Imaginary Number?

  • Thread starter Artman
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  • #1
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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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  • #2
Tom Mattson
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You think that what is an imaginary number?
 
  • #3
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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?
 
  • #4
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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.
 
  • #5
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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.

Thanks for your help.:smile:
 
  • #6
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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
 
  • #7
HallsofIvy
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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.

Artman is saying it awkwardly but there is an obvious point about this graph:

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!

NOW argue about whether it is "symmetric about the axes"!
 
  • #8
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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.

Artman is saying it awkwardly but there is an obvious point about this graph:

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!

NOW argue about whether it is "symmetric about the axes"!
That's where I am with this problem.
 
  • #9
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[vain hope]Maybe he meant y^2 = (-5/x)^2?[/vain hope]
 
  • #10
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[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.
 
  • #11
HallsofIvy
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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.
 
  • #12
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Agree, HallsofIvy.
 
  • #13
HallsofIvy
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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!
 
  • #14
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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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