1. Mar 14, 2014

### press

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

Copy-paste from my textbook:

Let S_1 be the sphere of radius 1, centered at the origin. Let a be a
number > 0. If X is a point of the sphere S_1, then aX is a point of the sphere of radius a, because
||aX|| = a||X|| = a. In this manner, we get all points of the sphere of radius a. (Proof?)

2. Relevant equations
3. The attempt at a solution

On another site I posted this below:

Suppose we have a sphere S of radius 1 centered at the origin. Let X be a point on S. Then ||X - 0|| = 1.

Since ||cA|| = c||A|| for any vector A and c > 0, we have ||cX|| = c||X|| =c that is if we stretch the vector X by a factor of c, then the length stretches also by that amount. So, cX is a point on a sphere S_2 of radius c.

How do we show all the points of Sphere S_2 of radius c are cX?

You have S1={ |X|=1 }, S2={ |X|=c }, and cS1 = { cX for some X in S1 }, and you want to show S2 = cS1. You show X in S2 implies X in cS1 and vice-versa.

If X in S2, then |X|=c, and |(1/c)X|=(1/c)c=1, so (1/c)X is in S1, and X=c((1/c)X) is in cS1.

The other way, starting with X in cS1, so X=cY for some Y with Y in S1, then |X|=|cY|=c|Y|=c*1=c, so X in S2.

Are we showing that if cS1 equals S2, then cS1 is a sphere of radius c and since |cX| = c|X| = c, cX is a point on cS1?

Didn't get any answer. At this point I am very confused and have no idea whats going on. Can anyone please elaborate on this problem?

Thanks.

2. Mar 14, 2014

### Simon Bridge

Lets see if I understand you - by rephrasing what you wrote:

If $\vec X$ is defined to be an arbitrary vector centered on the origin,
and R is a positive real number,
then the set $S_R=\{\vec{X}:X=R \}$ would be the set of all vectors that point to the surface of the sphere radius R, centered on the origin.

Thus - we could say that S_R "describes" a sphere radius R.

$S_1$ would be the set that describes the unit sphere.

You want to know if you have managed to prove that $RS_1=S_R$

Is this correct?

3. Mar 16, 2014

### press

Yes. Thank you.

Somehow managed to understand this :)