- #1

MHD93

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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

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- Thread starter MHD93
- Start date

In summary, the equation x^0.5 = -1 does not have a real number solution, but it does have a complex number solution. This is because in the real number system, the square root function is defined to be only non-negative, but in the complex number system, it can have multiple values. Thus, in the complex number system, there is a solution to x^0.5 = -1, which is x = 1. However, in the real number system, there is no solution.

- #1

MHD93

- 93

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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

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- #2

zgozvrm

- 754

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- #3

drag12

- 11

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sqrt(x)=-1

Does that help?

- #4

llamaprobe5

- 7

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You are right, no solutions exist.

- #5

zgozvrm

- 754

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llamaprobe5 said:You are right, no solutions exist.

Not true ... no

(There

- #6

llamaprobe5

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zgozvrm said:Not true ... noREALsolutions exist.

(Thereis, however, a complex solution).

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

- #7

berkeman

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llamaprobe5 said:Um, could you provide an example, I couldn't find one...

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

- #8

llamaprobe5

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berkeman said:Search Imaginary Number on wikipedia.org, for example...

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

- #9

zgozvrm

- 754

- 1

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

- #10

zgozvrm

- 754

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llamaprobe5 said:I know what an imaginary number is..

Apparently, you don't.

- #11

llamaprobe5

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- #12

llamaprobe5

- 7

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

- #13

jgm340

- 104

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This is what you are looking for: http://en.wikipedia.org/wiki/Exponentiation#Negative_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.

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.

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- #14

llamaprobe5

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zgozvrm said:Not true ... noREALsolutions exist.

(Thereis, however, a complex solution).

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...

- #15

jgm340

- 104

- 2

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.

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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- #16

HallsofIvy

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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.)

In the real number system, the square root function is "single valued" (as are all function in the real numbers) and we

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

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.)

Last edited by a moderator:

- #17

zgozvrm

- 754

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zgozvrm said:Not true ... noREALsolutions exist.

(Thereis, however, a complex solution).

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...

- #18

MHD93

- 93

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Thank you all for your replys

But the following one got me annoyed

No, it isn't fooling me!

Thank you too for your reply

But the following one got me annoyed

sqrt(x)=-1

Does that help?

No, it isn't fooling me!

Thank you too for your reply

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- #19

berkeman

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I may have misread the posts as well. Thank you to Halls for the great response!

- #20

theprofessor0

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- #21

giles123

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the solution is i1 ( the i shows its imaginary)

- #22

berkeman

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giles123 said:the solution is i1 ( the i shows its imaginary)

No. As Halls explained in post #16, the solution is 1.

For an equation to have a solution, it means that there is a value or set of values that can be substituted into the equation to make it true. In other words, it is the value(s) that satisfy the equation and make it a valid statement.

To determine if an equation has a solution, you can try to solve the equation by isolating the variable on one side and simplifying the other side. If you end up with a true statement, then the equation has a solution. If you end up with a false statement, then the equation has no solution.

Yes, an equation can have more than one solution. This is known as a system of equations, where there are multiple equations with multiple variables that can be solved simultaneously to find all the possible solutions that satisfy all the equations.

To solve an equation with multiple variables, you can use different methods such as substitution, elimination, or graphing. The key is to manipulate the equations to eliminate one variable at a time until you are left with a single equation and a single variable, which can then be solved to determine the values of the variables.

Yes, there are equations that do not have a solution. These are called inconsistent equations, where no value or combination of values can satisfy the equation. This can happen when the equation is contradictory, such as 2 = 3, or when there are no common solutions between two or more equations in a system.

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