Sqrt(x^2) and (Sqrt x)^2

  • #1
Please tell me
Sqrt (x^2)=______ For All Real Numbers
(Sqrt x^2)=______ For All Real Numbers
Thank YOu..!
 

Answers and Replies

  • #2
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Please tell me
Sqrt (x^2)=______ For All Real Numbers
(Sqrt x^2)=______ For All Real Numbers
Thank YOu..!
The first one is defined for all real numbers.
For the second one, I think you mean this: (sqrt(x))2. If so, that is defined for x >= 0.
 
  • #3
Thank You Mark44
Please also tell me whether they are equal to each other or not?
 
  • #4
22,089
3,293
The first one is equal to |x| by the way...
 
  • #5
22,089
3,293
Thank You Mark44
Please also tell me whether they are equal to each other or not?
The second one is not defined for negative numbers, so seeing whether they are equal has no sense.
The two are equal for POSITIVE numbers though...
 
  • #6
Sorry Second one was actually (sqrt x)^2.
Sir if this is not defined, then how we define iota?
 
  • #7
22,089
3,293
Doesn't matter, Sqrt(x) is only defined for negative numbers.

What is iota?
 
  • #8
22,089
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Oh, you mean i, right? Well, i is certainly not defined as [tex]\sqrt{-1}[/tex], that would be a very bad definition. Instead i is defined as a quantity such that [tex]i^2=-1[/tex]. But how do we define this rigourously?

Well, we can see the set of all complex numbers as [tex]\mathbb{R}^2[/tex] and we define an addition and a multiplication on that as

[tex](a,b)+(c,d)=(a+b,c+d)~\text{and}~(a,b)\cdot (c,d)=(ac-bd, bc+ad)[/tex]

Then i is defined as (0,1).

This is the simplest construction of the complex numbers, another one would be as the quotient [tex]\mathbb{R}[X]/(X^2+1)[/tex] or as a matrix group which consist of the numbers

[tex]\left(\begin{array}{cc} a & -b\\ b & a\end{array}\right)[/tex].

But i is certainly NOT defined as the square root of -1...
 
  • #9
Sir let (sqrt x)^2=x
And substitute x=-1, then we would have (i)^2=-1.
Is it right?
 
  • #10
MathematicalPhysicist
Gold Member
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i might also be defined as the root of x^2+1=0, in this case it is defined as the sqrt of -1.
 
  • #11
22,089
3,293
Sir let (sqrt x)^2=x
And substitute x=-1, then we would have (i)^2=-1.
Is it right?
No, this is not correct since [tex]\sqrt{-1}\neq i[/tex], for the simple reason that [tex]\sqrt{x}[/tex] is only defined for POSITIVE values of x.

You COULD make sense of things like [tex]\sqrt{-1}[/tex], but like I mentioned in the other thread, this is not standard notation.
 
  • #12
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I have seen such notation of \sqrt(-1) in textbooks which use for indices also i and j.

It's really a matter of notation.
For example Principles of Algebraic Geometry by Harris uses \sqrt(-1) instead of i.
 
  • #13
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well, the sqrt notation is "abused" all the time. Another example is writing sqrt(T) as a positive sqrt of a positive operator
 
  • #14
22,089
3,293
I have seen such notation of \sqrt(-1) in textbooks which use for indices also i and j.

It's really a matter of notation.
For example Principles of Algebraic Geometry by Harris uses \sqrt(-1) instead of i.
I see, if even the great Griffiths & Harris do it, then it appears that I'm incorrect. It's still not as widely used as i and j, but apparently it's been used quite a lot...

Thank you for your example!!
 
  • #15
HallsofIvy
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Yes, you will see that notation, but you should use it only after you have defined "i" more rigorously (and many elementary text,which are concerned with calculations, not rigor, don't do that). One difficulty with even writing "[itex]i= \sqrt{-1}[/itex]" is that while in the real numbers we can define [itex]\sqrt{a}[/itex] to be the positive number such that its square is a, the complex number are not an "ordered" field so we cannot distinguish between the two square roots of a number so easily.
 

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