# Is root over a^2=modolus of a?

1. Jun 9, 2007

### Ahmed Abdullah

Is "root over a^2=modolus of a?"

mod of a is always positive, but root over a^2 can both be positive or negative. So how these two can be equal to each other?
I have found this in a math textbook. But I can't convince myself about it.

It will be very helpful if you give a proof.

Thnx in anticipation.

Last edited: Jun 9, 2007
2. Jun 9, 2007

### matt grime

There are two square roots of any number, agreed, but the symbol x^{1/2} is explicitly chosen to be one of them, and for positive real numbers, the positive one is always chosen.

3. Jun 9, 2007

### Ahmed Abdullah

so root over (-5)^2=5
RIGHT?

4. Jun 9, 2007

### Ahmed Abdullah

Q:When x^{1/2} is a negative real number?
A: Never!
So the negative square root of x is expressed by the symbol -x^{1/2}.
Am I right? :grumpy:

5. Jun 9, 2007

Why not...

6. Jun 9, 2007

### uart

Yes that's correct. This way $$\sqrt{.}$$ is a function (single valued) and we can always refer to the positive root of $$x^2=1$$ as $$\sqrt{x}$$, or the negative root, $$-\sqrt{x}$$, or both roots, $$\pm \, \sqrt{x}$$ as we wish.

Last edited: Jun 9, 2007
7. Jun 10, 2007

### prasannapakkiam

This is a dilemma (I think I spelt it wrong...) I had when calculating ranges of certain functions. Because of the way computer have worked since the 70's functions have been defined (especially the SQRT function), to take only one value - the positive value...

8. Jun 10, 2007

### matt grime

Really? You think that functions are defined to be single valued owing to theinvention of computers in the 70s?

9. Jun 10, 2007

### prasannapakkiam

yes, and I have got a few people that agree strongly with this...

10. Jun 10, 2007

### matt grime

I doubt we can pin down the first time someone wrote down the formal definition of a fucntion, but it was many decades before the 1970s. One need only look at the notion of branch cuts and Riemann surfaces (c. 1900) to notice that.

11. Jun 10, 2007

### Ahmed Abdullah

:rofl:
It is great to get rid of every piece of misconceptions.
I am a happy man now. $$^2=1$$

Last edited: Jun 10, 2007
12. Jun 10, 2007

### Office_Shredder

Staff Emeritus
Next thing you know, arcsin(x) will only return a value between 0 and 2Pi!
:grumpy:

13. Jun 10, 2007

### prasannapakkiam

Well yes. This is one thing that really irritates me.

14. Jun 10, 2007

### Hurkyl

Staff Emeritus
You mean -pi/2 and pi/2.

15. Jun 11, 2007

### uart

Yes that was a typo , I meant to say :
... we can always refer to the positive root of $$x^2=a$$ as $$\sqrt{a}$$ ...

Last edited: Jun 11, 2007