1. ## prove the proposition

For every point P, there exists at least two distinct lines through P.

Thanks

2. Originally Posted by mandy123
For every point P, there exists at least two distinct lines through P.
Mandy, Do you realize that every course in axiomatic geometry has a slightly different set of axioms? We do not know what yours are.

That said, you should have axioms somewhat like these.
Axiom: There are three non-collinear points.
Axiom: Given two points there is a line containing the points.

So you could say something like this.
Given a point P, by the first axiom there is another point Q.
By the second axiom there is a line $\displaystyle \ell _1 \;\& \;\left\{ {P,Q} \right\} \subseteq \ell _1$.

Again by the first axiom $\displaystyle \left( {\exists R} \right)\left[ {R \notin \ell _1 } \right]$. Why is that true?

So $\displaystyle P\ne R$. Why is that true?

Can you find a second line $\displaystyle \ell _2 \;,\;\left\{ {P,R} \right\} \subseteq \ell _2 \;\& \;\ell _1 \ne \ell _2$.

3. Sorry about that, my axioms are as follows

Incidence Axiom 1: For every point P and for every point Q not equal to P, there exists a unique line l incident with P and Q

Incidence Axiom 2: For every line l, there exist at least two distinct points incident with l

Incidence Axiom 3: There exist three distinct points with the property that no line is incident with all three of them.

So would the proof be as follows??

Given a point P, by IA3 there is another point Q.
By the IA2 there is a line .

Again by IA3 because no line is incident with all three of them

So because IA1 says there must be a unique line and if P=R there is not two unique lines where P is not equal to R

To find the second line we would go through what we did above, only substituting R in for Q and vice versa right?