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Prove: For every pair of points $P, Q$
1. $d(P,Q)>0$
2. $d(P,Q) = 0$ if and only if $P=Q$
3. $d(P,Q) = d(Q,P)$

where $d(P,Q)$ is defined as the distance between $P$ and $Q$ and $d$ is a function

For the first part, I tried to assume the contrary, but I'm not sure where this gets me.

user141447
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    First show us what you have tried, and indicate what space you are working in and what distance function you are referring to. – K.Power Jan 24 '16 at 17:26
  • Um, those a re basic propositional axioms. They can't be proven unless you are working with a different set of equivalent propositions. ... Okay, what is the definition of a distance between two points? – fleablood Jan 24 '16 at 18:23
  • The only thing I'm given is that distance is defined as a funtion d: SxS -> R (where S is a point in an axiomatic system and R is the real numbers) – user141447 Jan 24 '16 at 18:25
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    There's nothing to prove. It is the definition of distance that distinct distances exist between points. That distances are positive between different points and independent of order. And that a point is 0 distance to itself. That's a definition. Nothing to prove. – fleablood Jan 24 '16 at 18:28
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    State the entire problem, word for word, and which book is this from and what class (title and discipline) this is for. And while you are at it, what is the chapter and section title and subject. Because right now, as stated, out of context, your question makes no sense. – fleablood Jan 24 '16 at 18:31
  • If "d: SxS -> R (where S is a point in an axiomatic system and R is the real numbers)" and "S is a point" then S is a single point and d maps one and only one ordered pair (S,S) to a single real number x. Any other pair of points in the axiomatic system doesn't have any function mapped to them. So ... context and precision matter. – fleablood Jan 24 '16 at 18:35
  • @fleablood, even assuming $S$ is a set, the question still doesn't make sense as you've noted. I'm expecting a line that says "for each of the functions $d$ defined below, show that they satisfy ...", but it seems like this part is missing. – Sherif F. Jan 24 '16 at 18:50
  • @user141447, clearly you must have been given some other info about $d$. Any arbitrary $d: S \times S \rightarrow \mathbb{R}$ will not necessarily satisfy these axioms. We can all easily think of counterexamples. So, only a class of $d$ that are valid distance functions will satisfy these axioms. – Sherif F. Jan 24 '16 at 18:53
  • I think I understand the question now and it's almost okay... but not quite. I'll post an answer. – fleablood Jan 24 '16 at 18:56
  • I don't think this should have been closed. The OP clearly has a problem s/he doesn't understand and wants help with. The OP doesn't state the problem well, because s/he doesn't actually understand it. We point out the limitations in the problem and point out using knowledge and a comfort with abstract concepts that the OP can't respond to simply because s/he doesn't have experience in the language of math. We should still work on trying to figure out what the question actually is. – fleablood Jan 24 '16 at 19:55

2 Answers2

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The three properties to prove that $d$ satisfies comprise the definition of a distance or metric. If $d$ is a distance, then it automatically satisfies these properties by definition. I think you're referring to a specific proposed $d$, which from the tag of "euclidean-geometry" I assume to be Euclidean distance. For the third property, exchange $P$ and $Q$ and mechanically show you obtain the same expression for $d(P,Q)$ as you get for $d(Q,P)$. For the second property,start with the assumption that $P = Q$ and evaluate $d(P,Q)$ directly. You'll notice this is exceptionally straightforward. Next, start with the assumption that $d(P,Q) = 0$. Then, directly derive a relationship between $P$ and $Q$. Finally, I think the first property should say either $d(P,Q) \geq 0$ (which is the standard definition) or it should say $P \neq Q \rightarrow d(P,Q) > 0$ (which also makes for an equivalent definition along with the second property), but it should not say $d(P,Q) > 0$.

Sherif F.
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    Yes, the first property should be d(P,Q)≥0. – user141447 Jan 24 '16 at 18:14
  • Which expression are you referring to for the third property? – user141447 Jan 24 '16 at 18:19
  • Explicitly evaluating $d(P,Q)$ and explicitly evaluating $d(Q,P)$, and showing they lead to the same expression. Please edit the question to indicate every word of the problem you're trying to solve and to specify the specific distance metric you're using (there are infinitely many possible distances - you need to specify which one). – Sherif F. Jan 24 '16 at 18:39
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In analysis or a topology or an abstract math course the idea of distance function is an abstract concept:

Let S be any set then a "metric" or a "distance" is a function, $d:S \times S \rightarrow \mathbb R$, a function that maps a pair of elements of the set S to a real number, such that.

1) $d(x,y) \ge 0$

1a) $d(x,y) = 0 \iff x = y$

2) $d(x, y) = d(y,x)$ for all $x,y \in S$

3) $d(x,z) \le d(x,y) + d(y,z)$.

Now this is an abstract definition. S could be a set of singing frogs and d could measure when their songs first start to differ, for all we know. (Assuming all songs start on the same note and all songs are eventually different.)

Now the fact that we call this function a "distance" should be a hint to what we have in mind.

So the question is (I'm assuming): prove that in Euclidean plane geometry the actual common sense use of distance between points fits the abstract definition of a metric.

The only problem is, that in our high school geometry class, we can't "prove" these statements about points; we were given them as axioms.

So:

1) Prove $d(P,Q) \ge 0$. Well this is one of Euclids basic postulates (the 1st, I believe) that between any two points there is a "length". In other words there is a distance between any two points; no two points can't be measured, distances are positive real numbers, distances between two points are consistant-- one day it won't suddenly be that New York is suddenly two and half miles from San Francisco; etc. But we can't prove these. This is a basic axiom.

2) Prove $d(P,Q) = 0 \iff P = Q$: Well, any two different points are always some positive distance apart. A point itself is no distance from itself. (Again, an axiom. As in ... well, it doesn't make sense otherwise..)

3) Prove $d(P,Q) = d(Q,P)$: Distance is measured between endpoints. It doesn't matter which end point we measure from. The is Euclid's definition of distance. It is defined as being between a pair; not an ordered pair-- an unordered pair.

At least I'm assuming that is what the question means.

fleablood
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