Inversion Through P in Euclidean Plane | Fix(Vp)

[Gail Brimeyer]

Homework Statement

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

Homework Equations

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)?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

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)?

 

[Gail Brimeyer]

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

 

[ystael]

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 V_p as a function or mapping, which is what you really need. If x = p, what is V_p(x)? What about if x \neq p; what is V_p(x), in a rough geometrical description?

It may help you to impose coordinates; in that case it is simpler to start by supposing p = (0, 0), 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 V_p 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 \pi/4).

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 \mathop{\mathrm{Fix}}(V_p)? It is the set of fixed points of V_p, that is, \mathop{\mathrm{Fix}}(V_p) = \{ x \in E : V_p(x) = x \}. You are right to consider whether p \in \mathop{\mathrm{Fix}}(V_p) separately from whether x \in \mathop{\mathrm{Fix}}(V_p) for x \notin p, but to build a correct argument, you need to think more carefully of V_p as a function, and what it does to each point.

 

[Gail Brimeyer]

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.

 

[ystael]

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 V_p, V_p(p) = p".

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 x \neq p".

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 V_p(p) = p, also V_p(V_p(p)) = V_p(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.

I think you mean to begin with "if x \neq p". The essence of your argument is correct. V_p(x) is the other endpoint of the segment having x as one endpoint and p as midpoint. V_p(V_p(x)) is the other endpoint of the segment having V_p(x) as one endpoint and p as midpoint. But these are one and the same segment; so V_p(V_p(x)) = x.

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