Has this equation got a solution?

Mohammad_93

I worked hard to figure out whether this could be solvable or not

x^0.5= -1

x belongs to any set in all of the mathematical world you prefer

This can't be a homework I think.

Note: it isn't x^2 = -1

zgozvrm

Read up on imaginary numbers:

http://en.wikipedia.org/wiki/Imaginary_numbers

drag12

I think the notation is just fooling you, x^0.5 is the same as sqrt(x), so your problem is:

sqrt(x)=-1

Does that help?

llamaprobe5

You are right, no solutions exist.

zgozvrm

Not true ... no REAL solutions exist.
(There is, however, a complex solution).

llamaprobe5

Um, could you provide an example, I couldn't find one...

berkeman

Search Imaginary Number on wikipedia.org, for example...

llamaprobe5

I know what an imaginary number is...I mean, could you provide an example of a solution for this equation.

zgozvrm

How do you give an example of a solution?

I guess "5" is an example of a solution, albeit not a very good one, because it is wrong!

zgozvrm

Apparently, you don't.

llamaprobe5

Apparently you don't either if you're example solution is 5... I simply asked for an example solution...I know that you think that you are extremely intelligent, but you have failed to answer a rather simple question...

llamaprobe5

Just to clarify, I am asking for a solution...

jgm340

This is what you are looking for: http://en.wikipedia.org/wiki/Exponen...tive_nth_roots

The solution is x = 1, but it depends on how you make your definitions. If you define the square root to be a relation instead of a function, you can say that sqrt(1) is both 1 and -1.

llamaprobe5

Oh ok thank you for the direct answer, that makes sense... I guess that does work if you say +/- 1 is the square root. So there ARE real solutions...

jgm340

Actually, let me clarify:

Let R be the space of real numbers. R x R (pronounced "R cross R") is the set of all ordered pairs (x,y) where x and y are both in R.

When you graph a function f(x) = ???, you are looking at every x and finding the unique y value that lies above it. In fact, the ability to graph f(x) like this is what makes it a function!

Another way to describe the graph of a function, however, is the set of ordered pairs (x,y) in R x R such that y = f(x).



The top diagram in this image is the graph of a function, since there is exactly one point in the graph for each x value. Notice that I'm using "graph" in a very specific sense: it refers to the set of blackened points (which in this case form a wavy line).

We can think of the color black as representing "TRUE" and white as representing "FALSE". So for any point (x,y), being in the graph (a.k.a. being black) is the same as it being TRUE that y = f(x). If (x,y) is not in the graph (in other words, it is white), then it is FALSE that y = f(x). So really, the graph of a function is just the set of points where it is TRUE that f(x) = y.

With this understanding of a graph, we can generalize the idea of a function. The middle diagram shows the graph of what's known as a "relation". A point (x,y) is blackened if and only if x ~ y (read, "x related to y"). In this case, each x value has TWO possible y values where x ~ y.

Just as a function can be defined by its graph, a relation can be defined by its graph. Just as you might ask "Does y = f(x)?", you can ask "Is x ~ y?" for any particular ordered pair (x,y). In fact, a function IS a special kind of relation. It's important to note that with most relations, order is importat! So just because x ~ y, doesn't mean y ~ x!

The bottom image (thrown in for good measure), shows that any graph can be a relation. In this case, there are an infinite number of y values such that x ~ y.

Now, we can define the square root to be a relation! We will say that x ~ y if x is "a square root of" y. And by this, we mean that x ~ y if and only if x^2 = y.

This is in fact a function of x, whether x is positive or negative (or imaginary, for that matter!), because for every x, there is exactly one y satisfying x ~ y.

It turns out that you are asking "Is there a y such that -1 is a square root of y?" Well, we just determined that for any x (including x = -1), we can find a unique such y! That is, we are looking for a y such that (-1)^2 = y. That y is just the number 1.

Now, the reason I brought up all this stuff about relations is that while the "is square root of" relation is a function of x, it does NOT have a natural inverse function. To create an inverse function, people just restrict the domain of the "is square root of" function to only non-negative x.

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HallsofIvy

This is an odd situation. Actually, there is a real number solution to [itex]x^{0.5}= -1[/itex], but you have to extend to the complex number system to find it!

In the real number system, the square root function is "single valued" (as are all function in the real numbers) and we define [itex]\sqrt{x}= x^{0.5}[/itex] to be the non-negative number, a, such that [itex]a^2= x[/itex]. By that definition, a square root is never negative and we cannot have real x such that [itex]x^{0.5}= -1[/itex].

However, in the complex number system, for a number of reasons, including the fact that the complex numbers cannot be made into an ordered field, we must allow "multivalued" functions. In the complex number system, [itex]\sqrt{x}= x^{0.5}[/itex] is any number, a, satisifying [itex]a^2= x[/itex].
In the complex number system, since [itex]1^2= 1[/itex] and [itex](-1)^2= 1[/itex], [itex]\sqrt{-1}[/itex] is both 1 and -1. Thus, [itex]\sqrt{x}= x^{0.5}= -1[/itex], in the complex number system, is satisfied by x= 1, which happens to be a real number.

(Although x= 1 is a real number, x= 1 would NOT be a solution to that equation in the real number system. In the real number system [itex]1^{0.5}= 1[/itex], only.)

zgozvrm

My bad ... I misread the question to mean that the OP was looking for [itex]x = \sqrt{-1}[/tex], or [itex]x^2 = -1[/tex]

It was explicitly stated that he was NOT looking for the latter:
"Note: it isn't x^2 = -1"


That's what I get for trying to help out when I've got a head cold ... I can't think straight!

My apologies...