# Root 2 is Irrational...

royalchicken
Posers and Puzzles 25 Jan '04 04:31
1. royalchicken
CHAOS GHOST!!!
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
2. Acolyte
25 Jan '04 18:31
Originally posted by royalchicken
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
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?
3. royalchicken
CHAOS GHOST!!!
25 Jan '04 23:17
Originally posted by Acolyte
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?
I leave the prime proof for the reader (writing sentences like that is apparently how to distinguish oneself as a true mathematician ðŸ˜›) 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.

It's called 'Meno II', which I think is Plato.

4. 25 Jan '04 23:25
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 &quot;larger&quot; of the two?
btw, i've seen the proof for the irrationality of 2^0.5.
5. royalchicken
CHAOS GHOST!!!
25 Jan '04 23:33
Originally posted by BarefootChessPlayer
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.
Right on--the algebraic numbers are countable. Can you give some indication why?
6. Acolyte
26 Jan '04 01:21
Originally posted by royalchicken
Right on--the algebraic numbers are countable. Can you give some indication why?
I can, but then I really ought to be able to, as I'm studying maths. Perhaps this should go in a new thread as a puzzle/'exercise for the reader'.
7. Acolyte
26 Jan '04 01:25
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?
8. royalchicken
CHAOS GHOST!!!
26 Jan '04 01:531 edit
Originally posted by Acolyte
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?
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 ðŸ˜€.

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 another problem, that I've been totally unable to solve, related to the transcendence of e. If you've got some time, I bet you're equal to the task of either solving it or showing it is worthless.

In any case, to prove transcendence first assume that there is some algebraic equation satisfied by the number in question, then use this to show that there is a natural number between 0 and 1, which does the trick.
9. royalchicken
CHAOS GHOST!!!
26 Jan '04 01:53
Originally posted by Acolyte
I can, but then I really ought to be able to, as I'm studying maths. Perhaps this should go in a new thread as a puzzle/'exercise for the reader'.
This was not directed at you. Sorry.
10. Acolyte
26 Jan '04 08:46
Originally posted by royalchicken
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 ðŸ˜€.

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.
I 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.
11. royalchicken
CHAOS GHOST!!!
26 Jan '04 19:51
Originally posted by Acolyte
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 what the theorem is about. Where the series converges:

log(1+x) = 1 - x + x^2/2 - x^3/3 + ...

The nth coeficient is (-1)^n/n, so the limiting ratio of successive coeficients is -1, which lies on the circle of convergence, so I don't know why the italicized bit in your post is relevant.

I don't know where you pulled the last paragraph from, but finding roots of polynomials using power series sounds interesting, it it is possible.

Well yes, the last paragraph of the problem contains an error. The number is transcendental if every possible sequence of which it is the limit fails to produce such a series. Also, my first sentence is very misleading; I didn't mean that having the sequence implies transcendence, I just mean that to decide transcendence using this method requires that one have all such sequences.
12. royalchicken
CHAOS GHOST!!!
26 Jan '04 21:364 edits
Originally posted by royalchicken
[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.
x - 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.

I don't know about it misbehaving at 0 either. The series is just 0 here, and log (0+1) = 0.
13. Acolyte
26 Jan '04 22:04
Originally posted by royalchicken
[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]
Oops, 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.

I see what you mean by the last paragraph now, though it still sounds quite ambitious. Do you have a way of systemically ruling out all possible sequences?
14. royalchicken
CHAOS GHOST!!!
26 Jan '04 22:17
Originally posted by Acolyte
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 relevant to that (note that the theorem discusses only the power series representations of rational functions).

However, your example does mean I need to be much more rigourous in the bit starting ''In addition to this result...'', namely that I need to account for source of the error in the last paragraph.

I have a nagging feeling that this is not what you're driving at and I'm not understanding though...

I see what you mean by the last paragraph now, though it still sounds quite ambitious. Do you have a way of systemically ruling out all possible sequences?

It is and, unfortunately, no ðŸ˜³.
15. Acolyte
26 Jan '04 23:28
Originally posted by royalchicken
[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 ðŸ˜³. [/b]
No, 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.

Have you worked out the answer to 'How smooth?' yet?