Denial of the 5th postulate of Euclid

Question

I am trying to recover the denial of the Playfair's axiom but it is logical a bit strange. "To a given line and a point not on it, there is only one line through this point parallel to it". This should become "To a given line and a point not on it, there is more than one line through this point parallel to it."

With the logical negation I never come on the second statement. I tried the following form which I suppose it is wrong.

$\forall l \forall P : P\notin l\exists! m:((P\in m)\wedge (m\parallel l))$

Any advice? thanks :)

Answer 1

Your expression is indeed wrong, since $\forall l\forall P:P\notin l$ means that "no point is on a line". What you want is in fact: $$\forall l\forall P\big(P\notin l\to\exists !m\left(P\in m\land m||l\right)\big)$$ which translates as: "for any line $l$, if a point is not on this line, then there is a unique line $m$ to which this point belongs and which is also parallel to $l$".

Now, for the contradiction of this statement, recall that $\neg\forall x\equiv\exists x\neg$ and $\neg(A\to B)\equiv A\land\neg B$. However, it is slightly more complicated to negate $\exists !x$, and I refer you to the (excellent) accepted answer on this post. The idea is that, to contradict "there exists a unique $x$", we want to say "either there is no $x$ or there is more than one". Thus the negation of the statement is: $$\exists l\exists P\big(P\notin l\land\neg\exists !m(P\in m\land m||l)\big)$$ $$\exists l\exists P\Big(P\notin l\land\forall m\big(P\in m\land m||l\to\exists m'(P\in m'\land m'||l\land m'\neq m)\big)\Big)$$ which is precisely what you want, since it states "there is a line $l$ and point not on it, such that if a line $m$ goes through that point and is parallel to $l$, then there is another such line".

From the point of view of logic, this works if there is no line going through $P$ at all, or if none can be found to be parallel to $l$. But of course, geometrical considerations remove the need for these cases and yield your desired statement of "To a given line and a point not on it, there is more than one line through this point parallel to it".

Answer 2

The logical negation of the parallel postulate is:

There exists a line L and a point P not on L such that $|\{$lines parallel to L passing through P$\}| \neq 1$.

However, the space we are trying to describe should be sufficiently homogenous that this is equivalent to:

Given a line L and a point P not on L, $|\{$lines parallel to L passing through P$\}| \neq 1$.

This is obviously equivalent to:

Given a line L and a point P not on L, either there are no lines through P parallel to L, or there is more than one.

By the same homogeneity argument, we can take the "either" outside:

Either, given a line L and a point P not on L, there are no lines through P parallel to L, or, given a line L and a point P not on L, there is more than one line through P parallel to L.

This gives the two possible "denials" of the parallel postulate - neither is the logical negation, but either contradicts it, which is the sense in which they are denials. (And, as I've shown above, they are in some sense the only such).