Problem with imaginary numbers

I think this is probably me not thinking, but:

VB Code:

If Sqrt( x ^ 4 ) = x ^ (4 * 0.5) = x ^ 2   'then shouldn't
 
Sqrt ( i ^ 4 ) = -1 ?  'since it would be
 
i ^ (4 * 0.5) = i ^ 2 = -1?

Mathematica returns 1 and I don't exactly want to doubt that.

Pardon my ignorance...but doesnt it depend entirely on the value of i?

On another point, how can anything ^2 be negative? I'm quite lost.

errr, the title indicates imaginary numbers, in this case, i, which is Sqrt(-1) (see edit)...

Surely you know about i, and how z = a + i b etc etc...

You know... complex and imaginary numbers....


Edit:
Whoops, got my definitions wrong. i is defined as; i^2 = -1.

sqrt( i ^ 4)
THe sqrt() part is worked out last.
=> sqrt( i ^ 4)
= sqrt( (i ^ ²)²)
= sqrt( -1²)
= sqrt(1)
= 1 (positive root worked out)

Ah I see... but really, aren't both those equations identical? ie, square root of 2 ^ 2 is the same as halving the power...

Remember that square root function has two possible values.

SquareRoot( x⁴ ) = +x²
SquareRoot( x⁴ ) = -x²

SquareRoot(i⁴) = +1
SquareRoot(i⁴) = -1

The root gives two answers, so it is best to find what you are square-rooting, and then make our final answer +/-
If you halve the power first, then find out the answer, you will only get one solution, so you must remember that the solution is +/- this number.

The answer you got, -1, is just the negative root, whereas mathematica, as a default, gives the positive root. i.e. they are both the same answer, just not complete. You and mathematica are both right.

(For more on this subject, see another forum of 2=3, or something like it. Actually, the debate there is rather "heated" )

i^4=1, since i^2=-1, which makes i^4=-1^2=1. so, sqrt of 1 may equal -1.

Sqrt is defined as the non-negative square root of a number. therefore the amswer to the example given is 1 as indicated by sql_lall. as for why it doesn't work, its because i is an imaginary number, and not all real operations are applicable to imaginary numbers. btw, the real defination of i is i^2=-1.

btw de moivre's theorem is most likely used to solve roots of imaginary numbers.

hope this helps! ^-^

Bugzpodder: The definition of i must include the possibility of both a positive and a negative square root of minus one.

(-i)² = -1

There might be some context in which square root is defined as a non-negative value. In an algebraic or calculus context, it is defined as either solution to the following equation.

x² - n = 0

In some circumstances, the negative root is preferred, due to additional conditions.

I retain my orginal ground on my point of view since:

(-i)^2 = (-1*i)^2=((-1)^2)*(i^2)=1*(i^2)=i^2=-1

and for x^2-n=0, you are confusing square roots with roots of square :D. One interesting point -- all squares of real numbers are positive (again this "rule" does not apply to imaginary numbers ---I should say complex-- just like square roots -- such as i^2=-1).

let's look at this equation:

x^2-n=0
x^2=n
x=sqrt(n) or -sqrt(n), n>=0

true x does have two roots, one positive, one negative. BUT since sqrt is a function, by defination of functions, it can only return one answer (hence the vertical line test, if you can draw a vertical line that intersect at two different points of the graph, then the graph is not a graph of a function). its always true just as when some one asks you the sqrt(9) is: you would only say 3 correct? and fyi, the plus and minus is there precisely because the sqrt(n) returns a positive value!

Remember, rules that apply to real numbers does not mean that it applies to imaginary and complex numbers!

Quote:

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Buszpodder:
Remember, rules that apply to real numbers does not mean that it applies to imaginary and complex numbers!

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True... but hey, I'm only learning the concept of complex numbers!

This might be a little off topic, but some people don't seem to get this...

Sqrt(x) ALWAYS has two opposite results (i.e. a and -a) regardless of the value of x (unless of course x = n/0).

NO! the SQRT(x) is a function. by definition of function, it only returns one result! Sqrt(9)=3. not -3. hence the vertical line test.

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

Old school maths people consider Sqrt(x) as a one-to-one function and x^(1/2) as a one-to-many function.

sqrt(x) is equilvalent x^(1/2) -- they are interchangable.
x^(1/2), exactly like sqrt(x) and every other function, gives only one answer. the definition of a function is to only have one answer. there is no such thing as one-to-many functions. for every x, you are going to get precisely one value for f(x). therefore restrictions for the domain values are often understood.

such as f(x)=sqrt(x). it is understood that x>=0.

an example of one-to-many "equation" (I can't think of a more proper term) would be like for example x^2+y^2=1
if you throw in any x, (restrictions here would be abs(x)<=1), you'll get two values for y (except when x=1), one positive, one negative. you can see by drawing a vertical line with the absolute value of x coordinate less than to 1, it'll intersect at two distinct points, thus the two y-values.

something similar to what you may have actually being talking about would be x^2, as take a look at this example:

x^2=9
abs(x)=sqrt(9)
abs(x)=3
x=3,-3

the *better* way to do the example is to do the difference of squares
x^2=9
x^2-9=0
(x-3)(x+3)=0
x=3,-3

Quote:

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hmm, sqrt(x) is every way same as x^(1/2).

---

You are mistaken.

sqrt(16) = +4
16^(0.5) = +/-4

no. i am 100% sure that sqrt(x) is equilvalent to x^(1/2). if you got a graphing calculator/software, plug in
f(x)=sqrt(x) and
f(x)=x^(1/2) you'll see the same graph.

if you don't have these handy, a weaker argument would be to just plug in 4^(1/2) into your calculator (or windows calculator). it is most definately give you 2 everytime. Why do you think is that? calculators are not suppose to be misleading and give you half of the correct answers. there is only one answer to x^(1/2), that is, the positive one. as mentioned in my last edited post, the restriction for the domain of f(X)=x^(1/2) is x>=0.

You must understand, that it is f(x)=x^(1/2), f(x) part precisely indicates that it is a *function*. as I have repeated several times, a function does *NOT* return many answers. if you are a student, speak with a math teacher or someone with authority. I can assure you that I am 100% correct. A side note, I take responsibility for my own posts as I usually do not provide [potentially] incorrect informations or infos that may mislead readers.


imagine functions that does actually return infinite amount of answers, what'll happen if you plug in that function in your calculator? which one is it going to display? does it display one, wait 1 second, and display another??

Now my head's spinning.. peoples confusion is making me confused.

I think some people are referring to (I would) "Sqrt(x)" == "The square root of x" and not a function within a programming language. I, myself, would define sqrt(x^2) = +/- x

It seems silly that mathematica would actually return only one value for this. I don't have Maple with me, but I believe it would return "-1, 1" or something like that.

However, all of this arguing is silly. I think it's probably confused people more than actually helping them learn something useful. God help us if stuff like the twin-paradox ever comes up - the solution will be mutilated beyond repair.

Oh well, I've ranted enough. :D

Destined

yea that reminds me, good point, sqrt(x) is not just "square root of x", it is *THE NON-NEGATIVE square root of x*. by looking at the mathematical definition of sqrt() FUNCTION, sqrt(x^2)=abs(x).
by the BEDMAS, you do x^2 first. thats going to get you a non-negative value. then you square root it to get abs(x). I can understand that sqrt(x^2)=+/-x is *seemed* to be logical and therefore right, but it isn't BECAUSE it is a function and functions only return EXACTLY one value.

1. functions only return EXACTLY one value.
2. sqrt(x) is the NON-NEGATIVE square root of x.
3. x^(1/2) is equilvalent to sqrt(x)

The only people that these posts are going to confuse is those people who insists that they are right but really does not know what they are talking about. i am not saying this to offend anyone, but merely making a point.

Quote:

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1. functions only return EXACTLY one value.

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Since when? I don't really remember this being defined anywhere, except in places like grade 8 math where they say stuff like, "just ignore sqrt(-1)" even though you can actually do it.

If you want a vb function that returns more than one value:

VB Code:

Private Type Point
   x As Long
   y As Long
   z As Long
End Type
 
Private Function SomeFn() As Point
' A fn that returns more than one value!
   SomeFn.x = 3
   SomeFn.y = -2
   SomeFn.z = 10
End Function

If you want a mathematical fn that returns more than one value "f:=x->x^0.5;"

Or am I missing something that others seem to know about?

Destined

Look: this all depends on whether we are talking about VBs intrinsic Sqrt function, or the mathmatical operator (notice I didn't say function because it's not one). VBs Sqrt(x) returns only the one value, +4 if x is 16. Therefore it is a function of x. The mathmatical operator (let's denote it Sq(x)), is not a function since it returns two values for every value of x, with absolutely NO exception.

THE definition of a function is: "Any relationship which takes one element of one set and assigns to it one and ONLY one element of a second set is said to be a function." And that's a quote.

So VB is right to call their Sqrt function a function, since it only returns the positive square root (it is therefore NOT equivalent to the similar mathmatical operator).

As for the mathmatical operator, Sq(x) (x with the line along the left hand side, top and the little tick coming off the bottom left), is defined as the positive square root of x. So that is a function. What is not a function is ±Sq(x), or x^1/2, since they both incorporate both square roots of x.

The bottom line here is that on it's own Sq(x) (with those lines) is not 'the square root of x', it's 'the positive square root of x'. x^1/2 is defined as both square roots of x, namely ±a and is therefore not a function.

I'm not sure if that came out clearly or not...

you claim that x^(1/2) is NOT a function but sqrt(x) -- the operator with tick thing --- is a function.

according to the fractional exponents law:

x^(a/b)=((b th-root)(x))^a
so x^(1/2)=((2 th-root)(x))^1=sqrt(x)
the equal sign indicates that the expressions are equilvalent.

here is the proof:

x^(1/2)*x^(1/2)=x(1/2+1/2)=x^1=x

by definition, sqrt(x)*sqrt(x)=x
so x^(1/2)=sqrt(x)

in reality there is nothing wrong with these either function notations, indicating that x^(1/2) is a function:
f(x)=x^(1/2) or y=x^(1/2)

as i said, plug it in a graph calculator and you'll see what I am talking about. if u still don't believe me, y don't you look up fractional exponents.

*corrected below*

*ditto*

The operator with the tick is a function because it only returns the non-negative square root, namely one value. However x^1/2 is not a function because it returns both the positive and negative square roots. Simple as that. Oh and I don't need someone explaining the 'fractional index law' or whatever name you came up with for it; it's such a basic concept of mathmatics that it doesn't need a pretenious name like that so you might want to complain to wherever you got that name from. That would be like saying 3 + 3 = 6 is a fine example of the recently discovered "Olbzmann's additional conjecture" or something silly like that.

lol. WHY is it not a function? because you think it returns two values? tell me one calculator that gives you both 2 and -2 if you put (4)^(1/2) and i'll believe you.

You're not listening. Calculators aren't perfect. On a calculator yes x^1/2 would be a function since the calculator isn't advanced enough to give both correct answers. That's a fault with calculators, not maths. In algebra, x^1/2 is not a function because it ALWAYS returns two values.

lol you have no proof of what you are saying. has it ever occured to you that you may be wrong? and don't blame it on the calculator. just use any calculator, or even your computer, or even better yet, any programming languages. try to raise a number to the exponent of 0.5 and see what you get. i mean, y a special rule to raise a number to 0.5? if you raise something to any other exponent it would give you one single answer, so y the special treatment?

Actually you're wrong. n^1/a will return two opposite values if a is an even number. And I don't need proof because I'm stating a fundamental law of mathematics: a function takes each value from one set, and assigns it to one and ONLY one value in a second set. Therefore f(x) = x^1/2 is NOT a function since it assigns TWO values in the second set for each value in the first set, wake up you fool!

You basic proof?

Here:

4² = 16, agreed?
So Root(16) = 4.

But WAIT!

(-4)² = 16 TOO!!

So Root(16) = -4 too!

If you don't understand why don't you go back to school and retake basic maths classes. I'm not blaming calculators, the makers of calculators don't include the second solution in the function because they assume you're CLEVER enough to know it exists and to know what it is for any given value!

You think I'm wrong, you give me two values for even a and any x (not including a = 0), for which x^1/a doesn't have two return values!

lol tell me to go back to school is your proof that n^(1/a) has two values??? you think attacking me is going to make your point?? and you should think that calculator assume things for you? well i am not smart enough for that kind of calculators then lol. mebbe the calculator is smart enough to know that the user (you) knows n^(1/2) returns 2 answer, but also assumes that the user (you) does not know that 1+1=2!! why don't you type into the calculator and tell me if it says "you are smart enough to know that!" btw find me one other piece of PROFESSIONAL evidence (website, book, etc) that says 4^(1/2)=2 and -2.

I have a few degrees in math, and I can tell you that

Sqrt(4) <> 4^(1/2)

Any number to a fractional power, that, once simplified, has an even denominator or an even numerator with a denominator of one, returns only a positive number.

Therefore,

Sqrt(4)=+-2, but
4^(1/2) = 2

no they are interchangable. sqrt(x) is a function and functions only return one value!

snakeeyes you telling me that (-4)^1 is positive and not -4?

lol if you are doing (-4)^(2/2) you are actually taking the square root first (fractional exponent law). you get 2i
square it gets -2

i don't care if you have a zillion math degrees. i'd be surprised if you can find any professional site that tells you sqrt(2) gives you two answers or that 2^(0.5) gives you two answers or that x^(1) is always positive! since you all believe what you say, so do that, find a professional source, i'd be interested to see one!

reread my post, i agree with you.

2/2 is simplified to 1/1, so please look into what you argue.

yea you are right, i misread, i apologize. but as i said sqrt(x) returns only 1 value because sqrt(x) is a function, as simple as that!

yes, the VB sqr() function returns only the positive number, and it is standard convention that when the radical sign is encountered, to take the positive value, but both are technically correct. The correct answer depends on the context of the question. In most questions, negative values can be disregarded because they aren't applicable to real problems. Take for instance, the following question.

A square has an area of 100 square units. What is the length of the side of this square?

Mathematically, both 10 and -10 are correct, but since -10 is not a possible answer, it is disregarded. Take for instance the opposite example, however much less common.

a^2 + 4a + 4=0

goes to

a(a+2)+2(a+ 2)=0

goes to

(a+2)^2=0

Therefore, a MUST equal -2.
You can see why negatives are most often left out, but not always.

i am not talking about the vb sqrt function! i am talking about the radical sign you call square root in math! it is also a function because you can right y=radical(x) (i used sqrt(x))

sqrt(x) (or radical (x) or whatyawantocallit) is defined mathematically to be the NON-NEGATIVE SQAURE ROOT of a number!

yes, you are right, it is defined that way. Therefore,

Sqr(100)

is taken at at face value to be

|Sqrt(100)|

Whenever you arrive at Sqrt(Expression) it is asking you for the positive one. What I am saying is that when it is not explicit, such as

X^2=4

X can be either 2 or -2. Therefore, 2 and -2 are both valid. Sorry about any confusion, my English is coming along though.

yes I COMPLETELY agree with you! a quadratic has two roots! x^2=4 is of course a quadratic equation. it has roots 2 and -2!

the roots of a square has two values, while sqrt(x) function only return the NON-NEGATIVE square root of the number

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?