Algebraic Test For Symmetry...1

  • Thread starter nycmathguy
  • Start date
  • #1
nycmathguy
Homework Statement:
Use the algebraic tests to check for
symmetry with respect to both axes and the origin.
Relevant Equations:
N/A
PUse the algebraic tests to check for
symmetry with respect to both axes and the origin.

x - y^2 = 0

Let x = -x

-x - y^2 = 0

Not symmetric with respect to the y-axis.

Let y = -y

x - (-y)^2 = 0

x - y^2 = 0

Symmetric with respect to the x-axis.


Let x = -x and y = -y

-x - (-y)^2 = 0

-x - y^2 = 0

Not symmetric with respect to the origin.
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
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397
Say f(x,y)=0 given
Symmetry respect to y-axis is
[tex]f(-x,y)=0[/tex]
Symmetry with respect to x-axis is
[tex]f(x,-y)=0[/tex]
Symmetry with respect to Origin is
[tex]f(-x,-y)=0[/tex]
 
  • #3
nycmathguy
Say f(x,y)=0 given
Symmetry respect to y-axis is
[tex]f(-x,y)=0[/tex]
Symmetry with respect to x-axis is
[tex]f(x,-y)=0[/tex]
Symmetry with respect to Origin is
[tex]f(-x,-y)=0[/tex]
Good to know but is my work correct?
 
  • #4
anuttarasammyak
Gold Member
883
397
If you are interested in post #2 for
Homework Statement:: Use the algebraic tests to check for
symmetry with respect to both axes and the origin.
Relevant Equations:: N/A

x - y^2 = 0
apply
[tex]f(x,y)=x-y^2[/tex]
and check it as another homework.
 
  • #5
nycmathguy
If you are interested in post #2 for

apply
[tex]f(x,y)=x-y^2[/tex]
and check it as another homework.
How is f(x, y) applied to the equation?
 
  • #7
nycmathguy
##f(x,y)=x-y^2=0##
##f(-x,y)=-x-y^2=-2x+f(x,y)=?##
If f(x,y) = 0, then -2x + f(x,y) is -2x + 0 = -2x.
 
  • #8
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6,876
##f(x,y)=x-y^2=0##
##f(-x,y)=-x-y^2=-2x+f(x,y)=?##
I don't know what you're doing in the 2nd line above.
##f(-x, y) = -x - y^2 \ne f(x, y)##, so not symmetric about the y-axis.
 
  • #9
nycmathguy
I don't know what you're doing in the 2nd line above.
##f(-x, y) = -x - y^2 \ne f(x, y)##, so not symmetric about the y-axis.
Someone else got it wrong? Happy to know it's not me this time. Precalculus is way over our heads, right? Mark44, when I think about precalculus, abstract algebra aka modern algebra does not come to mind. Understand?

I am not intimidated by a course I took and passed with an A minus in 1993. I take it day by day. I can do without belittling comments from several members like THIS IS BABY MATH or YOU SHOULD ALREADY KNOW THIS STUFF, etc. Get it? Not very encouraging at all. Either provide help or skip my "easy" problems.

P. S. My upper case letters are for emphasis only. Sorry but I am not a millennial. I was raised in a society that was not "politically correct" as the new saying goes.
 
  • #10
35,129
6,876
Mark44, when I think about precalculus, abstract algebra aka modern algebra does not come to mind. Understand?
Abstract algebra/modern algebra is an advanced subject that you have never seen before. It should not be confused with algebra as taught in high school. Modern algebra deals with things like groups, rings, and other mathematical structures.
 
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  • #13
nycmathguy
##=-2x+0=-2x##
So ##f(-x,y)\neq 0##. f is not symmetric wrt y axis.

Abstract algebra/modern algebra is an advanced subject that you have never seen before. It should not be confused with algebra as taught in high school. Modern algebra deals with things like groups, rings, and other mathematical structures.
You missed my point. When I think about precalculus abstract algebra does not come to mind. By this I mean that precalculus is not so bad if guided correctly, calmly and patiently.
 
  • #14
nycmathguy
##=-2x+0=-2x##
So ##f(-x,y)\neq 0##. f is not symmetric wrt y axis.
Is my answer -2x wrong?
 
  • #15
35,129
6,876
You missed my point. When I think about precalculus abstract algebra does not come to mind.
No, you missed my point. Abstract algebra is different from "ordinary" algebra as taught in high school. I gave a brief description of some of the topics that are included in abstract algebra AKA modern algebra.
Is my answer -2x wrong?
It's incomplete in the context of this problem, which is to determine whether the function ##f(x, y) = x - y^2 = 0## is symmetric about the y-axis. What do you conclude when you determine that ##f(-x, y) = -x - y^2 = -2x + 0##?
 
  • #16
nycmathguy
No, you missed my point. Abstract algebra is different from "ordinary" algebra as taught in high school. I gave a brief description of some of the topics that are included in abstract algebra AKA modern algebra.

It's incomplete in the context of this problem, which is to determine whether the function ##f(x, y) = x - y^2 = 0## is symmetric about the y-axis. What do you conclude when you determine that ##f(-x, y) = -x - y^2 = -2x + 0##?
You know, Mark44, I am not going to bother arguing about this again and again. Precalculus is easier than abstract algebra. This is my point. The answer to your question is: IDK. Good night.
 
  • #17
35,129
6,876
Precalculus is easier than abstract algebra.
Yes, which is why abstract algebra is often taught as a 400-level course in two semesters in the senior year of a mathematics undergrad degree.
 
  • #18
nycmathguy
Yes, which is why abstract algebra is often taught as a 400-level course in two semesters in the senior year of a mathematics undergrad degree.
Thank God I never have to face that course. My goal: To learn precalculus through calculus 3. I may try linear algebra after calculus 3 but not sure at this time. I still have a long way to go.
 
  • #19
jbriggs444
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Homework Statement:: Use the algebraic tests to check for
symmetry with respect to both axes and the origin.
Relevant Equations:: N/A

PUse the algebraic tests to check for
symmetry with respect to both axes and the origin.

x - y^2 = 0

Let x = -x

-x - y^2 = 0

Not symmetric with respect to the y-axis.

Let y = -y

x - (-y)^2 = 0

x - y^2 = 0

Symmetric with respect to the x-axis.


Let x = -x and y = -y

-x - (-y)^2 = 0

-x - y^2 = 0

Not symmetric with respect to the origin.
As I understand your approach to the problem...

You have an equation in x and y. You substitute -x for x in this equation and test whether the result is an equivalent equation. You judge that it is not. So you conclude that there is no symmetry about the y axis.

Correct conclusion.

You repeat, substituting -y for y and test whether the result is an equivalent equation. After simplifying, you see that the resulting equation is indeed identical. So you conclude that there is symmetry about the x axis.

Correct conclusion.

You repeat again, performing both substitutions this time. Again you test whether the resulting equation is equivalent to the original. You judge that it is not.

Correct conclusion.


However, the approach that you have selected has at least two potential pitfalls.

1. It depends on your being able to correctly recognize when two equations are or are not equivalent.

In this case your judgement was sound.

2. Strictly speaking, symmetry does not require that the substituted equation be formally equivalent to the original. It is enough that the solution sets are identical.

Say, for instance, we have the equation x2 + 2x + 100 = -y2 and that we are testing for symmetry about the y axis. We replace x and -x and see that the new equation: x2 - 2x + 100 = -y2 is not equivalent. We [falsely] conclude that there is no symmetry.

But the solution set to both equations is empty. So the graphs of the solution sets are both empty and, therefore, identical. So there is symmetry.
 
  • #20
nycmathguy
As I understand your approach to the problem...

You have an equation in x and y. You substitute -x for x in this equation and test whether the result is an equivalent equation. You judge that it is not. So you conclude that there is no symmetry about the y axis.

Correct conclusion.

You repeat, substituting -y for y and test whether the result is an equivalent equation. After simplifying, you see that the resulting equation is indeed identical. So you conclude that there is symmetry about the x axis.

Correct conclusion.

You repeat again, performing both substitutions this time. Again you test whether the resulting equation is equivalent to the original. You judge that it is not.

Correct conclusion.


However, the approach that you have selected has at least two potential pitfalls.

1. It depends on your being able to correctly recognize when two equations are or are not equivalent.

In this case your judgement was sound.

2. Strictly speaking, symmetry does not require that the substituted equation be formally equivalent to the original. It is enough that the solution sets are identical.

Say, for instance, we have the equation x2 + 2x + 100 = -y2 and that we are testing for symmetry about the y axis. We replace x and -x and see that the new equation: x2 - 2x + 100 = -y2 is not equivalent. We [falsely] conclude that there is no symmetry.

But the solution set to both equations is empty. So the graphs of the solution sets are both empty and, therefore, identical. So there is symmetry.
Interesting reply. I thank you.
 
  • #21
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Say, for instance, we have the equation x2 + 2x + 100 = -y2 and that we are testing for symmetry about the y axis. We replace x and -x and see that the new equation: x2 - 2x + 100 = -y2 is not equivalent. We [falsely] conclude that there is no symmetry.
It's true that the equation in your example has symmetry, but the kinds of symmetry in the multiple problems posted here so far are specifically limited to symmetry about the x- or y-axis or the origin. The equation you showed is symmetric about the x-axis and about the line x = -1, but not about the y-axis.
 
  • #22
jbriggs444
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It's true that the equation in your example has symmetry, but the kinds of symmetry in the multiple problems posted here so far are specifically limited to symmetry about the x- or y-axis or the origin. The equation you showed is symmetric about the x-axis and about the line x = -1, but not about the y-axis.
On the contrary. It is symmetric about the y axis as well. The graph is not a parabola. It is empty.

However, you may be working one step beyond me, speaking of a formal symmetry of the equation rather than a graphical symmetry of the solution set. If one extended the domain of x and y to include the complex numbers for instance, then the graphical symmetry would vanish and the formal symmetries (if any) would remain.
 
Last edited:
  • #23
35,129
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On the contrary. It is symmetric about the y axis as well. The graph is not a parabola. It is empty.
I didn't say it was a parabola, but I mistakenly thought it was a circle.

If we write ##f(x, y) = 0 = x^2 + 2x + y^2 + 100##, then ##f(x, -y) = 0##, so the equation is symmetric about the x-axis (whatever that means for an equation with an empty graph), but ##f(-x, y) \ne f(x, y)##, so the equation is not symmetric about the y-axis.

In any case, your example would not be a good choice for a problem in the first chapter of a precalculus textbook.
 
  • #24
jbriggs444
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I didn't say it was a parabola, but I mistakenly thought it was a circle.
Ahh, yes. A funny sort of circle with a negative radius. I hadn't visualized it that way. Quite apt.
 
  • #25
nycmathguy
Interesting reading. Thank you everyone.
 

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