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Artman

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- Thread starter Artman
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Artman

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- #2

quantumdude

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You think that *what* is an imaginary number?

- #3

Artman

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Does this sound correct?

- #4

arcnets

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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

Artman

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Thanks for your help.

- #6

mighty2000

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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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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

Artman

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Originally posted by HallsofIvy

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

FZ+

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

- #10

Artman

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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 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

arcnets

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Agree, HallsofIvy.

- #13

HallsofIvy

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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

Artman

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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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