Is root over a^2=modolus of a?

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Discussion Overview

The discussion revolves around the relationship between the square root of a squared number and the modulus of that number. Participants explore whether the expression "root over a^2" is equivalent to the "modulus of a," considering the implications of positive and negative values.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the equality of "root over a^2" and "modulus of a," noting that the modulus is always positive while the square root can be both positive and negative.
  • Another participant asserts that the square root symbol is conventionally defined to yield the positive root for positive real numbers.
  • A participant confirms that the square root of (-5)^2 is indeed 5, implying agreement with the positive outcome of the square root operation.
  • There is a discussion about the definition of functions, particularly the square root function, being single-valued and always returning the positive root.
  • Some participants express a belief that the single-valued nature of functions, such as the square root, is influenced by computational practices established in the 1970s.
  • Another participant challenges this view, suggesting that the formal definition of functions predates the 1970s and references historical concepts like branch cuts and Riemann surfaces.
  • There is a light-hearted exchange about misconceptions and typos related to mathematical expressions.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the implications of defining the square root function and its relationship to the modulus. While some agree on the positive nature of the square root, others challenge the historical context of function definitions.

Contextual Notes

The discussion includes various assumptions about the definitions of mathematical functions and the historical context of these definitions, which remain unresolved.

Ahmed Abdullah
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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.
 
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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.
 
so root over (-5)^2=5
RIGHT?
 
matt grime said:
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?
 
Why not...
 
Ahmed Abdullah said:
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 [tex]\sqrt{.}[/tex] is a function (single valued) and we can always refer to the positive root of [tex]x^2=1[/tex] as [tex]\sqrt{x}[/tex], or the negative root, [tex]-\sqrt{x}[/tex], or both roots, [tex]\pm \, \sqrt{x}[/tex] as we wish.
 
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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...
 
prasannapakkiam said:
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?
 
yes, and I have got a few people that agree strongly with this...
 
  • #10
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.
 
  • #11
:smile:
uart said:
Yes that's correct. This way [tex]\sqrt{.}[/tex] is a function (single valued) and we can always refer to the positive root of [tex]x^2=1[/tex] as [tex]\sqrt{x}[/tex], or the negative root, [tex]-\sqrt{x}[/tex], or both roots, [tex]\pm \, \sqrt{x}[/tex] as we wish.

It is great to get rid of every piece of misconceptions.
I am a happy man now. :approve:[tex]^2=1[/tex]
 
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  • #12
prasannapakkiam said:
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!
 
  • #13
Well yes. This is one thing that really irritates me.
 
  • #14
Office_Shredder said:
Next thing you know, arcsin(x) will only return a value between 0 and 2Pi!
You mean -pi/2 and pi/2. :wink:
 
  • #15
Ahmed Abdullah said:
:smile:It is great to get rid of every piece of misconceptions.
I am a happy man now. :approve:[tex]^2=1[/tex]

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