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Graphmastur

Joined: 25 Mar 2009
Posts: 360

 Posted: 26 Sep 2009 03:14:45 am    Post subject: 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

Elite

Joined: 22 Jan 2007
Posts: 770

 Posted: 26 Sep 2009 07:56:07 am    Post subject: 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
Authentic INTJ

Super Elite (Last Title)

Joined: 25 Nov 2004
Posts: 3976

 Posted: 26 Sep 2009 08:16:34 am    Post subject: Mapar007 wrote:"Common Sense" is my main argument Math is stuffed to the brim with counterintuitive concepts. Moreover, common sense would have us regard the earth as flat.
Mapar007

Joined: 04 Oct 2008
Posts: 365

 Posted: 26 Sep 2009 11:31:26 am    Post subject: I knew that one. Yeah, algebra utterly nullifies intuition... (sometimes)
simplethinker
snjwffl

Active Member

Joined: 25 Jul 2006
Posts: 700

 Posted: 26 Sep 2009 12:28:55 pm    Post subject: 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

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 26 Sep 2009 02:25:59 pm    Post subject: 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!Last edited by Guest on 26 Sep 2009 02:38:02 pm; edited 1 time in total
Graphmastur

Joined: 25 Mar 2009
Posts: 360

 Posted: 26 Sep 2009 08:54:32 pm    Post subject: 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

Joined: 22 Aug 2008
Posts: 472

 Posted: 26 Sep 2009 09:46:41 pm    Post subject: 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.Last edited by Guest on 26 Sep 2009 09:55:15 pm; edited 1 time in total
polarBody

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 26 Sep 2009 10:25:38 pm    Post subject: 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

Joined: 25 Mar 2009
Posts: 360

 Posted: 27 Sep 2009 09:09:46 am    Post subject: @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

Joined: 19 Apr 2009
Posts: 51

 Posted: 27 Sep 2009 10:57:27 am    Post subject: 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
snjwffl

Active Member

Joined: 25 Jul 2006
Posts: 700

 Posted: 27 Sep 2009 11:32:18 am    Post subject: 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|. Graphmastur wrote:This is basic algebra, don't try to make it so complicated. Some things are not as simple as they appear: -By basic algebra, 0^0 = 0 since zero to anything is 0. -By basic algebra, 0^0 = 1 since anything to zero is 1.
DarkerLine
ceci n'est pas une |

Super Elite (Last Title)

Joined: 04 Nov 2003
Posts: 8328

 Posted: 27 Sep 2009 11:33:35 am    Post subject: polarBody wrote: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 Statements about limits are much stronger over the complex numbers. The limit of (√(2-x))² as x approaches i would be 2-i only if this were true for any direction of approach to i (including your (0.9*i,0.99*i,0.999*i, ...) but also other things like (i+1,i+1/2,i+1/4,i+1/8,...) or even completely weird spiraling stuff). Alternatively, there's the universal statement where for every real ε>0, there is a real δ such that if |x-i|<δ, |√(2-x))²-(2-i)|<ε. This works for limits over arbitrary metric spaces. FloppusMaximus wrote: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.This.
Graphmastur

Joined: 25 Mar 2009
Posts: 360

 Posted: 27 Sep 2009 01:38:03 pm    Post subject: Builderboy2005 wrote: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! (5/0)*0=(5/0)(0/1)=0/0=undefined The square root of a real number is an imaginary number. The square of the square root of a negative number is a negative number. Ya know what. I am not even gonna push this issue anymore. Quertior, the Domain is all real numbers.
polarBody

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 27 Sep 2009 01:45:49 pm    Post subject: 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

Joined: 25 Mar 2009
Posts: 360

 Posted: 27 Sep 2009 02:27:49 pm    Post subject: polarBody wrote: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? That is the problem, no one is thinking. I = sqrt(-1). Say we input 5 for x. That means that it is (sqrt(-3))^2. We can't take the square root of a negative number, without using i. (You can use I so I will demonstrate using both methods.) Without i, the square root and square cancel out. With i, that means that it is (i(sqrt(3)))^2. I^2 is -1. So you have -1(sqrt(3))^2. The square root and the square cancel, because you square everything inside the parenthesis, and that leaves -1(3). How is that a non-real number?
polarBody

Newbie

Joined: 02 Sep 2009
Posts: 30

 Posted: 27 Sep 2009 02:38:34 pm    Post subject: 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?Last edited by Guest on 27 Sep 2009 02:52:40 pm; edited 1 time in total
Builderboy2005

Joined: 19 Apr 2009
Posts: 51

 Posted: 27 Sep 2009 03:37:32 pm    Post subject: polarBody wrote: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? Exactly, that's all it boils down to, whether or not his teacher defines The square root of a negative number to be defined or not.
Quertior

Newbie

Joined: 30 Aug 2009
Posts: 12

 Posted: 30 Sep 2009 08:24:19 pm    Post subject: 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!
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