Question

What is the mistake in the computation of: -
\[\color{blue}{1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1} \times \sqrt{-1}=i \times i =i^2=-1.}\]

Answer 1

-1 * -1 = 1

Answer 2

Look at the fourth step

Answer 3

\(\Rightarrow \sqrt{(-1)(-1)} = \sqrt{1} \)

Answer 4

No need to write \(\sqrt{-1} \times \sqrt{-1}\)

Answer 5

When you write something in this form, it all falls apart.

Answer 6

it is nt the answer @ParthKohli :/

Answer 7

\(\sqrt{(-1) \times (-1)} = \sqrt{1}\)

Answer 8

No, Parth, it is like this:
you can only write
sqrt(a)*sqrt(b)=sqrt(ab)
if and ONLY if a & b are not negative at the same time.

Answer 9

\(\sqrt{a} \times \sqrt{b} = \sqrt{ab}\)

Let's try it when both a and b are negative.

\(\sqrt{-25} \times \sqrt{-4}\)
\(\Rightarrow \sqrt{(-25)(-4)} \)
\(\Rightarrow \sqrt{100} \)
\(\Rightarrow10\)

But this is wrong

Answer 10

Yes, rajath

Answer 11

I sure wish I knew Latex :'(

Answer 12

There's another way to do this:

\(\sqrt{-25} = 5i\)
\(\sqrt{-4} = 2i\)

\(5i \times 2i = 10i\)

Answer 13

\(10i^2\)*

Answer 14

\(10 \times -1\)
\(\Rightarrow -10\)

Answer 15

hey hey going so long
i want only ur conclusion about the mistake in the question.

Answer 16

The fourth step is wrong as \(\sqrt{a} \times \sqrt{b} \ne \sqrt{ab}:a,b=-ve\)

Answer 17

so it would be \[\sqrt{a}\times \sqrt{b}=-\sqrt{ab} \space ; \space a,b=-ve.\]
so can u give me the whole & correct solution again plz:)

Answer 18

Okay, so we were at this step:

\(\sqrt{(-1)(-1)}\)

But wait, we can simplify the radical.
-1 * -1 = 1

\(\sqrt{-1 \times -1} = \sqrt{1} = 1\)

Answer 19

no, parth.

Answer 20

Yes, Rajat.

Answer 21

bt it is not \[-\sqrt{ab}\] @ParthKohli

Answer 22

Actually it is plus or minus 1, but minus one is an extraneous solution.

Answer 23

What do you mean by \(-\sqrt{ab}\)?

Answer 24

it is the correct solution of this question:)

Answer 25

take sqrt(-1)=i & then proceed. So, sqrt(-6) for example will be sqrt(6)*i

Answer 26

Yeah, of course. But when you could simplify the thing inside the radical, then why not?

Answer 27

No, you are going against your own answer.

Answer 28

-1 * -1 = 1 right? so \(\sqrt{(-1)(-1)}= \sqrt{1}\)

Answer 29

But not inside a radical.

Answer 30

Can you be more specific, Rajath?

Answer 31

\[\color{blue}{1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1} \times \sqrt{-1}=i \times i \sqrt{1} =i^2=-1.}\]well then also i m getting this @rajathsbhat

Answer 32

Just refer to my first response, Parth.

Answer 33

I know, that is what I am saying.

Answer 34

You can't split the '1' INSIDE the radical as (-1)(-1).

Answer 35

You need not stretch a problem like \(\sqrt{60.5 \times 2}\) to \(\sqrt{60.5} \times \sqrt{2}\), as that may cause problems.

Answer 36

You could compute 60.5 * 2 which is 121 so \(\sqrt{121} = 11\)

Answer 37

well it is an IIT question.
bt u r nt giving me satisfactory answer @ParthKohli @rajathsbhat :/
Plz give me stepwise solution
i wanna see wt do u wanna say actually.

Answer 38

The point is, if you have a +ve numer inside the radical, it can only have come from multiplying 2 +VE numbers.

Answer 39

Can you please be more specific as to what we can do, Mahesh?

Answer 40

We can split it into (-1)(-1), but we can't make it go like \(\sqrt{-1} \times \sqrt{-1}\)

Answer 41

ya just as i did the whole solution.
\[\color{blue}{1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1} \times \sqrt{-1}=i \times i \sqrt{1} =i^2=-1.}\]i want the correct one from u @ParthKohli @rajathsbhat

Answer 42

Okay, I explained it already.
\(\sqrt{a} \times \sqrt{b} \ne \sqrt{ab}:a,b\in -ve\)

Answer 43

ok

Answer 44

Do you need any more explanation of this?

Answer 45

I'm telling you this exact "paradox" was raised by my lecturer when I was undergoing IIT JEE coaching. The answer to this problem IS MY FIRST RESPONSE.

Answer 46

wait a minute
i have the solution at the end of my book
i m just typing that.

Answer 47

No need to gobble this all up from your book. Do not believe in these fake IIT books.

Answer 48

You see, Parth, your 3rd latest response just doesn't allow for you to write sqrt(−1)(−1)=sqrt1

Answer 49

lol parth The answer is: -
The formula \[\sqrt{a}\sqrt{b}=\sqrt{ab}\]is applicable only when at least one from a & b is positive.
otherwise \[\sqrt{-a}\sqrt{-b}=-\sqrt{ab} \space ; \space a,b>0\]Well i want answer acccording to this @ParthKohli & @rajathsbhat

Answer 50

This is exactly what I've been trying to sat from the beginning.(My first response=just worded differently). D:
I think Parth was saying the same thing too....

Answer 51

say* not sat.

Answer 52

bt actually parth was asking \[-\sqrt{ab}=?\]well that is the answer
can u give me now the correct method of the question.?

Answer 53

can you explain that again? I don't understand your ques.

Answer 54

@maheshmeghwal9 Isn't that what I said?

Answer 55

i wanna say that
are u seeing the process of sqrt{1} in our original question
can u do it now correctly ?
& if yes plz give me full solution:)

Answer 56

There is not just one notation in this world

Answer 57

actually correct the mistake & give the question in correct form again
omg

Answer 58

The original question is flawed in going from step 2 to 3.

Answer 59

No actually step 2 -> 3 is still acceptable, but then 3 -> 4 was applied wrong.

Answer 60

? getting more confused aarrghhhhh.
@ujjwal Plz help:)

Answer 61

NO.NO.NO....

Answer 62

1=sqrt(1)
That's all. You can't do any thing furthur.
If you want you can only write
sqrt(1)=sqrt[(1)(1)]=sqrt[(1)(1)(1).....]
That's all & that's it.

Answer 63

so i gt u now clearly @rajathsbhat but the main problem is that why can't we do it by taking -1 x -1 =1.

Answer 64

Let me write something in paper & post it> I don't know Latex & I'm finding it hard to explain my thoughts.

Answer 65

k find my mistake now
it is now improved version lol @rajathsbhat
\[\color{blue}{1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1} \times \sqrt{-1}=i \times i \sqrt{1 \times 1} =i^2\sqrt{1}=-1.}\]

Answer 66

wait for 5.

Answer 67

ok:)

Answer 68

\[1=\sqrt{1}\neq \sqrt{(-1)(-1)}\]

Answer 69

do you get it?

Answer 70

At least i don't get it. @rajathsbhat

Answer 71

is there any reason
why it is nt like this?

Answer 72

@rajathsbhat

Answer 73

I'm sorry terrible mistake.\[1=\sqrt{1}= \sqrt{(-1)(-1)}\neq \sqrt{-1}\sqrt{-1}\]

Answer 74

Is that just a postulate which can't be questioned or it has some logical reasoning?

Answer 75

then it is like this \[1=\sqrt{1}= \sqrt{(-1)(-1)}= \sqrt{1}i\sqrt{1}i=1i^2=-1\]

Answer 76

because the opposite is not true. ie..\[\sqrt{-1}\sqrt{-1}\neq \sqrt{(-1)(-1)}\]

Answer 77

no; it is a sensible question & correct too from IIT @ujjwal

Answer 78

Now that I can use Latex, I'll explain one big thing that summarizes everything I said. Is that OK? Sorry for being the dumb-retricethat I am on Latex --.--

Answer 79

retrice not retrice.

Answer 80

retrice

Answer 81

spell-check problem. Trying to type retrice

Answer 82

its ok:)

Answer 83

And you are typing retrice, which is not a big issue here. just waiting for your latexed answer..

P.S. 'latexed' is also a spelling mistake... :P ..
Infact there is no such word. :P :D

Answer 84

Now, \[\sqrt{a}\sqrt{b}= \sqrt{(a)(b)} \: \: \: \: iff\; \; a\; and\; b\; are \; not \; both\; negative.\]

Answer 85

Therefore, the reverse is also true.

Answer 86

If a& b are both negative, \[\sqrt{(a)(b)}\neq \sqrt{a}\sqrt{b}\]

Answer 87

\[\sqrt{ab}=\sqrt{(a)(b)}= \sqrt{ab}\] if both a & b are -ve. There is nothing more you can do about it.

Answer 88

\[{\color{DarkOrange} {Thanks -Rajathsbhat}}\]

Answer 89

Sure thing. :)