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June 30

Geometry

Two lines A and B are situated transversal to a line C. If the interior angles formed by the intersection of A and B with C lines are summed to less than 90 degrees, then AB and CD will intersect at some other point in Euclidean space.

Correct?

Then it follows that for a line D there is only one line through a point x not on D that exists parallel to D since there is only one summed angle (90 degrees) of the aforementioned statement that supports two lines that do not intersect.

Euclid and Playfair's equivalent axioms respectively, I was asked to demonstrate that one implies the other in Roger Penrose's book "The Road To Reality" and I was wondering if the above demonstration was adequate? Could you show me some elegant proofs? I am not a mathematician. Dogmorfmocion 18:49, 30 June 2007 (UTC)[reply]

I'm not familiar enough with the phrasings of the axioms to answer your question (though what you have looks right), but I wanted to point out that you probably want to say "the interior angles sum to less than 180 degrees". You may also want to consider the case when the angles sum to more than 180 degrees as well. Tesseran 19:09, 30 June 2007 (UTC)[reply]

180 is right, I completely misinterpreted that. I was under the assumption that the text meant "interior" to mean the angles that faced in; that is, if on the other side, the two angles summed to greater than 180, we would consider the interior angles to be on the otherside and therefore less than 180. Hurf Dogmorfmocion 01:31, 1 July 2007 (UTC)[reply]

OK, let me give this a shot. I'm going to switch notation a bit here and use lowercase letters for lines and uppercase letters for points (which is a bit more standard). To show that the two statements are equivalent we need to show that Euclid implies Playfair and that Playfair implies Euclid. So there are actually two proofs.

Euclid

\to

Playfair. We have a line 'l' intersecting line 'm' and a point 'A' not on 'l' or 'm'. If we draw a line 'n' through 'A' which intersects, we know that the interior angles must sum to less than 180°. What happens if the interior angles sum to more than 180°? (Draw a diagram, use the basic axioms about supplementary and opposite angles.) So what possibility is there for a line to not intersect? (Use the contrapositive of Euclid). (Be careful about what

\lnot (x>y)

is).

I'll let you see if you can do Playfair

\to

Euclid on your own.

It's also worth noting that when you draw your diagrams that you really have multiple possibilities for what you can draw. Looking at the angles of intersection on the top of the line left to right, you might go actute-obtuse-acute-obtuse (with parallel and non-parallel cases), acute-obtuse-right-right, acute-obtuse-obtuse-acute, right-right-acute-obtuse, etc. It kind of sounds like you were only envisioning the obtuse-acute-acute-obtuse possibility. Donald Hosek 16:54, 4 July 2007 (UTC)[reply]

Martingale

After my Markov Chain professor alluded to Martingales as some far-off piece of mathematics (as my primary school teachers once alluded to calculus, etc. etc.) I finally looked them up, and they just seem to be a special type of Markov Chain with some restrictions based on the expectation of X_n+1 yada yada. Is that all there is to it? Rawling4851 23:05, 30 June 2007 (UTC)[reply]

I assume that you have found the articles martingale (probability theory) and Markov chain? As I understand it, a martingale is not necessarily a Markov chain, and a Markov chain is not necessarily a martingale. A martingale is a process X(t) where the expectation value of X(t₂), given observations up to time t₁<t₂, equals x(t₁). A Markov chain is a process X(t) where the entire probability distribution of X(t₂), given observations up to time t₁<t₂, depends only on x(t₁), not on earlier observations. Thus, for example, a Gaussian process for which X(t₂)) has an expectation value that equals x(t₁) but a standard deviation that also depends on earlier observations would be a martingale but not a Markov chain. Conversely, a Gaussian process for which X(t₂)) has an expectation value that equals, say, 7 x(t₁)² and a unit standard deviation would be a Markov chain but not a martingale. --mglg_(talk) 00:37, 1 July 2007 (UTC)[reply]

Aah yes, I see the finer distinction now! Many thanks :-) 213.208.94.205 11:53, 2 July 2007 (UTC) (Rawling)[reply]