Question

What is the mistake in the computation of $1=\sqrt{1}=\sqrt{\left( -1 \right)\left( -1 \right)}=\sqrt{-1}\times \sqrt{-1}=i\times i=-1$ ?

Answer 1

Hint: We have given the expression and are asked to pinpoint the mistake in that expression. There is a property when we multiply two square root terms: $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$. This property holds true when one of $a\And b$ are non-negative integers and this property won’t hold true when both $a\And b$ are negative integers. Now, to locate the mistake we are going to use this property.

Complete step by step answer:
In the above problem, we are asked to find the mistake in the following expression:
$1=\sqrt{1}=\sqrt{\left( -1 \right)\left( -1 \right)}=\sqrt{-1}\times \sqrt{-1}=i\times i=-1$
Now, to locate the mistake, we are going to move from left to right in the above expression.
First of all, we are investigating the first equality:
$1=\sqrt{1}$
This equality is perfectly correct. Because the square root of 1 is 1.
Now, we are going to scrutinize the second equality which we have shown below:
$\sqrt{1}=\sqrt{\left( -1 \right)\left( -1 \right)}$
The above expression is also correct because we know that if we multiply two negative signs then it becomes positive so multiplying -1 with -1 will give 1 which is absolutely correct.
After that, moving on to the next equality sign which says:
$\sqrt{\left( -1 \right)\left( -1 \right)}=\sqrt{-1}\times \sqrt{-1}$ …… (1)
The above expression is incorrect because there is a property which says that if we multiply the square root of two integers then they follow some conditions to be multiplied:
$\sqrt{ab}=\sqrt{a}\times \sqrt{b}$
The above equality holds when at least one of the $a\And b$ must be non-negative integers. But as you can see that in eq. (1) this property is violated because none of $a\And b$ (-1 and -1) is a non-negative integer.
Hence, this equality $\sqrt{\left( -1 \right)\left( -1 \right)}=\sqrt{-1}\times \sqrt{-1}$ is not correct in the given expression.

Note: You can check the remaining equalities too but as in the question we just have to pinpoint the mistake so we have not gone further. Let us examine the remaining equalities which we have not examined in the above solution:
$\sqrt{-1}\times \sqrt{-1}=i\times i$
The above equality is correct because we can write $\sqrt{-1}=i$.
The last equality given in the above problem is as follows:
$i\times i=-1$
The above equality is also correct because there is a property that says that multiplication of iota with iota will give -1.