Question

Prove the distance from a point to itself is zero?

Answer 1

Basically it's just travelling from one point to another, and since you are travelling from the point A to point A (the same) you don't travel at all, making it zero.

Answer 2

I'm too lazy to prove it mathematically. (think)

Answer 3

you could use the distance formula


d=√[(x_2-x_1)^2+(y_2-y_1)^2]

Answer 4

where the point is (x_1, y_1) = (x_2, y_2)

Answer 5

so it's like saying point A = AB = BA =B then the distance must be zero

Answer 6

Well, metrics are defined such that any point has 0 distance from itself. Otherwise it isn't a metric. The only way to really prove this rigorously otherwise is to have a definition of distancel.

Answer 7

how do i get a line out of a point cuz A B would be = nothing lies between AB i'd need a third point there must be a better way to prove it....def of distance requires a ruler right so i have to have line before measurement

Answer 8

so it's better to do it by contradictions saying that AB != 0 right

Answer 9

The level or rigor is established by first establishing the definitions and theorems you are allowed to use. You haven't provided these, so there is no point in even trying to prove it.

Answer 10

i'm not given anything but the theorem above ..i can't assume zero i know that cuz that is what i'm proving

Answer 11

You can't prove anything without a definition of terms.

Answer 12

i can use axioms and definitions not sure if that helps

Answer 13

does axiom 0 allow me to form my line with a third point that way with a line i can use a ruler then

Answer 14

but before i use axiom 0 can i say A= B and before that say let A and B be points

Answer 15

what axioms and definitions do you have?

Answer 16

axiom 1 is the point line incidence

Answer 17

the ruler function to is used to define betweenness for points on a line and use betweenness to define segments and a ray

Answer 18

What is point line incidence?

Answer 19

Okay, I think I could help you come up with a proof.

Answer 20

Are there any more axioms besides 1 and 2?

Answer 21

So just axiom 0, 1, and 2?

Answer 22

yes but is it easier to do it by contradiction

Answer 23

But would axiom 0 just state that there is no line, and thus no distance between the points? Because that wouldn't make sense would it?

Answer 24

Would you like to hear my attempt of a proof?

Answer 25

sure

Answer 26

I thought I had it, but I didn't.

I was thinking about considering all points which are distance d from A. In this case we'd assume that AA = d and that d != 0. We'd try to find a contradiction.

Answer 27

A would have to be among the points which are distance d from A. There would be some other point B which is distance d from A, and we'd have to show that somehow AB = d and yet AB != d

Answer 28

Okay, I have to talk to you a bit about axiom 2 for a moment



```
axiom 2 For every line L there is a one to one onto function f:L --> IR such that fro any point A B on L I f(B) - f(A) I = AB
```
You are saying that f maps from L to IR.
Now we know L would be the set of all points on L, so maps from points to elements in IR
IR would be what exactly? Real number, correct?

In this case, f maps from points to real numbers.
We know that f(A) = k, and it can't equal anything else otherwise f would not be a function.
This means AA = |f(A) - f(A)| = |k-k| = 0

Answer 29

This would be a direct proof relying on axiom 1 as well as the definition of a function.

Answer 30

Also relies on the fact that any number minus itself is 0.

Answer 31

I don't understand your proof, and it doesn't proof what you are trying to prove.

Answer 32

It was proving the converse.