25 Jan '04 04:31>
Proving that sqrt(2) is irrational is considered a high achievement of the Greeks, and rightly so. This is amusing:
http://www.gutenberg.net/etext95/meno210.txt
http://www.gutenberg.net/etext95/meno210.txt
Originally posted by royalchickenIndeed. A fine demonstration of the methods of exhaustion and contradiction. Throw in an inductive proof that there are infinitely many primes, and you can demonstrate the basics of virtually any mathematical argument. When was the book, from which this story is taken, written? Is it meant to be in any way inspired by the real-life Socrates, or by Plato's character Socrates?
Proving that sqrt(2) is irrational is considered a high achievement of the Greeks, and rightly so. This is amusing:
http://www.gutenberg.net/etext95/meno210.txt
Originally posted by AcolyteI leave the prime proof for the reader (writing sentences like that is apparently how to distinguish oneself as a true mathematician 😛) because it is in Euclid. However, have you seen Euler's proof of the same statement? It did give us the original connection between the zeta function and primes, but it seems a bit like overkill to me.
Indeed. A fine demonstration of the methods of exhaustion and contradiction. Throw in an inductive proof that there are infinitely many primes, and you can demonstrate the basics of virtually any mathematical argument. When was the book, from which this story is taken, written? Is it meant to be in any way inspired by the real-life Socrates, or by Plato's character Socrates?
Originally posted by BarefootChessPlayerRight on--the algebraic numbers are countable. Can you give some indication why?
a side light:
for those who didn't know, sqrt(2) belongs to the family of algebraic irrational numbers as opposed to transcendntal.
did you know that the latter class is the "larger" of the two?
btw, i've seen the proof for the irrationality of 2^0.5.
Originally posted by AcolyteErm, Lindemann proved the transcendence of pi in the 1880s, so you can put away the ruler and compass and generally get to bed earlier 😀.
On a similar note: I've seen a proof that e is transcendental, but no proofs that pi is transcendental, or even that it is irrational. Given how long people spent trying to 'square the circle', it doesn't sound like an easy matter. Any suggestions?
Originally posted by royalchickenI don't know where you pulled the last paragraph from, but finding roots of polynomials using power series sounds interesting, it it is possible. Mind you, consider this: the power series for log(1+x) has radius of convergence 1, but the limit of its coefficients is zero; also, while it does have a discontinuity at -1, it doesn't do anything funny at 1, unlike its power-series analogue, nor does it misbehave at 0.
Erm, Lindemann proved the transcendence of pi in the 1880s, so you can put away the ruler and compass and generally get to bed earlier 😀.
FWIW, I proved the transcendence of e this past autumn, and I suspect it is much harder to do for pi, as I still am not clever enough to do it. However, if you go to my site, there is an excerpt in which I pose a ...[text shortened]... then use this to show that there is a natural number between 0 and 1, which does the trick.
Originally posted by AcolyteMind you, consider this: the power series for log(1+x) has radius of convergence 1, but the limit of its coefficients is zero; also, while it does have a discontinuity at -1, it doesn't do anything funny at 1, unlike its power-series analogue, nor does it misbehave at 0.
Originally posted by royalchickenx - x^2/2 + x^3/3 - ... is the real powers series of log(x+1). What I had before was the power series of 1-log(x+1), although I don't know why I put that. But it doesn't matter, becasue the limit of the ratio of the coefficients is still -1. Additionally, this power series doesn't do anything funny at x = 1 at all; it converges to log 2, as expected. The real problem is at -1, where the series converges to -log 2, but the function itself is undefined.
[b] Mind you, consider this: the power series for log(1+x) has radius of convergence 1, but the limit of its coefficients is zero; also, while it does have a discontinuity at -1, it doesn't do anything funny at 1, unlike its po ...[text shortened]... dence using this method requires that one have all such sequences.
Originally posted by royalchickenOops, my bad. Just goes to show I hadn't read your work very carefully. Still the point remains that log(1+x) doesn't misbehave at 1, so you can't say that r is a root because it lies on the frontier.
[b] Mind you, consider this: the power series for log(1+x) has radius of convergence 1, but the limit of its coefficients is zero; also, while it does have a discontinuity at -1, it doesn't do anything funny at 1, unlike its power-series analogue, nor does it misbehave at 0.
However, the limit of the ratio of consecutive coefficients is w ...[text shortened]... t mean that to decide transcendence using this method requires that one have all such sequences.[/b]
Originally posted by AcolyteStill the point remains that log(1+x) doesn't misbehave at 1, so you can't say that r is a root because it lies on the frontier.
Originally posted by royalchickenNo, I admit I haven't thought of a counter-example (and your statement may well be correct), I was just trying to suggest that anything involving what happens on the radius of convergence needs to be handled with care.
[b]Still the point remains that log(1+x) doesn't misbehave at 1, so you can't say that r is a root because it lies on the frontier.
The theorem, however, says that if r is the limiting value of the ratio of the coefficients, then r is a root of the denominator of a rational function represented by the power series. So the log(x+1) bit isn't r ...[text shortened]... way of systemically ruling out all possible sequences?[/b]
It is and, unfortunately, no 😳. [/b]