(-i) = i {not 1/i}.
sir statement is false because 2nd step is wrong you can't write -a/b=a/-b for any random number logic behind it was multipling -1 at numrator and denomirator in 2nd step multiple by i
Sir this question is wrong because the property of a / - b = -a/b
Is only valid for real n.os not for iota. As the √-1 is not a real n.o the formula can't be applied. Thus the step of √(1/-1) is not equal to √(-1/1) as √-1 is not considered as a real n.o
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,
This is false because the quotient property of square root i.e.,√a/√b=√a/b is true for a>=0 and b>0.
But here from 1st to 2nd step it is used for a
The rule (-a/b=a/-b ) fisible where a and b are real numbers.but this is a complex number
The step 2 is wrong where we take tha quantity in whole root because it can only be done when either both nos are non negative or negative
2step not equal to 3step
Let iota=c(any quantity)
Then, iota and 1/iota are not equal
Therefore, √iota is not equal to √(1/iota)
Sir yeh false hai as:
if √-1/1=√1/-1
then, (√-1)(√-1)=1(1)
=>(-1)=1
which is not possible
hence, the above equation is wrong
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
√(-1)/√1
= √(-1)/√1*√(-1)/√(-1)
= √[(-1)^2]/√(-1)
Now the question arises if √[(-1)^2]= +-(-1) or not.....
Here since it is positive square root, it should be +(-1)
So finally the expression becomes -1/√(-1)
What the question does is that it takes the negative square root [-(-1)] which is incorrect, therefore it is a false statement....
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.
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
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.
√-1/√1
= i / 1 = i
√1/√-1
= 1 / i
Multiplying both numerator and denominator by i
= i / i²
= i / -1
= - i
Therefore, false
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.
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
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
Property √a/b = √a/√b only for real integers gm
root a . root b = root ab is true when both are positive. Therefore it is false.
sqrt(-1)/sqrt(1): This step is incorrect. You can't simply take the square root of -1 and the square root of 1 and equate them in this manner. The square root of 1 is 1, but the square root of -1 is "i," not 1. So, this step should be i/sqrt(1), which is still i.
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
-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.
It should be false because, if we assume 1st and the last expression to be equal i.e
√-1 / √1 = √1 / √ -1 we get
i² = 1 that gives
-1 = 1 , which contradict our facts, hence our assumption was wrong, and the answer to the question is. "False"
Considering for intermediate steps , we are not allowed to make operations under square root, if expression within are not all positive real.
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
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.
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
It is true because √-1/√1=i/1 and √1√-1=1/i
If ,1st step = last step
Then,i/1=1/i
Therefore,i×i=1 = i^2=1
False
i,(1÷i) are opposite in sign