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The attempted proof:

Given $P$ not on line $l$, line $PQ$ perpendicular to $l$ at $Q$, line $m$ perpendicular to $PQ$ at $P$ and point $A \neq P$ on $m$. Then, let $PB$ be the last ray between rays $PA$ and $PQ$ that intersects $l$, $B$ being the point of intersection. There exists a point $C$ on $l$ such that $Q * B * C$ (B is between $Q,C$). It follows that the ray $PB$ is not the last ray between rays $PA$ and $PQ$ that intersects $l$, and hence all rays between $PA$ and $PQ$ meet $l$. Thus $m$ is the only parallel to $l$ through $P$.

I can't seem to find it. I'm not sure if it is wrong because he is starting off with right angles, or something else. I feel that because he starts off with right angles, and this clearly only works for right angles, he is basically stating the parallel postulate for all the rays between the two lines as reasons for those not working. By stating and using the parallel postulate, he is being inconsistent. Am I on the right track?

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I'm trying to parse "last ray (??) between...", but I honestly can't: I've no idea what this means. –  DonAntonio Nov 18 '13 at 18:10
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@Don: I assume he means something that is equivalent to "Every ray between $PB$ and $PQ$ (inclusive) intersects $\ell$, and every ray between the angle $PB$ and $PA$ (exclusive) does not. –  Hurkyl Nov 18 '13 at 18:14
    
A diagram would help a lot for this question. –  Greg Hill Nov 18 '13 at 20:33

2 Answers 2

up vote 4 down vote accepted

"$PB$ is not the last ray between rays $PA$ and $PQ$ that intersects $l$, and hence all rays between $PA$ and $PQ$ meet $l$"

The "hence" here is asserted without proof. In fact, it does not follow. All we have shown is that given any point $B$ there is a point $C$ beyond it.

For example, given any real number $b$ less than 2, there is another real number less than 2 but greater than $b$. Does it follow that all real numbers are less than 2?

Similarly, given any ray through $P$ intersecting $l$ at $B$, we have shown there is another ray through $P$ intersecting $l$ at $C$. Logically, it does not follow that all rays therefore intersect $l$.

There's not need for a counterexample; we just need to show the reasoning is flawed.

Like many of these attempted proofs that popped up throughout history, it's just a clever way to subtly hide the assumption of the parallel postulate by cloaking it in "geometric intuition". The "hence" is an appeal to the imagination; it is hard to imagine that such a ray between $PA$ and $PQ$ can exist without intersecting $l$, but that's not proof that it doesn't exist.

Edit: Ironically, people are downvoting my answer because they fell into the same trap Gergonne fell into. These would probably be the people (many of whom were otherwise skilled mathematicians) who accepted Gergonne's proof back in the day. What Gergonne proves by contradiction is that there is no last ray between $PA$ and $PQ$ that intersects $l$. That is not the same as saying there is no ray between $PA$ and $PQ$ that does not intersect l. They are not the same thing. With a little more structure you can think of it like this: he proves that the set of all the angles between $PQ$ and the rays that intersect $l$ has no greatest element. Logically it's still possible that it's bounded from above by some angle less than a right angle, so he hasn't proved what he set out to prove.

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Would you mind showing me a counterexample? I understand it does not follow. I suppose that is the crux that I shall elaborate on. –  David Nov 18 '13 at 18:17
    
But can't we show that if two rays emanating from $P$ intersect $l$, then all rays between them also intersect $l$? –  Keshav Srinivasan Nov 18 '13 at 18:45
    
The part you object to is actually perfectly correct. The proof sketch made an assumption that there was a ray between PA and PQ that doesn't intersect $\ell$, and from there it derives a contradiction. Thus, it is correct to infer that every ray between PA and PQ intersects $\ell$. The actual problem occurs earlier in the proof, when $PB$ is introduced. –  Hurkyl Nov 18 '13 at 19:51
    
@Hurkyl please read the argument more carefully. The assumption is that there is a LAST ray. do not downvote my answer because you do not understand the question, especially when i upvoted yours –  TBrendle Nov 18 '13 at 19:54
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@Hurkyl No, he doesn't. In that sentence he passes from the proof that there is no last ray to the conclusion that all the rays inside the right angle meet $l$. If you can't keep track of the points please draw a diagram for yourself. I won't clutter the comments further by replying to your made-up objections. I am familiar with the history of this argument and its logical flaws. –  TBrendle Nov 18 '13 at 20:07

There is no "last ray PB"... even in Euclidean geometry.

The boundary between "rays from $P$ that intersects $\ell$" and "rays from $P$ that do not intersect $\ell$" turns out to be a ray that does not intersect $\ell$ affinely. However, it does meet $\ell$ "at infinity".

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Interesting. I agree because we haven't proposed that there is such a thing as "last ray $PB$". If there was, we would have to state such a proposition and prove what it means, or at least take it as a definition. What do you mean by affinitely? –  David Nov 18 '13 at 18:15
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@David: "Affine", here, is the opposite of "at infinity": it means a point of the ordinary plane. –  Hurkyl Nov 18 '13 at 18:17
    
How would such a ray meet $l$ at infinity? –  David Nov 18 '13 at 18:20
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@David: It's clear once you have a definition of "at infinity". :) In projective Euclidean geometry, the points at infinity are defined so that every pair of (ordinary) lines intersects at a unique point. And if you have three parallel lines, they all intersect in the same point. Finally, the projective plane adds one extra line, which contains all of the points at infinity (and no other points). For hyperbolic geometry, the points at infinity are clear if you're familiar with the disk model. In both cases, a keyword if you want to go searching is "ideal point". –  Hurkyl Nov 18 '13 at 18:56
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@Hurkyl I see how it helped you (I was helped by the same intuition), and i agree that it is informative, but it would be circular if offered as a refutation of Gergonne's argument. and by advacned i meant advanced in the development of the theory. Here we are still on basic postulates from Euclid. (This is synthetic geometry, btw) –  TBrendle Nov 18 '13 at 19:45

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