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**TARAJS****Member**- Registered: 2011-12-01
- Posts: 19

If there exists a unique point p sch that p lies on both l and m, then l and m are distinct, non parallel lines.

Prove using Incidence geometry axioms. A full proof would be wonderful.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,307

Proof by contradiction:

If l and m were the same then they would have more than one points that lie on both.

If l and m were parallel or skew, then there would be no point that lies on both lines.

The only option left is that they intersect and are thus distinct and non-parallel.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**TARAJS****Member**- Registered: 2011-12-01
- Posts: 19

ok i need to use IG axioms.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,363

Hi TARAJS;

What are IG axioms?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,294

hi all,

http://www.math.washington.edu/~lee/Cou … idence.pdf

and

http://web.math.isu.edu.tw/yeh/2008spri … cture1.pdf

Axioms of Incidence Geometry

Incidence Axiom 1. For every pair of distinct points P and Q there is exactly one line such that P and Q lie on .

Incidence Axiom 2. For every line there exist at least two distinct points P and Q such that both P and Q lie on .

Incidence Axiom 3. There exist three points that do not all lie on any one line.

Definitions :

meet (intersect) Instead of saying that the intersection of two lines is a nonempty set, we shall simply say that the two lines meet, or pass through a contain point.

lie on When a point belongs to a line or a plane, we say it lies on the line or the plane, or we say that the line or plane passes through the point. When a line is a subset of a plane, we say it lies on the plane.

concurrent Two or more lines or planes that meet at the same point are called concurrent.

collinear Two or more points that lie on the same line are said to be collinear.

determine Two points are said to be determine a line if there exists a unique line passing through them,

and three points determine a plane if there exists a unique plane passing through themparallel Two lines are parallel if they have no points at all in common. (They do not meet.)

To prove:

If there exists a unique point p such that p lies on both l and m, then l and m are distinct, non parallel lines.

If p exists then the lines are not parallel (definition of parallel)

Suppose l and m are not distinct.

=> There exists at least two points on both l and m =><= p is unique.

=> l and m are distinct.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,307

Hi Bob

The definition of parallel lines you quoted is flawed. Two lines are parallel iff they have no common points or have all points in common ie. are the same line.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,294

hi Stefy,

That set of definitions also doesn't properly cover skewed lines.

But, when setting up a system of axioms, you can make any rules you like. incident geometry isn't the same as Euclidean geometry, so that definition may be OK for IG. The point is, you have to work within the rules you are given; not alter them to suit yourself.

Pure mathematicians like to think that a precise, self consistent, set of axioms makes the topic perfectly rigorous, but the definition problem shows the flaw in that thinking. You have to define your terms; and that means using words; and that means you have to define those; which means you have to use more words; which .........

Sometimes the definitions go round in circles.

(a) First law: If an object experiences no net force, then its velocity is constant; the object is either at rest (if its velocity is zero), or it moves in a straight line with constant speed (if its velocity is nonzero).

So, what is a force?

(b) A force is that which can cause an object with mass to change its velocity

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,307

Hi Bob

We were doing axiomatic geometry in the 9th grade. We always said that two lines are parallel iff they are the same or have no common points, but lie in the same plane. If it wasn't for that, then we would lose some interesting properties of parallelism. It wouldn't be an equivalence relation. We would lose the property that any two lines that lie in a common plane and are perpendicular to the same line are parallel. There is more of that. It is like defining x^0=1 for x<>0. It is not necessary for it to hopld, but if it does, some interesting things arise.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,294

Ok.

But it doesn't alter the 'proof' as I 'dismiss' the equality case at the start.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,307

I have nothing against the proof. Just the definitions you quoted.

Here lies the reader who will never open this book. He is forever dead.

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**TARAJS****Member**- Registered: 2011-12-01
- Posts: 19

I know this is super late but thank you so much for your help bob!

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