Is root over a^2=modolus of a?

[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.

 

[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.

 

[Ahmed Abdullah]

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

 

[Ahmed Abdullah]

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.

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?

 

[Werg22]

Why not...

 

[uart]

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?

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.

 

[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...

 

[matt grime]

T 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...

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

 

[prasannapakkiam]

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

 

[matt grime]

I doubt we can pin down the first time someone wrote down the formal definition of a function, 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.

 

[Ahmed Abdullah]

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.

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

 

[Office_Shredder]

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

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

 

[prasannapakkiam]

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

 

[Hurkyl]

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

You mean -pi/2 and pi/2.

 

[uart]

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

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