# Imaginary Number?

#### Artman

I have an Algebra problem giving me trouble. I am supposed to determine without graphing, if the following equation is symetrical to the x or y axis or to the origin. The equation is y^2 = -5/x^2. The problem is, I think that this is an imaginary number. Am I right about that?

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#### Tom Mattson

Staff Emeritus
Gold Member
You think that what is an imaginary number?

#### Artman

In order to discover whether or not the axis are symetrical you replace the x or the y or both with -x or -y if you rearrange the equation to (-y)^2 * (-x)^2 =-5 the only way to solve for that -5 is to use the imaginary number "i" (as far as i can tell). Because replacing the y with -y or the x with -x still gives a positive number when squared.

Does this sound correct?

#### arcnets

Artman, I think it's good to first find out what the problem actually is. OK, you have an equation containing variables x and y. There will be a set of points (x,y) whose coordinates x and y fulfill that equation. This set is called the graph of that equation.
Now you want to know if that graph shows any symmetries. For example, if the graph is mirror-symmetric WRT the y-axis, then for any point (x,y) in the graph, the point (-x,y) also belongs to the graph. This means, when you substitute -x into x, the equation stays the same.
There's no need to 'solve' the equation. Just check if it stays the same.

#### Artman

I did that part, but I wasn't sure if this possibly being an imaginary number problem would throw a wrench into the works. I always tend to overcomplicate what should otherwise be a simple problem.

#### mighty2000

In order to discover whether or not the axis are symetrical you replace the x or the y or both with -x or -y if you rearrange the equation to (-y)^2 * (-x)^2 =-5 the only way to solve for that -5 is to use the imaginary number "i" (as far as i can tell). Because replacing the y with -y or the x with -x still gives a positive number when squared.
Isn't this just when you find the inverse of the equation.

I know that this is a shortcut to do so, just swap the x and the y and solve for y, and that will give you the inverse of the function. But as far as the imaginary number thing, I don't know what you are talking about bro.

peace out

M2k

#### HallsofIvy

Homework Helper
I started to write a rather sarcastic response, echoing Tom's original response "You think that what is an imaginary number?"
but then thought better of it.

y^2= -5/x^2 CAN'T be true for any non-zero value of x because then the right hand side is negative and the left hand side can't be. On the other hand this can't be true for x=0 because we can't divide by 0.

The "graph" of this relation is the EMPTY SET!

#### Artman

Originally posted by HallsofIvy
I started to write a rather sarcastic response, echoing Tom's original response "You think that what is an imaginary number?"
but then thought better of it.

y^2= -5/x^2 CAN'T be true for any non-zero value of x because then the right hand side is negative and the left hand side can't be. On the other hand this can't be true for x=0 because we can't divide by 0.

The "graph" of this relation is the EMPTY SET!

That's where I am with this problem.

#### FZ+

[vain hope]Maybe he meant y^2 = (-5/x)^2?[/vain hope]

#### Artman

[vain hope]Maybe he meant y^2 = (-5/x)^2?[/vain hope]
No. The problem was written as y^2 = -5/x^2. It could be written y^2 = -5/(x^2) without changing the meaning.

Isn't it true that if I was to find a solution set for this equation I would have to use the unit imaginary number i? Like this (iy^2)*(x^2) = -5 or (y^2)* (ix^2) = -5.

#### HallsofIvy

Homework Helper
The original post said "I am supposed to determine without graphing, if the following equation is symetrical to the x or y axis or to the origin. The equation is y^2 = -5/x^2."
The whole point is that the graph is the empty set! (By the way, the question of whether or not the "equation" is symmetrical is non-sense. SETS OF POINTS, such as graphs, are, or are not, symmetrical so I assume this meant "determine if the graph of the equation is symmetrical".)

If you are using as your definition of symmetrical:

"A set of points A is symmetrical to the x axis if and only if whenever (x,y) in in A then (x,-y) is also"

"A set of points A is symmetrical to the y axis if and only if whenever (x,y) in in A then (-x,y) is also"

"A set of points A is symmetrical to the origin if and only if whenever (x,y) in in A then (-x,-y) is also"

Then the empty set is symmetrical in all three ways because the condition is vacuously true. That is, the statement "if a then b" is true if a is always false as in the case of the empty set.

#### arcnets

Agree, HallsofIvy.

#### HallsofIvy

Homework Helper
Of course, the original problem said "without graphing" and talking about the empty set is dangerously like "graphing"!

I suspect that what the original poster's instructor wanted was something like:

If you replace x with -x in y^2= -5/x^2, since the x is squared the equation is not changed: it is symmetric with respect to the y-axis.

If you replace y with -y in y^2= -5/x^2, since the y is square the equation is not changed: it is symmetric with respect to the x-asxis.

Of course, if you replace both x and y with -x and -y, respectively, then the equation is not changed: it is symmetric with respect to the origin.

That is the type of argument that would be accepted by someone who talks about a symmetric "equation" rather than a symmetric graph!

#### Artman

Originally posted by HallsofIvy...

If you replace x with -x in y^2= -5/x^2, since the x is squared the equation is not changed: it is symmetric with respect to the y-axis.

If you replace y with -y in y^2= -5/x^2, since the y is square the equation is not changed: it is symmetric with respect to the x-asxis.

Of course, if you replace both x and y with -x and -y, respectively, then the equation is not changed: it is symmetric with respect to the origin.
The other problems in the chapter were looking for this type of response. This matches the answer I got when I worked it out. I'll let you know if it was what he was looking for after I get my paper back in a week or two (correspondence course).

Thank you all.

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