Question

Use complete sentences to describe why √-1 ≠ -√1

Answer 1

Therefore
`√(1) \\eq -√(1)`

Explanation:

`√(-1) \\` cannot be calculated as no two identical numbers that are multiplid could give -1. However -
`√(1)` means -1 x
`√(1)` = -1 x 1 = -1 . Therefore
`√(1) \\eq -√(1)`

Answer 2

Well let's say that to compare these two numbers, we have to start with the definition first.

Definition

`\displaystyle \large{ {y}^(2) = x} \\ \displaystyle \large{ y = \pm √(x) }`

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

Definition II

`\displaystyle \large{ {a}^(2) = a * a = |b| }`

Where b is the result from a×a. Let's see an example.

Examples

`\displaystyle \large{ {2}^(2) = 2 * 2 = 4} \\ \displaystyle \large{ {( - 2)}^(2) = ( - 2) * ( - 2) = | - 4| = 4}`

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

`\displaystyle \large{ {y}^(2) = - 1}`

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

`\displaystyle \large{ \sqrt{ {y}^(2) } = √( - 1) }`

Then where does plus-minus come from? It comes from one of Absolute Value propety.

Absolute Value Property I

`\displaystyle \large{ \sqrt{ {x}^(2) } = |x| }`

Solving absolute value always gives the plus-minus. Therefore...

`\displaystyle \large{ y = \pm √( - 1) }`

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

`\displaystyle \large{ {y}^(2) = 1} \\ \displaystyle \large{ \sqrt{ {y}^(2) } = √(1) } \\ \displaystyle \large{ y = \pm √(1) }`

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

Conclusion

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.