Domain of a function

[Graphmastur]

Floppus, where do you see a complex number in the domain or range? What real x value can you put in that gives a complex output? You see the square root of a negative, and just stop there, and walk away. The domain and range are all real numbers.

[calc84maniac]

But if you're exclusively working with the set of real numbers and the square root of a negative number is undefined, how can you define the square of that?

That's like saying (5/0)*0=5. It's just not a valid argument.

[Weregoose]

[Mapar007]

I knew that one. Yeah, algebra utterly nullifies intuition... (sometimes)

[simplethinker]

There is a difference between the limit of a function at a point and its value at that point. For any A>2, the limit as x->A of (√(2-x))² is 2-A. The value, however, is undefined.

A similar situation is the point x=2 in (x²+x-6)/(x-2). Just because the discontinuity is removable it doesn't mean the discontinuity isn't there.

[polarBody]

I don't think limits apply here. If you define complex numbers, then both the value of (√(2-A))² and the limit of (√(2-x))² as x approaches A is 2-A for any A. Again, it all comes down to whether you've defined complex numbers or not.

EDIT: Oops. Limits only work with real-valued functions. Still, it comes down to the blasted complex numbers! Are they defined or undefined? As a mathematician, it's your choice!

[Graphmastur]

What you people keep ignoring is the fact that a square root and and a square negate each other. Give me one input value that outputs a complex number. Just one number. There is none. Just try to find one input point that gives me a bad output point.

THE DOMAIN AND RANGE ARE ALL REAL NUMBERS.

Try any number.

[FloppusMaximus]

polarBody: You certainly can take limits when complex numbers are involved!

Graphmastur: A square root and a square do not "cancel each other out." Algebraic simplification rules are all well and good, but they are only valid insofar as you can prove that they do not make any difference to the result of the problem. In this case, your "simplification" changes the problem dramatically.

Anyway, it was a stupid problem to begin with. The domain of a function is not something you can derive from its definition - the domain is part of the definition to begin with, or you haven't defined it properly.

(edit: having a bit of trouble with my keyboard there, sorry)

In any case, Graphmastur, what is keeping you from using i? Surely if your square-root function knows of a square root of -1, it knows of a square root of that number as well. And your "simplification" - which is valid in when working in C, though not in R - ought to tell you that any complex number will work just as well.

[polarBody]

So you're saying that the limit of (√(2-x))² as x approaches i is 2-i? I'm having trouble seeing how a number can approach i. Like this?
0.9*i, 0.99*i, 0.999*i

[Graphmastur]

@polarbody
i is the 3d Z plane if memory serves.

@Floppus

It does not change the function. A square root is the same thing as raising something to the 1/2 power. By the basic rules of powers, they cancel out. Since we are in the 2D plane, get rid of complex numbers. What I am saying, is that there is no real input that will give a non-real output. This is basic algebra, don't try to make it so complicated.

[Builderboy2005]

Graphmastur, you can't just cancel out the square root and the square, if imaginary numbers are undefined. Like calc84 said if you had (5/0)*0 by the 'simple' rules of algebra, you would be able to cancel out both the zeros and end up with 5, even though one of the most important rule IN algebra is to not divide by Zero!

You are not allowed to cancel out the square root (or ^1/2 if you want) if you are not working with complex numbers. Even if the input and output are both real (lets discard imaginary input ) you still need to go into imaginary mathematics if you want to get an answer.

[simplethinker]

There is one key point in this debate that I don't believe anyone has addressed: we are dealing with a function. A function must satisfy the property that if b=f(a) and c=f(a), then b=c. Because of this fact, we must restrict the range of the square root so that it is single-valued at a point. The accepted convention is that you take the positive root (e.g. √1 = 1, not -1). However, what's the "positive root" of 3-4i? Once you go into the realm of complex numbers things that are considered elementary in the real numbers get a whole lot messier.

Also, consider this: what's the range of the function √(x²)? Once you square the x, it doesn't matter if it was initially positive or negative. Taking the square root then gives |x|.

[DarkerLine]

[Graphmastur]

[polarBody]

The square root of i is approx. 0.707+0.707i. The square of the square root of i is i. What's preventing you from including i or any complex number in the domain of the square root function?

[Graphmastur]

[polarBody]

I think you mis-read my post. Let's do a little math.

Here's our function: f(x)=(√(2-x))²

Why not plug in i?

f(i)=(√(2-i))² = 2-i

Do you see that I plugged in a complex number and ended up with a complex number? In other words, do you see how the domain includes complex numbers, and so does the range? You keep saying that the domain of this function is all reals, but I've just shown that it's not.

EDIT: Also, you're again making the mistake of arbitrarily canceling exponents. If complex numbers are not defined in the context of the problem, then you CANNOT cancel the square root function by squaring it. Since I defined complex numbers in the example above, then it was okay to make that cancellation. But I ask you, if √(-1) is undefined within a problem, then how do you square it? What is the square of an undefined number?

[Builderboy2005]

[Quertior]

I checked with my teacher. He is perfectly fine using complex numbers. The only thing is, in the context of domain and range, we are supposed to find the domain such that the range is a subset of R. For this reason the domain of my function is R . By the way thanks for all the responses!