Why is a equal to negative a in this scenario?

[noahsdev]

I've been messing around with numbers (as you do) and I'm wondering why this occurs..
lets let a = b-c.
√a
= √(b-c)
=√(-(c-b))
=i√(c-b)
=i√(-(b-c))
=i²√(b-c)
=-√(b-c)
=-√a
For example if you let a = 1, b = 2, and c = 1.

 

[Mark44]

I've been messing around with numbers (as you do) and I'm wondering why this occurs..
lets let a = b-c.
√a
= √(b-c)
=√(-(c-b))
=i√(c-b)

The step above is where the problem is. You're using the property that $\sqrt{ab} = \sqrt{a}\sqrt{b}$
There are restrictions on this and some of the other square root properties - both a and b have to be nonnegative.

=i√(-(b-c))
=i²√(b-c)
=-√(b-c)
=-√a
For example if you let a = 1, b = 2, and c = 1.

 

[noahsdev]

The step above is where the problem is. You're using the property that $\sqrt{ab} = \sqrt{a}\sqrt{b}$
There are restrictions on this and some of the other square root properties - both a and b have to be nonnegative.

That makes sense. Thanks.

 

[FactChecker]

$\sqrt[2]{-1} = \pm i.$ The choice of the '+' or '-' depends on the situation. So your result is '$a = \pm a$' where you must decide which sign is correct.