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597
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Why can't I take (-1)^1/2

 Nov 4, 2015

Best Answer 

 #2
avatar+95284 
+10

Why can't I take (-1)^1/2

 

lets think about this.

This means

 

\(\sqrt{-1}\)

 

Now

 

\(\sqrt{25}=5\\ because\;\; 5*5 = 25\\ so\\ think\;\;about\;\; \sqrt{any\;negative\;number} \)

 

a positive * a positive = a positive

a negative * a negative = a positive

 

Mmm

you cannot multiply any real number by itself and get a negative answer.  That is impossible !!

So

\(\sqrt{-1}\)

has no real solution!!!

That is why you cannot do it!!

 

 

\(\sqrt{-1}\)

is the basis of the imaginary number system.  It is called i (or sometimes j)

 

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

.
 Nov 4, 2015
 #1
avatar+56 
0

Try (-1)*-0.5

 Nov 4, 2015
 #2
avatar+95284 
+10
Best Answer

Why can't I take (-1)^1/2

 

lets think about this.

This means

 

\(\sqrt{-1}\)

 

Now

 

\(\sqrt{25}=5\\ because\;\; 5*5 = 25\\ so\\ think\;\;about\;\; \sqrt{any\;negative\;number} \)

 

a positive * a positive = a positive

a negative * a negative = a positive

 

Mmm

you cannot multiply any real number by itself and get a negative answer.  That is impossible !!

So

\(\sqrt{-1}\)

has no real solution!!!

That is why you cannot do it!!

 

 

\(\sqrt{-1}\)

is the basis of the imaginary number system.  It is called i (or sometimes j)

 

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

Melody Nov 4, 2015
 #3
avatar
0

Okay super I understand, but what about \({(-1)}^{1/3}\)

 

I think that will be \(\sqrt[3]{-1}\)

You can say (-1)*(-1)*(-1) = -1, but it still says error?

 Nov 4, 2015
 #4
avatar+95284 
0

Yes you are right.

It is my experience that even very sophisticated calcs can't handle roots of negative numbers. 

I think Alan went into more detail on this once before. Maybe he will comment later. :/

 Nov 4, 2015

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