## SolomonZelman one year ago where are the properties faulty?

1. SolomonZelman

$$\large\color{black}{ \displaystyle i=i^{1}=i^{4\times(1/4)}=\left(i^4\right)^{1/4}=\sqrt[4]{1}=1 }$$

2. mathmate

It's like $$-1=\sqrt{(-1)^2}=\sqrt 1= 1$$, whenever we square an expression, we need to get rid of extraneous roots by back substitution. There are 4 fourth-roots of 1, $$\pm 1~ and~ \pm i$$.

3. SolomonZelman

Oh, in the example you proposed, i think $$\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}$$ doesn't work if √a and √b are both complex. that is what i heard

4. SolomonZelman

but yeah it is a similar accomplishment

5. SolomonZelman

but even if i²=±1, then i still get i=1

6. SolomonZelman

i think i got what you are trying to say

7. SolomonZelman

the even root of a constant is presumed to be positive, but that doesn't apply to i, so you get ±1. But, how would then (i^4)^(1/4)= 1 not be true?

8. anonymous

$\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}$ is true if $$a,b>0$$ otherwise it is not

9. mathmate

There are 4 fourth-roots of 1, ±1 and ±i. +1 is one of them. The choice depends on the context.

10. SolomonZelman

Alright, yes, thanks:)

11. mathmate

You're welcome! :)

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