P is a point in E

1. The problem statement, all variables and given/known data

If P is a point in E (euclidean Plane) Then the "Inversion through P" is
Vp = {(x,y)| x,y in E and either,
1. x=y=p, or
2. p is the midpoint of segment xy}


What's Fix(Vp)

Show that Vp composed of Vp = The identity


2. Relevant equations



3. The attempt at a solution
Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)

I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?

1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 
Last edited:

HallsofIvy

Science Advisor
41,626
821
1. The problem statement, all variables and given/known data

If P is a point in E (euclidean Plane) Then the "Inversion through P" is
Vp = {(x,y)| x,y in E and either,
1. x=y=p, or
2. p is the midpoint of segment xy}
so Vp is the set of all pairs of points satisfying those conditions?


What's Fix(Vp)
What does "Fix" mean? My first thought was "fixed point" but you find fixed points for a function from a set to itself and here V takes points to sets of pairs of points- it can't have any fixed points. And, further, it appears you are applying "Fix" to Vp, not V. Again, what is "Fix"?

Show that Vp composed of Vp = The identity
The way you have defined it, Vp is a set of pairs of points, not a function- you compose functions, not sets.


2. Relevant equations



3. The attempt at a solution
Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)

I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?

1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 
so Vp is the set of all pairs of points satisfying those conditions?
Yes, Vp satisfies those two conditions,
Given a point p, the inversion through p will give either,
1. x=y=p
2. P which is the midpoint of the line segment XY


What does "Fix" mean? My first thought was "fixed point" but you find fixed points for a function from a set to itself and here V takes points to sets of pairs of points- it can't have any fixed points. And, further, it appears you are applying "Fix" to Vp, not V. Again, what is "Fix"?

Here's the definition I found that might help you out,
Let T be a mapping.
A point B in the euclidean plane is called fixed for T if (B,B) is an element of T.


The way you have defined it, Vp is a set of pairs of points, not a function- you compose functions, not sets.
The only way I can help you with this question is to give you an example of another composition

given a point P not lying on a line L, Reflect point P over line L to give you point P' which is the Rl(p) (the reflection of P.)
Then Rl(p) composed of Rl(P) will bring you right back to the beginning point P and therefore
Rl composed of Rl is the identity
 
352
0
If P is a point in E (euclidean Plane) Then the "Inversion through P" is
Vp = {(x,y)| x,y in E and either,
1. x=y=p, or
2. p is the midpoint of segment xy}
OK, you have the formalities here, but you don't seem to be thinking of [tex]V_p[/tex] as a function or mapping, which is what you really need. If [tex]x = p[/tex], what is [tex]V_p(x)[/tex]? What about if [tex]x \neq p[/tex]; what is [tex]V_p(x)[/tex], in a rough geometrical description?

It may help you to impose coordinates; in that case it is simpler to start by supposing [tex]p = (0, 0)[/tex], and then figure out how to generalize.

Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)
It is true that [tex]V_p[/tex] is a rigid motion of the plane, but it is not true that a rigid motion composed with itself always equals the identity (think of translation by any nonzero distance in any direction, or rotation by say [tex]\pi/4[/tex]).

I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?
What is [tex]\mathop{\mathrm{Fix}}(V_p)[/tex]? It is the set of fixed points of [tex]V_p[/tex], that is, [tex]\mathop{\mathrm{Fix}}(V_p) = \{ x \in E : V_p(x) = x \}[/tex]. You are right to consider whether [tex]p \in \mathop{\mathrm{Fix}}(V_p)[/tex] separately from whether [tex]x \in \mathop{\mathrm{Fix}}(V_p)[/tex] for [tex]x \notin p[/tex], but to build a correct argument, you need to think more carefully of [tex]V_p[/tex] as a function, and what it does to each point.
 
Ok I'm going to write what I came up with, I'm not the best with notation so see if this makes sense to you...
Describe Fix(Vp) and proof - I now say that there's only one point fixed in Vp and that's when X=Y=P.
Proof:
Part 1. Given an element X and and a point P which is X, The inversion of element X through P will result in X therefore Vp(x)=x

Part 2. Given a element X and a point P that's not X in E. The inversion through P will produce a point X' where P is the midpoint of XX'. I say that it's impossible to have the inversion through P = P because if X isn't = P then the inversion will result in another element that's equal distance from P that X is.


Vp composed of Vp = Identity
Here's what I came up with, again sorry for the notation.
In the definition of Vp if x=y=p then we know that if x is equal to p then the inversion will give us p', then once again we take the inversion of p' it'll take us back to P. Where P=P'.

Now if X=Y=P, then you have your element x and a point p. the inversion through P will result in a point X' where XP congruent to X'P. because p is the midpoint. So by taking the inversion of X' will result back to point X where p is once again the midpoint.
 
352
0
Most of this is right in essence; I'll show you how to phrase some of it more carefully.
Part 1. Given an element X and and a point P which is X, The inversion of element X through P will result in X therefore Vp(x)=x
You could just say "by the definition of [tex]V_p[/tex], [tex]V_p(p) = p[/tex]".
Part 2. Given a element X and a point P that's not X in E. The inversion through P will produce a point X' where P is the midpoint of XX'. I say that it's impossible to have the inversion through P = P because if X isn't = P then the inversion will result in another element that's equal distance from P that X is.
This is correct. You might end by observing "and this distance is nonzero, because [tex]x \neq p[/tex]".
Vp composed of Vp = Identity
Here's what I came up with, again sorry for the notation.
In the definition of Vp if x=y=p then we know that if x is equal to p then the inversion will give us p', then once again we take the inversion of p' it'll take us back to P. Where P=P'.
This is correct, but it is simpler to say that since [tex]V_p(p) = p[/tex], also [tex]V_p(V_p(p)) = V_p(p) = p[/tex].
Now if X=Y=P, then you have your element x and a point p. the inversion through P will result in a point X' where XP congruent to X'P. because p is the midpoint. So by taking the inversion of X' will result back to point X where p is once again the midpoint.
I think you mean to begin with "if [tex]x \neq p[/tex]". The essence of your argument is correct. [tex]V_p(x)[/tex] is the other endpoint of the segment having [tex]x[/tex] as one endpoint and [tex]p[/tex] as midpoint. [tex]V_p(V_p(x))[/tex] is the other endpoint of the segment having [tex]V_p(x)[/tex] as one endpoint and [tex]p[/tex] as midpoint. But these are one and the same segment; so [tex]V_p(V_p(x)) = x[/tex].

Therefore [tex]V_p(V_p(x)) = x[/tex] for all [tex]x\in E[/tex], whether [tex]x = p[/tex] or [tex]x \neq p[/tex]; that is, [tex]V_p \circ V_p[/tex] is the identity map.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top