How many solutions does the equation x^2= a have?

[Ali Asadullah]

1^2=(-1)^2
taking square root on B.s
1=-1..?

 

[micromass]

This demonstrates that it is not true that $$\sqrt{x^2}=x$$. It is a very popular mistake with students. The reason that it is not true is that the square is not a bijection, thus defining an inverse operation is tricky, and will give you consequence which are not so nice (like this one).

The rule can be fixed however, and the correct formula is $$\sqrt{x^2}=|x|$$. Note, that it is however true that $$(\sqrt{x})^2=x$$, but you can only do this when x is positive.

 

[Edi]

square root of (-1)^2 (witch is really same as sqrt 1 )= -1 and 1
and sqrt 1^2 (witch is .. you know.. same as sqrt 1 )= 1 and -1 ..
sooo... something to do with this.
but its quite fascinating.

 

[micromass]

square root of (-1)^2 (witch is really same as sqrt 1 )= -1 and 1
and sqrt 1^2 (witch is .. you know.. same as sqrt 1 )= 1 and -1 ..
sooo... something to do with this.
but its quite fascinating.

This is not true. The square root only yields ONE answer. Thus the square root of (-1)^2 is 1. It is NOT 1 and -1.
The square root of a number x is defined as the unique POSITIVE number y such that $$y^2=x$$. It is not a multi-valued operation.

 

[Ali Asadullah]

Note, that it is however true that $$(\sqrt{x})^2=x$$, but you can only do this when x is positive.

Then Sir what will be the square of i?

 

[micromass]

Then Sir what will be the square of i?

The square of i is -1...

 

[Ali Asadullah]

Bur Sir, Sqrt(x^2)=x is valid only when x is positive. How can it be valid for -1?

 

[micromass]

Bur Sir, Sqrt(x^2)=x is valid only when x is positive. How can it be valid for -1?

Yes, it is valid only when x is positive, it is not valid for -1. That's what I said in my previous post (or did I make a type somewhere?)...

 

[olivermsun]

Note, that it is however true that $$(\sqrt{x})^2=x$$, but you can only do this when x is positive.

I believe that it's conventionally taken to hold for negative numbers as well, since for positive x,
$$\sqrt{-x} = i\sqrt{x}$$
so that
$$(\sqrt{-x})^2 = (i\sqrt{x})^2 = -1 \cdot x = -x.$$

 

[micromass]

I believe that it's conventionally taken to hold for negative numbers as well, since for positive x,
$$\sqrt{-x} = i\sqrt{x}$$
so that
$$(\sqrt{-x})^2 = (i\sqrt{x})^2 = -1 \cdot x = -x.$$

If you could provide me of a reference that does this, I would be pleased. But I don't think people do this.

In fact, the square root is only defined for positive real numbers. The square root is not defined for negative numbers. I.e. the value $$\sqrt{-1}$$ does not exist. The problem is of course that both i and -i qualify as being the square root, and there is not good reason to choose one above the other.
You could of course define it to be i (or -i), but I haven't seen any author do this. I did see definitions for things like $$(-1)^{1/2}$$ however, but I didn't see anybody using the square root operator on anything but positive numbers yet...

EDIT: I actually DID see notations like $$\sqrt{-1}$$ in math books once. The point is that all these books were from 1800 and so. I haven't seen modern authors do this yet...

 

[HallsofIvy]

Bur Sir, Sqrt(x^2)=x is valid only when x is positive. How can it be valid for -1?

It isn't and he did not say it was. Sqrt(x^2)= |x| which is equal to x only when x is positive.

It is also true that Sqrt((-1)^2)= Sqrt(1)= 1 so that Sqrt(x^2)= x is NOT valid for x= -1. Sqrt((-1)^2)= |-1|= 1 is true.

(The question, and thread, has nothing to do with the square root of negative numbers and I believe introducing complex numbers into it is a mistake.)

 

[Ali Asadullah]

Another Question Sir,
Is it right that sqrt(x^2)=(sqrtx)^2 for all real x?

 

[micromass]

Another Question Sir,
Is it right that sqrt(x^2)=(sqrtx)^2 for all real x?

For positive x, yes. For negative x, the notation $$\sqrt{x}$$ is not defined, thus your equation has no meaning for negative x...

 

[olivermsun]

In fact, the square root is only defined for positive real numbers. The square root is not defined for negative numbers. I.e. the value $$\sqrt{-1}$$ does not exist. The problem is of course that both i and -i qualify as being the square root, and there is not good reason to choose one above the other.

But if you view -x in the complex plane, then there is a well-defined way to choose a principal square root, and hence it's convenient to interpret $$\sqrt{-x}$$ in this way. If you're interested, I suppose you might find this in Brown and Churchill, Complex Variables and Applications, or something similar (but I'd have to double check).

 

[micromass]

But if you view -x in the complex plane, then there is a well-defined way to choose a principal square root, and hence it's convenient to interpret $$\sqrt{-x}$$ in this way. If you're interested, I suppose you might find this in Brown and Churchill, Complex Variables and Applications, or something similar (but I'd have to double check).

Yes, I agree that there is a convenient way to choose a principal branch of the square root and that it would be well-defined. But I argue that authors don't do this, and certainly won't assign the square root sign to that. But I could be wrong, and I'm going to check that book you recommended right now...

But anyway, as Halls pointed out, this is a question about real variables, and when working over $$\mathbb{R}$$, then $$\sqrt{-1}$$ is certainly not defined...

 

[micromass]

I've checked the entire book of Brown and Churchill, and it appears that they never define something called $$\sqrt{-1}$$ or something similar. They do define square roots of -1 and denote it as $$(-1)^{1/2}$$, like I've mentioned. They also define a principal branch of the square root and they denote it as $$z^{1/2}$$, but they never use the radical symbol for it. Instead, they specifically mention on p23, that the radical symbol will always indicate the unique positive square root of a positive real number.

So notations like $$\sqrt{-1}$$ do not occur in the book...

 

[olivermsun]

I've checked the entire book of Brown and Churchill, and it appears that they never define something called $$\sqrt{-1}$$ or something similar.

*Shrug* I freely admitted that I needed to double check to see if that particular work used the notation. Evidently it does not, but a quick Google search turns up lots of appearances of the square root over the negative, one amusing title being: http://press.princeton.edu/titles/6388.html" . It's just some people's convention though, not really worth arguing about, is it?

Back to the OP: the squares/square root confusion is the whole thing that makes graphing circles fun.
A usual way to define a unit circle around the origin is x^2 + y^2 = 1

On the x-axis, where y=0, the equation reduces to x^2 = 1, which is similar to your original equation. Taking square roots on both sides gives one solution x=1, but the circle crosses the x-axis at two points, right?

 

[dalcde]

This is not true. The square root only yields ONE answer. Thus the square root of (-1)^2 is 1. It is NOT 1 and -1.
The square root of a number x is defined as the unique POSITIVE number y such that $$y^2=x$$. It is not a multi-valued operation.

The square root of a function has two values. The square root function yields only one unique number.

 

[Mark44]

This is not true. The square root only yields ONE answer. Thus the square root of (-1)^2 is 1. It is NOT 1 and -1.
The square root of a number x is defined as the unique POSITIVE number y such that $y^2=x$. It is not a multi-valued operation.

The square root of a function has two values.

No. Assuming the function evaluates to a positive real number, the square root of that number is a single positive real number, just as micromass says above.

The square root function yields only one unique number.

Yes.

 

[HallsofIvy]

The square root of a function has two values. The square root function yields only one unique number.

No, you are wrong on this one.
"The square root of ... " is the square root function.

The equation $x^2= a$ has two solutions, $x= \sqrt{a}$ and $-\sqrt{a}$. The reason we need to write the "-" is because $\sqrt{a}$ by itself means the positive solution only.