Structure R^k and midpoints of vectors (rudin's PMA chapter 1 problem 16)

In summary, the conversation is about a self-study of mathematical analysis using Rudin's Principles of Mathematical Analysis. The person is stuck on problem 16 in the first chapter, which involves proving statements about the norm in k-dimensional space. They have made progress on parts a and b, but are struggling to find a general solution for part a and to prove that the only solution for part b is x+y=2. They are looking for help in finding a clean and rigorous proof using only definitions and theorems.
  • #1
SiddharthM
176
0
First this is NOT a homework problem. I am undertaking a self-study of mathematical analysis by following Rudin's Principles of Mathematical analysis. I have done a course in analysis before but this is a high-powered review of sorts.

So I'm currently on the first chapter's problem set and I've gotten stuck on problem 16, which asks:

Let x,y in R^k, k>/=3 (at least 3-space), norm[x-y] = d>0 prove:

a)If 2r>d There are infinitely many z in R^k s.t. norm[z-y]=norm[z-x]=r
b)If 2r=d there is exactly one such z.
c)If 2r<d there are no such z.

That is the question as stated, to clarify the norm I speak of is the tradition euclidean k-space norm (i.e. root of sum of squares of components).

Part a) is the one I've made the least progress on, b) I'm half done and c) is a simple proof by contradiction using the triangle inequality.

Geometrically (for part a)) consider the line between x and y, there is a perpindicular plane at the midpoint (x+y)/2 (perpendicular to the line connecting x and y) and the set of infinite z that a) asks for is the circle of radius (r^2 - (d/2)^2)^(1/2) which is nonzero b/c of the hypothesis lying on the tangent plane centered at the midpoint (x+y)/2. The thing is, I presume rudin wants me to construct a general z that admits infinitely vectors, but I've found this very difficult to do using the definitions and theorems given in the chapter. Any ideas?

Part b) the only such z is x+y=2 but I can't for the life of me prove that it is the ONLY solution with rigour.

Help would be much appreciated.

Cheers,
Siddharth M.

PS: obviously the problem I'm having is providing a clean and neat proof strictly using definitions and theorems as is required of an analyst but such a solution has thus far escaped me.
 
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  • #2
$|x-z|+|z-y| =2r \Rightarrow |x-y| \leq 2r \forall z$ by the triangle equality, the rest pretty much falls in your lap.
 

1. What is the definition of "Structure R^k"?

"Structure R^k" refers to the concept of a k-dimensional Euclidean space, also known as a Cartesian space. It is a mathematical structure that represents points in space using coordinates, and allows for the measurement of distances and angles between points.

2. What is the significance of midpoints in vectors?

In vectors, the midpoint refers to the point that lies exactly halfway between two given points. It is significant because it allows for the calculation of other important vector properties, such as the magnitude and direction, as well as the addition and subtraction of vectors.

3. How are midpoints of vectors calculated?

The midpoint of two vectors can be calculated by adding the corresponding components of the two vectors and dividing the result by 2. This gives the coordinates of the midpoint, which represents the average of the two points.

4. What is the relationship between midpoints and the structure R^k?

The concept of midpoints is closely related to the structure R^k, as it is a fundamental concept in Euclidean geometry and vector spaces. The structure R^k allows for the precise representation and manipulation of points and vectors, including the calculation of midpoints.

5. How are midpoints used in Rudin's PMA chapter 1 problem 16?

In this problem, Rudin uses the concept of midpoints to prove that the distance between two points on a line is equal to the absolute value of the difference between their coordinates. This is an important result that is used in many other mathematical proofs and applications.

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