Problem with imaginary numbers

First up, let's not post in this thread about how many mathematics qualifications we have. There is another thread for that, and assurances of qualifications are not proof. In any case, I have the most and best qualifications. So don't go there.

Second up, posts are easier to read if you can stretch over to that shift key to use capital letters at the beginning of sentences. (I had to read this post about forty times to check for my capitals :D)

Now onto the main part. Above I stated this:

Quote:

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Sqrt(x) only returns the +ve result. x^(1/2) returns both +ve and -ve.

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Or more explicitly:
The mathematical function Sqrt is defined to return the +ve results only. x^(1/2) returns both +ve and -ve roots.

Bugz, although I find the logical structure of your posts difficult to follow, you seem to agree that Sqrt is defined to return only the +ve results. Thus, we disagree only on x^(1/2), namely I believe it to have 2 roots, and you believe it to have 1 root.

You may not trust a calculator.
You may not trust your computer.
You may not trust your intuition.
You may only trust proof.

By the way, the vbforums server keeps going down so I can't get on very often. I'll read subsequent posts eventually...

While having a little break from my programming, I found this link that's related to the original post, if anyone was still wondering about it. There's a link on the bottom of the page for the solution.

BrainTeaser.html

Enjoy.

BTW, DHooper, I was wondering when you'd show up. ;) hehee..

Destined

this is where you are wrong

we are NOT (at length i am not) arguing how many roots x^(1/2) has. i know it has two roots. its a totally different question

we are arguing a simiple question: evaluate the following expression

4^(1/2)=?
some say 4^(1/2)=2 and -2
i say 4^(1/2)=2

or in general: for the funtion f(x)=x^(1/2), how many values does f(5) give? how many values does f(x) give, for all x>=0, x is an element of Real number set? does it give a positive value and a negative value?

so my question is:
f(x)=x^(1/2)
is there anything wrong with that?
if not, then you agreed it is a function.
function by definition returns only 1 value. or it is not a function. do you agree?
so if u agree, then how does x^(1/2) returns +ve and -ve values since it only allowed to return one single value?

From what I have read through this, this is an interesting debate going on here. For reference to the topic:

http://mathworld.wolfram.com/SquareRoot.html.

They know what they are talking about. The basic arguement is that 4^0.5 = only +2 because of the convention of defining it as the principal square root. But if you were to solve the equation x^0.5 = a for values of a, you would return a = ±x^0.5, which to holt all confusion means plus or minus that, not both at the same time.

Quote:

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so my question is: f(x)=x^(1/2)
is there anything wrong with that?
if not, then you agreed it is a function.
function by definition returns only 1 value. or it is not a function. do you agree?
so if u agree, then how does x^(1/2) returns +ve and -ve values since it only allowed to return one single value?

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1) no there is nothing wrong with that, since x^(1/2) can be expressed as a function because it is defined as only being the principal (positive) square root and will therefore only assign one element in the codomain to each element in the domain.
2) yes it is a function because of that so Gordon is wrong
3) x^(1/2) on its own only returns a positive value, since it only defines the principal square root of x. However, in an equation 4(1/2) = x, x = ±2, since when taking the square root in an equation you must include both correct answers. In that form (±x^0.5), it's not a function of x because it returns both values.

It seems to me this entire thread is mostly understood but the thing that's keeping people apart on this is notation so let me clear this up:

x^0.5 = Sqrt(x); e.g. if x = 16, x^0.5 = Sqrt(x) = 4 and ONLY 4

In an equation: say x² = 36, x = ±36^0.5 = ±6

Does that help?

i agree with everything else you said. but could you clear up this bit on your 3rd point:

if i understand you right, you said the solution to 4^(1/2)=x would be x=2 or -2?


if i were given this algebriac equation, i'd simiplify the LHS. i take the principal square root of 4, which gives me 2. so i'd take x=2 and go.

so if i have a really complex LHS, I'd have to consider both values of the square roots instead of just taking the square root and go?

such as this nested square root question: x=sqrt(7+sqrt(3+sqrt(1)))
i'd have to consider sqrt(1)=-1 as well? and same with the second and third sqrt and can't just take LHS=3, so x=3?