- #1

- 99

- 0

Sqrt (x^2)=______ For All Real Numbers

(Sqrt x^2)=______ For All Real Numbers

Thank YOu..!

- Thread starter Ali Asadullah
- Start date

- #1

- 99

- 0

Sqrt (x^2)=______ For All Real Numbers

(Sqrt x^2)=______ For All Real Numbers

Thank YOu..!

- #2

Mark44

Mentor

- 34,545

- 6,248

The first one is defined for all real numbers.

Sqrt (x^2)=______ For All Real Numbers

(Sqrt x^2)=______ For All Real Numbers

Thank YOu..!

For the second one, I think you mean this: (sqrt(x))

- #3

- 99

- 0

Thank You Mark44

Please also tell me whether they are equal to each other or not?

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

The second one is not defined for negative numbers, so seeing whether they are equal has no sense.Thank You Mark44

Please also tell me whether they are equal to each other or not?

The two are equal for POSITIVE numbers though...

- #6

- 99

- 0

Sorry Second one was actually (sqrt x)^2.

Sir if this is not defined, then how we define iota?

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?

What is iota?

- #8

- 22,089

- 3,293

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

- 99

- 0

Sir let (sqrt x)^2=x

And substitute x=-1, then we would have (i)^2=-1.

Is it right?

And substitute x=-1, then we would have (i)^2=-1.

Is it right?

- #10

MathematicalPhysicist

Gold Member

- 4,341

- 234

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

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.Sir let (sqrt x)^2=x

And substitute x=-1, then we would have (i)^2=-1.

Is it right?

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

MathematicalPhysicist

Gold Member

- 4,341

- 234

It's really a matter of notation.

For example Principles of Algebraic Geometry by Harris uses \sqrt(-1) instead of i.

- #13

- 371

- 1

- #14

- 22,089

- 3,293

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...

It's really a matter of notation.

For example Principles of Algebraic Geometry by Harris uses \sqrt(-1) instead of i.

Thank you for your example!!

- #15

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 961

- Last Post

- Replies
- 19

- Views
- 19K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 12

- Views
- 15K

- Last Post

- Replies
- 3

- Views
- 966

- Last Post

- Replies
- 13

- Views
- 8K

- Last Post

- Replies
- 9

- Views
- 7K

- Replies
- 9

- Views
- 2K

- Replies
- 6

- Views
- 30K

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 2K