Why is Radical i Squared Not Equal to -1?

[grad]

I am sure that it's a stupid question that has already been answered in the past, but I couldn't find the solution anywhere. Well I want to know why the following attached equality is not true:

 

[HallsofIvy]

This has been said over and over again. a^maⁿ= a^m+n is NOT true for non-real numbers. In particular, a¹= a^1/2a^1/2 is not true for non-real numbers.

 

[rock.freak667]

because

$$\sqrt{i^4}=\pm i^2$$


you have two roots.

 

[grad]

Thank you! I later found out that $$\sqrt[]{a*b}$$=$$\sqrt{a}$$*$$\sqrt{b}$$ is NOT correct for NON-REAL numbers.

 

[D H]

$$1=\sqrt{i^4}=\sqrt{i^2i^2}=\sqrt{i^2}\sqrt{i^2}=i*i=-1$$

What you wrote is invalid:

This has been said over and over again. a^maⁿ= a^m+n is NOT true for non-real numbers.

It's not even true for all real numbers. If m and n are real numbers, a^maⁿ=a^m+n is only true for positive real numbers a and only when a^m is interpreted to mean the principal value.[/quote]
I was assuming, as is standard, that the real valued aⁱ is only defined for a> 0 and that it meant the principal value.

Trouble arises as soon as one starts allowing solutions other than principal values (e.g., -1 as a solution to $x^2-1=0$). For example, the same invalid mathematics as used in the original post can be used to show -1=1 without resorting to imaginary numbers:

$$1=\sqrt{(-1)^2}=\sqrt{(-1)^21^2}=\sqrt{(-1)^2}\sqrt{1^2}=(-1)*1=-1$$

The error here arises from writing $\sqrt{(-1)^2}=-1$.

 

[prasannaworld]

I hope this helps...

"1" can be written as 1*cis(0 + 360n)

where:
- cis(x) = cos(x) + i*sin(x)
- I am working in degrees...
- n is any integer

one can see that there may be an infinite no. ways of writing i depending on what n is...

Now square rooting; the De Moivre's Theorem is used...

SQRT(1) = cis(360n / 2) = cis(180n)

you see, this gives +/- 1 (since the argument can be 0 or 180)...

You in fact have taken 2 different solutions and equated them... Which is what the replies have said already...

 

[yuiop]

Here is a similar "puzzle":

$$\frac{-1}{1}=\frac{1}{-1}$$

$$\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$

$$\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}$$

$$\frac{i}{\sqrt{1}}=\frac{\sqrt{1}}{i}$$

$$i^2 = \left(\sqrt{1}\right)^2$$

$$i^2=1$$

$$-1 = 1$$

 

[yuiop]

$$1=\sqrt{i^4}=\sqrt{i^2i^2}=\sqrt{i^2}\sqrt{i^2}=i*i=-1$$

In this problem posted by grad the error is due to the fact that:

$$\sqrt{i^2i^2}\ne\sqrt{i^2}\sqrt{i^2}$$

The correct way to evaluate the problem is:

$$1=\sqrt{i^4}=\sqrt{i^2i^2}=\sqrt{(-1)(-1)} = \sqrt{1}= 1$$

In the problem I posted in my last post the error is this step:

$$\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$ ==> $$\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}$$

because:

$$\sqrt{\frac{1}{-1}} \ne \frac{\sqrt{1}}{\sqrt{-1}}$$

The correct way to evaluate that problem is

$$\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}$$ ==> $$\sqrt{-1}=\sqrt{-1}$$==> $$i = i$$

...
Trouble arises as soon as one starts allowing solutions other than principal values (e.g., -1 as a solution to $x^2-1=0$). For example, the same invalid mathematics as used in the original post can be used to show -1=1 without resorting to imaginary numbers:

$$1=\sqrt{(-1)^2}=\sqrt{(-1)^21^2}=\sqrt{(-1)^2}\sqrt{1^2}=(-1)*1=-1$$

The error here arises from writing $\sqrt{(-1)^2}=-1$.

D H has correctly identified the error in the example he posted above.
The correct way to evaluate his example is:

$$1=\sqrt{(-1)^2}=\sqrt{1}= 1$$

The pattern that emerges from all these examples is that the errors arise when bracketed terms are evalauted from the outside first and then working inwards. All the errors can be avoided by evaluating the expressions in the inner brackets first and working outwards.

Generally speaking $$(a^ba^c)^d$$ is not guaranteed to be the same as $$a^{bd}a^{cd}$$ or even $$a^{d(b+c)}$$ and it is less risky to evaluate $$(a^b*a^c)^d$$ as $$(a^{(b+c)})^d }$$