Talk:Irrational number

Request

It would be helpful to have a list of irrational numbers, probably as a seperate entry. I have not made this edit as I owuld not be able to go much beyond the obvious, but I am sure there are others that could.

That might be a long list, pal. ;^> DavidCBryant 18:19, 27 November 2006 (UTC)

Moved from article page

I moved this from the main page:

[I regret that I cannot carefully edit this paragraph at this time, but in good conscience I must question whether any responsible historian of mathematics ascribes the argument below to Pythagoras. An algebraic argument for the irrationality of the square root of 2 observes that if √2=m/n, then √2=(2n-m)/(m-n), so that a fraction in lowest terms is reduced to yet lower terms. That is a contradiction, completing the reductio ad

absurdum. To me it is plausible that Pythagoras or someone of his school discovered a geometric argument showing that if n and m are respectively the leg and the hypotenuse of an isosceles right triangle, then m-n and 2n-m are respectively the leg and the hypotenuse of a smaller isosceles right triangle. An ancient Greek geometer would have constructed the smaller triangle from the larger one, rather than doing algebra, as we do today. I would recommend that any mathematician editing this page look at Thomas Heath's translations of the writings of ancient Greek geometers before ascribing anything to Pythagoras.]

and changed the paragraph about Pythagoras's discovery accordingly. AxelBoldt 03:30 Oct 23, 2002 (UTC)

I suspect the Greeks' argument might also have used Euclid's own version of Euclid's algorithm, involving repeated subtraction rather than the division used in today's optimised variant. PML.

About the Irrationality of the squareroot of 2. My math teacher said today that Pythagoras believed that sqrt(2) actually WAS a rational number and that that was a thought that his followers the Pythagoreans also thought. He also said that someone during the Middle Ages proved that sqrt(2) = irrational and that that guy subsequently was murdered. BL 22:58, 16 Sep 2003 (UTC)

That is ignorant nonsense; just look at Euclid's Elements and you will see that irrationality was know to the ancient Greeks. Michael Hardy 22:12, 17 Aug 2004 (UTC)

Euclid was born about a hundred years after pythagoris' death. The proof of the irrationality was discovered by one of pythagoris' followers, but if I remember my reading correctly he was banished, not killed. --Starx 01:09, 18 Aug 2004 (UTC)

In other words, as I said, it is ignorant nonsense to say it was not done until the middle ages. Michael Hardy 01:52, 18 Aug 2004 (UTC)

If Euclid was around after pythagoras, then the fact that he knew of the irrationality of the square root of 2 is not surprising, considering it was during pythagoras' time that it was first proven. Do you have any sources? Cause there are plenty documented sources saying it was, in fact, a follower of pythagoras. --Starx

So you and I both agree that it was known to the Pythagoreans and therefore to Euclid, who came later. And we both agree therefore that it is ignorant nonsense to say that it was not done until the middle ages. Right? As for sources, I've read some of Thomas Heath's books, but it's been a while, so I cannot cite chapter and verse. On another matter, why do you keep deleting my assertion in the article that the conventional algebraic argument is not the one that the Pythagoreans used? Michael Hardy 21:40, 18 Aug 2004 (UTC)

Because everything I've read has said that that was the proof. The only one I can think of off the top of my head is the golden ratio by mario livio, which has quite a bit on the history of math. If you have a better source that says otherwise then I'll concede, but all you've done so far is claim that it's ignorant nonsense. If we both agree that it was known to the pythagoreans, and I'm saying it was the pythagoreans who first discovered it, where do you get the middle ages?? Who brought that up? --Starx 01:10, 19 Aug 2004 (UTC)

I did not "get middle ages"!! That is what I called "ignorant nonsense". I never said that it is "ignorant nonsense" to say that the relatively recent algebraic proof of irrationality is how the Pythagoreans did it. It is not how the Pythagoreans did it; it is how many mathematicians believe (and write) that the Pythagoreans did it; I never said that that error is "ignorant nonsense" -- only that it is an error. Michael Hardy 02:31, 19 Aug 2004 (UTC)

I'm not debating about anything that happened during the middle ages. I'm debating about whether or not the proof displayed on the page was done by one of pythagoras' followers. That's what our recent edits have concerned so I think it would be fairly obvious that that is what the discussion is about. I don't understand why you're still bringing up the comment another user made on the middle ages, that's not the subject of the debate and that's why I want to know where you're getting that from. I'm sorry if I was unclear. I'm asking what referances do you have pertaining to what proof pythagoras used to determine the irrationality of the square root of two. Because I have referances that say that what's displayed is the correct proof. I said this in my above post and I'll say it again: If you have a better source that says otherwise then I'll concede. --Starx 03:41, 19 Aug 2004 (UTC)

I will get the references.

What I called "ignorant nonsense" was the statement about the middle ages. Then you attacked me for calling your statements about the Pythagoreans and Euclid "ignorant nonsense". That's why I brought up the matter of the middle ages. Michael Hardy 18:35, 19 Aug 2004 (UTC)

Ok, to add my understanding (based on lecture notes given to me by a lecturer of Mathematical History). Firstly, it wasn't irrational numbers which were discovered, but incommensurable magnitudes i.e. a relationship, not a kind of number. The way the Greeks would have described it is roughly as follows;

a:b::c:d "a enjoys a relationship in size with b that is equal in size to the relationship shared by c and d" 

It's looking at it from a very modern viewpoint to see these relationships as actually being irrational numbers if c and d are not both integers, and not at all how the ancients would have viewed it. They wouldn't have thought of these relationships as occupying a space on a number line for example. He then goes on to say that the length of the diagonal compared to the side of a square (i.e. ${\sqrt {(}}2)$ in modern notation) wasn't really talked about until quite far into the 4th century BC. The first relationship found to be incomensurable was probably that of the diagonal of a pentagon in the 5th century BC - not much earlier than 410-420 BC (based on research by Wilbur Knorr). He also mentions that it wasn't really until the late 16th century AD that what we'd now call an irrational number was beginning to be discussed properly Richard B 00:06, 2 December 2005 (UTC)

Complicated proof?

Isn't the first proof for the irrationality of ${\sqrt {2}}$ overly complicated? It basically states that when you transform $\left({\frac {p}{q}}\right)^{2}=2$ to $p^{2}=2q^{2}$, the multiplicity of prime factor 2 is even on the left side, and odd on the right side -> contradiction.

Aragorn2 21:00, 17 Sep 2003 (UTC)

No, because the proof builds on other proofs that has to be explicitly stated. Like that the square of an even number also is even. As it is on the page is how my math teacher described it. BL 21:27, 26 Sep 2003 (UTC)

Aragorn, you're assuming that a number has only one prime factorization. But that's much harder to prove than the special case that says the product of two odd numbers is odd, which is all that this proof needs. Michael Hardy 21:56, 13 October 2006 (UTC)

While that's all the proof needs, it is needlessly wordy thus obscuring its essential simplicity. My hope in tightening it up as I did just now is that it is easier to read than the far longer version it replaced without being any less accessible. --Vaughan Pratt (talk) 02:38, 17 March 2008 (UTC)

---

The recent posting on the history is directly taken from Article 3 of a 1906 book at www.gutenberg.net/etext05/hsmmt10p.pdf .

I'll leave it there for the present; but in any case it would need a thorough edit.

Charles Matthews 16:50, 29 Jan 2004 (UTC)

---

BL: a root of a natural number m (i.e. a positive/non-negative integer) is either a natural number or an irrational: Suppose we are looking at m^(1/n) and this was a/b (i.e. rational with a,b integers), so a^n=m*b^n. Then write m in terms of a product of powers of prime numbers (m=p^x * q^y * r^z * ...). Do the same with a and b, and then match exponents on each side.

If all of x,y,z,... are multiples of n, we will be able to take the n-th root of m and get a natural number. If any of them are not, then we will not even be able to get a rational number because the LHS of a^n=m*b^n will be a product of powers of primes where all the exponents are multiples of n while the RHS will not be, which based on the fundamental theorem of arithmetic leads to a contraction of the hypothesis that m^(1/n) is rational. --Henrygb 23:28, 13 Feb 2004 (UTC)

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The irrational numbers are precisely those numbers whose decimal expansion never ends and never enters a periodic pattern

I know that is true but there is no need to invoke decimal when describing irrational numbers. I have witnessed confusion when irrational numbers are defined thus. People think that the set of irrational numbers are different in base-2 than they are in base-10 because of definitions like that. Paul Beardsell 05:03, 20 Feb 2004 (UTC)

Thank you, Paul. I think you just answered a question of mine before I even got around to asking it. To be sure though, are some (or all) irrational numbers simply artifacts of the decimal system? That is, could a number which is irrational in base 10 be expressed rationally in, for example, base 9 or base 17? --Zaklog 05:56, 21 March 2006 (UTC)

No. Irrational numbers are irrational in every natural base. The definition of irrational number has nothing to do with how this number is written by humans.  Grue  06:39, 21 March 2006 (UTC)

The relevant theorem here is, for all integers r,s ≥ 2 and real x, x is ultimately periodic in radix r if and only if it is ultimately periodic in radix s. Vaughan Pratt 03:05, 7 May 2006 (UTC)

---

I didn't really like this line in the proof: "Since a:b is in its lowest terms, b must be odd."
Can't it be replaced with "Assume b is odd?" —Preceding unsigned comment added by 76.172.43.73 (talk) 07:27, 12 June 2008 (UTC)

Grammar

From the article: (because none of its prime factors is 2) Factors is plural, so shouldn't it be are instead of is? --Starx 01:51, 20 Dec 2004 (UTC)

No. "Its factors" is the object of the preposition "of". If I wrote "Not even one of its factors is prime", obviously it would be grossly wrong to write "are". Similarly if I wrote "Just one of these factors is prime", would you say I should have written "are", when I'm writing about only one, on the grounds that "factors" is plural? Traditionally, "none" is singular. Of course, recently many people have used "none" as plural, but even so, there can hardly be a grammatical objection to using a singular "none". (And somehow the misspelling of "grammar" in the edit summary doesn't inspire confidence either.) Michael Hardy 23:24, 20 Dec 2004 (UTC)

... and also, when you say "because factors is plural", I almost fear that next you'll write something like "One of these are correct". I actually hear people say that from time to time; it's as if the fact that these is plural means that the phrase one of these is plural. Obviously the phrase one of these is singular and should be followed by is, not are. Michael Hardy 23:57, 20 Dec 2004 (UTC)

• It's nice to see you take an honest question and be a dick about answering it. You sounded like a decent human being right up until the parenthetical remark in your first responce. But that one remark wasn't enough, you had to go back for a second responce. Kudos. --Starx 06:16, 21 Dec 2004 (UTC)

I am in fact a decent human being. And I stand by what I wrote above: misspelling grammar two times running doesn't inspire confidence; it may be useful for you to know that. If you disagree with that or any of the other points above, you could argue the point instead of engaging in name-calling. What, specifically, do you object to in the second response? Writing "One of these are correct" is in fact grammatically parallel to the usage you raised a question about. Michael Hardy 21:44, 21 Dec 2004 (UTC)

It may be grammatically parallel in a technical sense, but the original case is far more obscure. Your example states one of these factors, it's obviously singular. The snippet from the article isn't so clear. In either event you very nicely explained things to me and should have stopped there. But instead you chose to make remarks about my spelling not inspiring confidence and how you "fear" I'll do something even stupider. I stand by what I wrote above, you're being a dick for no reason I can see other then possibly a superiority complex. --Starx 22:44, 21 Dec 2004 (UTC)

In discussions of politics or scientific controversies a rhetorical device such as "Since you're advocating X's theory, next I expect you'll be saying the Big Bang didn't happen" is not generally construed literally; people aren't so touchy. But when the topic is grammar, it seems they are. I don't understand why the difference. Let me rephrase my comment that was found offensive. Originally I wrote:

... and also, when you say "because factors is plural", I almost fear that next you'll write something like "One of these are correct". I actually hear people say that from time to time; it's as if the fact that these is plural means that the phrase one of these is plural. Obviously the phrase one of these is singular and should be followed by is, not are.

Here is a rephrasing:

... and also, the phrase "because factors is plural", is syntactically parallel to, "One of these are correct". I actually hear people say that from time to time; it's as if the fact that these is plural means that the phrase one of these is plural. Obviously the phrase one of these is singular and should be followed by is, not are.

If I had not thought that was obviously what was meant, I would have phrased it in that literal way originally. Michael Hardy 23:22, 30 Dec 2004 (UTC)

Repitend thing

Michael Hardy wrote:

I think this should say either "the period is 3" or "the length of the repitend is 3", but NOT "the length of the period is 3".

You are right. My spell-checker gave me "repitend" as an option. I should have looked up a dictionary and confirm this is correct. I instead chose to replace it with "period" assuming it will be the same thing. I would actually appreciate a bit of clarification here, if it would not take too long. Oleg Alexandrov 02:51, 4 Apr 2005 (UTC)

Actually, I think I figured it out myself. All it took is reading what you wrote, and actually thinking about it. Thanks! Oleg Alexandrov 02:56, 4 Apr 2005 (UTC)

I would appreciate in general more feedback with my spelling. And sorry for "polluting" your watchlist, I believe quite a bit of my bot changes show up there. Oleg Alexandrov 03:04, 4 Apr 2005 (UTC)

I believe that the mathematical notation of the set of irrational numbers is not mentioned anywhere in the text, although it is quite obviously under this title, is it not?

I feel rather foolish for asking this...

...but I can't figure out the logic behind the statement, "if √2=m/n, then √2=(2n-m)/(m-n)." Can someone derive that, or point me to another site that has the derivation? --Jay (Histrion) 16:50, 26 October 2005 (UTC)

One way is by algebra: If √2 = m/n, then n√2 = m, and then we have

${2n-m \over m-n}={2n-n{\sqrt {2}} \over n{\sqrt {2}}-n}={n{\sqrt {2}}({\sqrt {2}}-1) \over n({\sqrt {2}}-1)}={\sqrt {2}}.$

Well, I feel silly now — a straightforward substitution. I was trying to derive the expression, when I could have just simplified it. As Strong Bad might say, "Holy crap!" --Jay (Histrion) 20:17, 26 October 2005 (UTC)

Another way is by geometry: if m is the diagonal, and n the side, of an isosceles right triangle, then by a simple ruler-and-compass construction one creates a smaller isosceles right triangle in which the the respective lengths of diagonal and side are 2n − m and m − n. Someone's added a diagram showing this to the article titled square root of 2. Michael Hardy 18:59, 26 October 2005 (UTC)

Says we have a fixed point of the mapping x → (-x + 2)/(x - 1). Well, easy to check the fixed points are √2 and −√2. Charles Matthews 19:32, 26 October 2005 (UTC)

Actually, that brings up a good question -- how much of the material in this entry is repeated in that √2 entry? Should we look at merging them -- or, alternately, moving some text from this entry to the other? -- written and unsigned at 13:21, 26 October 2005 by Histrion

I would oppose the merger. The main article for irrational numbers is this one. As for the square root of two, it is an important enough example of irrational numbers to keep its own article. Oleg Alexandrov (talk) 23:28, 26 October 2005 (UTC)

(Whoops, yeah, that last one was me. Thanks, Oleg.) OK, then is there any stuff in this article that might be best moved to square root of two (or cut from this article because it's duplicated in that one, like the algebra proof that √2 is irrational)? --Jay (Histrion) (talk • contribs) 00:03, 27 October 2005 (UTC)

Why do you like to be so efficient? :) I think square root of two was the first real number to be proved to be irrational. The proof is also very simple. It makes sense to me that it would show up in both articles, it greatly clarifies (in my view) what irrationals are all about. Oleg Alexandrov (talk) 02:47, 27 October 2005 (UTC)

Efficiency? Can't help it. Copy editor for my high school newspaper, and I tutor at the local community college, so I proofread papers now and then. Proofread enough beginning writers, you start to look for redundancy. Anyway, I dunno, I just think it might be better if, rather than having the entire proof in both entries, one of them just pointed to that subsection of the other. But I'm not feeling bold enough to do it myself. Maybe tomorrow, when I've slept some. ;) --Jay (Histrion) (talk • contribs) 03:57, 28 October 2005 (UTC)

I read both Irrational number and Square root of 2 now. I would say that there are too many proofs of irrationality of square root of two at irrational number. I would agree with removing all proofs of that except for the first and referring for more detail to Square root of 2. I would disagree with removing all the proofs of irrationality of square root of two from there. I believe that proof is important enough in illuminating the article that it better be inline rather than referring the reader to a different article. Oleg Alexandrov (talk) 05:18, 28 October 2005 (UTC)

Drowning of Hippasus

Can someone provide a cite for "he [Pythagoras] sentenced Hippasus to death by drowning."? The Wikipedia article on Hippasus calls the story a rumor, other sites use the word "legend". Others say that Hippasus died accidentally and the Pythagoreans were guilty only of tactless amusement at the fact. About.com says that there are many legends, and no one knows for sure. In any case, I haven't seen any claims that Hippasus was "sentenced", in the sense of receiving some kind of process.

Here is my version of the accident. Pythagoras was the smartest guy of the bunch, so it was him who proved the irrationality of sqrt(2). He told his disciples and Hippasus snitched the result to general public. That's why he was killed.  Grue  06:57, 27 May 2006 (UTC)

I reworded that part, to keep the story and at the same time make clear it's just that - a story. And also that Pythagoras is only :-) a mathematician and is in no position to sentence anybody. Haonhien 15:53, 22 September 2006 (UTC)

I changed "drowned to death" to simply "drowned", since the former is really quite tautologous. You drown, you die. :-) 81.159.229.181 10:39, 24 September 2006 (UTC)

Pythagoras was not only a mathematician, he was the founder of a fanatical religious cult mixed with science. Maybe he did it as a joke, then got rich, and couldn't stop the joke. Oops, that's another guy.13:05, 23 October 2006 (UTC)

Galois theory

The following I removed on the grounds that it is irrelevant and sometimes erroneous:

Paolo Ruffini (1799) first proof, (largely ignored) of Abel–Ruffini theorem that the general quintic or higher equations cannot be solved by a general formula involving only arithmetical operations and roots. Évariste Galois (1831) sends a memoir to the French Academy of Science: On the condition of solvability of equations by radicals, later developed into Galois theory which has been central to the proof that π and e are transcendental. Joseph Liouville (1840) showed that neither e nor e² can be a root of an integral quadratic equation. Niels Henrik Abel (1842) partially proves the Abel–Ruffini theorem. Gene Ward Smith 00:36, 13 July 2006 (UTC)

Maybe the removal of Galois Theory left a hole in the History part; sometime during Math History people thought that every irrational number was expressible by radicals, then transcendals were found, then the unsolvability of the 5th degree polynomial was found. Albmont 13:11, 23 October 2006 (UTC)

Decimal expansions

In the numerical example, the old version ends with this:

$A={\frac {715.5}{999}}={\frac {7155}{9990}}={\frac {135\times 53}{135\times 74}}={\frac {53}{74}}.$

(The "135" above can be found quickly via Euclid's algorithm.)

I omitted the reduction of the fraction and also the reference to Euclid's algorithm, because the proof ends when we have a fraction - any fraction; and also because introducing the new concept of Euclid's algorithm unduly complicates what we're discussing here, and that is only about the repeating/terminating decimal expansion. (If you disagree you can copy the above back into the text.) Haonhien 15:46, 22 September 2006 (UTC)

Formula in the lead

I don't understand. Why is it so important that it should be on its own line? It's not a definition of irrational number, in fact it's exact opposite. If there should be a formula, it should be something like that: $x\in \mathbf {I} \iff \not \exists a,b\in \mathbf {Z} :x={\frac {a}{b}}$. I can very well imagine a situation when someone wants to look up a quick definition of irrational number, sees a huge ${\frac {a}{b}}$ in the lead paragraph and then walks away thinking that he got it.  Grue  20:22, 13 October 2006 (UTC)

"Irrational number" is inherently a negatively defined concept: an irrational number is a real number that is not rational. And if you can't understand why the formula you propose to display is inferior to expressing the same thing in words, then you'll probably never be any good at expressing things in words. It is absurd to think anyone will see the fraction and think they've got it---they'd need to read the words "where a and b are integers", and by then they'd have read the preceeding words too. What audience are you contemplating here? Those who readily understand such symbols as $\not \exists a,b\in \mathbf {Z} :x={\frac {a}{b}}$ will not be easily confused, and those who don't already know what an irrational number is will not understand such a thing as what you propose. And look: your proposed displayed formula requires the reader to understand such things as sets and membership, which is subtler and less basic---and obviously FAR newer---than the concept of irrational number. Would you seriously propose that only those who understand such modern (19th-, 20th-, and 21st-century) concepts should be told what irrational numbers (an ancient Greek idea) are? Haven't you heard that the so-called "new-math" pedagogical approach of the 1960s is discredited, for excellent reasons? If not, what planet have you been on for the last four decades? Michael Hardy 21:54, 13 October 2006 (UTC)

"And if you can't understand why the formula you propose to display is inferior to expressing the same thing in words..." Huh? I was in fact arguing for making the formula inline or removing it altogether (replacing it with words). There is absolutely no reason for that formula to stay on its own line. It doesn't define irrational number. My formula does, but it's "new math" so you don't like it. That's fine with me, as long as the wrong formula is not displayed in such a prominent place.  Grue  07:30, 14 October 2006 (UTC)

I agree with Michael here. The formula does not define the irrationals, but comes as close as one can to that. Oleg Alexandrov (talk) 17:53, 14 October 2006 (UTC)

It comes as far as one can get from the formula defining the irrational numbers. It defines rational numbers, which are exactly opposite (wouldn't have thought I'll have to explain that to the two mathematicians!)  Grue  11:57, 15 October 2006 (UTC)

But the concept of irrational number is defined by saying what it is the opposite of! Michael Hardy 00:06, 16 October 2006 (UTC)

Definition independant of a real number

I think we should find a definition that is independant of the definition of a real number - Since the definition of a real number depends on the definition of an irrational number. The current definition is circular, but according to the editors that reverted my edit, the second paragraph in the intro can't be used as a definition. We need a better one anyway. Fresheneesz 21:50, 25 October 2006 (UTC)

A reasonable definition of real number does not depend on the concept of irrational number. A reasonable definition of irrational number does depend on the concept of real number: an irrational number is any real number that is not a rational number. The suggested circularity needs to be avoided by giving a good definition of "real number"---one that does not presuppose knowledge of what an irrational number is. Michael Hardy 02:34, 26 October 2006 (UTC)

PS: the correct spelling is "independent". Michael Hardy 02:34, 26 October 2006 (UTC)

SHOULD THIS BE INCLUDED ?

Someone has removed [this weblink to the proof that Richard Palais relates] from this article because (he says) : "I think that this inline reference to an external proof is not as helpful, and kind of distracting." Where could it be included? Should it go on another page? It is the simplest and nicest proof by descent I have ever seen. Robert2957 16:21, 27 October 2006 (UTC)

Possibly in the "External links" section of this article, or that of the article titled square root of 2. But it seems you need to fix the URL: it should be <http://www.mathacademy.com/pr/prime/articles/irr2>. Michael Hardy 19:35, 27 October 2006 (UTC)

Would it be possible to reproduce a version of this proof in the article without violating copyright ? Robert2957 20:33, 27 October 2006 (UTC)

It should be possible, but credit should be given. There is a question of whether that proof is actually novel. Michael Hardy 22:37, 27 October 2006 (UTC)

General cleanup

Hi, all!

Overall this is a very nice article, but it needs a bit of cleanup (poorly constructed sentences, overuse of the verb "to see", too many id ests, etc). I've put that on my schedule of things to do this week, but thought I'd put this note up here first, to give fair notice to those who may have excessive emotional capital invested in the existing verbiage. DavidCBryant 19:36, 27 November 2006 (UTC)

Don't Use The Word Obvious

"Therefore a² is even because it is equal to 2 b² which is obviously even."

How is it obvious? I didn't know it was even. If you say something is obvious and the reader didn't know it, it makes them feel stupid.

^{(b·talk·contribs)} 22:57, 4 December 2006 (UTC)

It's obvious because it's 2 times another integer. That's what "even" means. Michael Hardy 23:46, 4 December 2006 (UTC)

Informal statement in introduction

I had considered your point that you should include m<>0, but have decided to leave it out to make the introduction clearer for non-mathematicians. Here is my reason. It is true that a rational number is of the form n/m with m<>0, and something that is NOT rational is NOT of the form n/m with m<>0. However, the mathematical definition is an irrational number is a REAL number that is NOT RATIONAL. I feel that by adding m<>0, you are clarifying what is meant by a rational number, but there is no mention of what a real number is. So if you were to put "m<>0", a non-mathematician might think "what if m=0?". By leaving out the m<>0 condition, you are defining an "irrational number" as a number that is not of the form n/m for ANY integers. I might not be getting my message across clearly but I hope you understand that I have kept it this way for clarity. AbcXyz 13:06, 4 January 2007 (UTC)

I understand your purpose, but I think your edit makes the text more confusing than illuminating. I will leave it they way you put it, but if somebody else reverts you also, then please don't revert. Let us see what others say. Oleg Alexandrov (talk) 16:42, 4 January 2007 (UTC)

`The set of all irrationals' is badly written

I find these last two paragraphs to be badly written and obscurely organized.

1) Better would be: "Since the rationals are countable and the reals uncountable, the irrationals are uncountable."

2) This does not seem to be the right place to mention the uncountability of transcendental reals.

3) "form a metric space" is better than "become a metric space"

4) "a homeomorphism", not "the homeomorphism" -- there is more than one!

5) "This shows that the Baire category theorem applies to the space of irrational numbers." But this was not in doubt: the complement of a countable set of closed points in a Baire space is, immediately from the definition, again a Baire space.

6) Rather than "Whereas..." why not just say, if necessary, that the space of irrationals is totally disconnected.

7) "If removing the rationals from the continuum...one might imagine that...would connect it even better than with one copy." This is a horrible sentence: I am a mathematician and I can't quite parse it. (What is "it"?) Ditto the following sentence: "just as totally disconnected"?!?

The paragraph doesn't get any better from here, and I gave up. The sentiments expressed here, if they can be written so as to make sense (even) to a mathematical audience, would be more suitable on a topology page. 22:51, 27 January 2007 (UTC)Plclark

Cantor displaces Liouville??

"the proof being subsequently displaced by Georg Cantor"

The space formally occupied by Liouville's proof was, starting in 1873, occupied by Georg Cantor??

Presumably what is trying to be discussed here is that Liouville's proof constructs explicit transcendental numbers, whereas Cantor's proof shows, more easily, that all but countably many real (or complex) numbers are transcendental. I don't know what it means for one proof to displace the other: these are the first two theorems in transcendence theory. Plclark 23:00, 27 January 2007 (UTC)Plclark

History to the end

I hesitate to suggest at the bottom of such a passionate talk page, but could the history section be moved to the end as it is not core to the explanation. Diggers2004 07:15, 11 April 2007 (UTC)

Regarding the history section, the following sentence is curious: "Weierstrass's method has been completely set forth by Salvatore Pincherle in 1880,[10] and Dedekind's has received additional prominence through the author's later work (1888) and the recent endorsement by Paul Tannery (1894)." All of the sources listed in the References section are either from 1945 and later or 1880 and earlier. So I don't know which source could have referred to a "recent endorsement" in 1894. It sounds like something from the Encyclopedia Britannica, but that is not listed as a source for the article. —Preceding unsigned comment added by 75.3.22.86 (talk) 05:12, 7 December 2007 (UTC)

<math>

Should this page be converted to use <math> tags rather than radical symbols? IMO, the radicals with no overline look really ugly. If nobody objects within a few days, I'll switch it over. --Simetrical 01:13, 30 Dec 2004 (UTC)

Please note that there is no consensus about using <math> tages, i.e. TeX, for inline symbols, rather than displayed formulae. It is generally preferred that inline MathML is left as such, until there is more agreement. How it appears may well be browser-dependent, so that changing it to suit one user may not have a good effect for another. It is often reported that inline TeX looks odd. Charles Matthews 07:14, 30 Dec 2004 (UTC)

Hmm. Well, I very much prefer it even for inline, but I can see how some might have problems with it. Is there a talk page for discussing the use of inline TeX? --Simetrical 00:16, 31 Dec 2004 (UTC)

I think there is; I don't know the URL at this moment. Formerly it failed to get centered and looked terrible. Now, when it does get centered, you have things like $2^{x}\,$ getting centered that should not -- the bottom of the "2" should be at the same level as the bottoms of the letters, as in 2^x. Also, in $M+N\,$ the variables appear much bigger than they should, whereas in M + N they look good. Various other problems like those, too. Michael Hardy 00:25, 31 Dec 2004 (UTC)

I just found the URL myself: Wikipedia:WikiProject_Mathematics. Anyway, maybe we should at least overline the roots—√2. Unless that shows up funny in some browsers, I don't see any reason not to. --Simetrical 00:38, 31 Dec 2004 (UTC)

Using "Bitstream Vera Sans" in Firefox 2.0.0.3, it shows up weird. There's space between the overline and the 2. Inline MathML is better IMHO. However currently both is used in the article. --Ysangkok 14:35, 17 May 2007 (UTC)

reductio ad absurdum is inappropriate

The use of the term reductio ad absurdum in the discussion of proofs that the square root of 2 is irrational is inappropriate. Those are simply examples of proof by contradiction. For an example of reductio ad absurdum, see Schrödinger's CatHetware 23:24, 4 June 2007 (UTC)

I have always understood reductio ad absurdum to be the same thing as proof by contradiction. What do you understand the difference to be? Michael Hardy 00:39, 5 June 2007 (UTC)

Question

I would like to know whether e raised to any natural number is irrational? —Preceding unsigned comment added by Sumitagarai (talk • contribs)

Yes. — Loadmaster 23:33, 5 October 2007 (UTC)

I seem to recall a statement from Ivan Niven's book Irrational Numbers that the only rational point on the graph of y = e^x is (0, 1). Consequently e raised to any rational power except 0 is irrational. Michael Hardy 03:03, 6 October 2007 (UTC)

This follows from the trancendence of e. If k is a rational number, and e^k=l is rational (apart from the special case given), then e would be the l^(1/k) which would imply e is algebraic. —Preceding unsigned comment added by 81.153.227.62 (talk) 23:19, 27 July 2008 (UTC)

Complex and imaginary numbers

Since an irrational is a number that is not a rational, are imaginary and non-real complex numbers considered irrational? — Loadmaster 16:32, 2 October 2007 (UTC)

You'll notice the definition says an irrational number is any real number that is not rational. However, if one were to speak of rational and irrational complex numbers, then rational would probably generally be taken to mean a quotient of two Gaussian integers, so many non-real complex numbers would, by that definition, be rational. If you were to say that 2 + 3i is irrational, I doubt you'd find many mathematicians who would think that was weird. Michael Hardy 03:11, 6 October 2007 (UTC)

I ask because of a recent thread on sci.math. In particular, this post states that:

A Google Books search quickly confirms that this modern usage is widespread among eminent mathematicians, e.g. Conway, Gelfond, Manin, Ribenboim, Shafarevich, Waldschmidt (esp. in diophantine approximation, e.g. Thue-Siegel-Roth theorem, Gelfond-Schneider theorem, etc).

I specifically wanted to know whether i was considered irrational or not among mathematicians. Apparently some theorems involving complex numbers are easier to state if it is. — Loadmaster 16:04, 9 October 2007 (UTC)

I find it hard to imagine anyone considering i irrational. That doesn't mean it's included within the usual meaning of "rational number". It is, however, a Gaussian integer and a fortiori it is a "Gaussian rational". Michael Hardy 04:46, 10 October 2007 (UTC)

.(9)

I keep hearing that .999... is a rational number, i cannot come up with 2 numbers that when divided by each other equals .999... 71.74.154.252 16:14, 11 October 2007 (UTC)

See 0.999... (talk page). — Loadmaster 17:12, 11 October 2007 (UTC)

That's easy: 1/1 = 0.99999... Michael Hardy 19:28, 11 October 2007 (UTC)

Famous irrational numbers

The "most famous" or "best known" irrational numbers have been added to the lead, but now that they're sourced, the sources don't support the list. I think the list should go. — Arthur Rubin | (talk) 15:38, 24 October 2007 (UTC)

Please explain how "the sources do not support the list".

Please also explain why the list should go. (Because "the sources don't support the list", or because of some other reason? Perhaps we just need better sources?) My personal view is that some good, commonly known examples should be in the article, identified as such, and be placed near the top.

Also, please know that I view your use of italics on the word "now" as unnecessarily antagonistic to me, as were your reversions, as I do not believe they were in line with the guidelines at Help:Reverting. I prefer constructive criticism, please, and I am actually surprised that an administrator such as yourself would engage in such behavior. Please let me know if I've done something to personally offend you. Thank you in advance.

JonathanFreed 16:46, 24 October 2007 (UTC)

The two sources I moved to "perhaps" are separate lists of ten or fifteen "famous" irrational numbers, which include π and ${\sqrt {2}}$, but do not identify a "most famous". I really don't think the list should be in the lead, but regardless of the where the list occurs, e is also a likely candidate for "most famous", and the list needs to have a reliable source. As for now, you inserted the text twice without a respectable source. The "current" sources which I moved to "perhaps" don't fully support the text. I still doubt a WP:RS can be found, without adding a full section, including φ. — Arthur Rubin | (talk) 17:11, 24 October 2007 (UTC)

Explaining the name "irrational"

I contributed two explanations today: The first was to make it clear that "irrational" numbers are Not "numbers lacking in rational reasoning". The other contribution was in a similar vein to make it clear that "imaginary" numbers are Not "numbers lacking meaning in the real world".

Both contributions were hastily reverted. Here was the reason given for the revert:

"irrattional means not rational, and imaginary numbers are useful -- but don't exist. And also, unsourced."

Any encyclopedia entry that does a thorough job in explaining what these types of numbers are will make those points perfectly clear. Regarding the complaint that these fundamental points are not referenced, the entire 'history' section talks about how the set of numbers were thought to be non-rational (outside of the realm of sound logic) and therefore doomed to be excised out of the discipline of mathematics.

If after this anyone still has a problem with the comment being unsourced, then just google ["irrational number" misnomer] and you will find plenty of sources that make the exact same point.

If anyone has a substantial rebuttal to these points, we can all scrutinize that point of view for merit. However, if all objections are found to be lacking, then the proper action for improving the Wikipedia articles in question would be reinstatement of the contributions that were reverted today.

ChrisnHouston 19:13, 29 October 2007 (UTC)

Google search is not a viable sourcing technique -- see WP:V and WP:RS. But obviously, that's besides the point. I can't find where it states that irrational means "outside rational reasoning," rather than just "non-rational," but if it does then please point out the sentences -- I may have missed it. But if this is supposed to make the reader understand that the concept that irrational numbers are outside reasoning is a misnomer, it belongs after the point it's refuting -- i.e. the end of the history section, or preferably weaved into it -- not as a part of the lead. Gscshoyru 19:53, 29 October 2007 (UTC)

A point of confusion here is that there are two distinct meanings to the word "irrational". One means "non-logical". The other means "non-ratio-able" (if you will). I don't know the full etymology of the first (more common) meaning of the word, but it is quite possible that it came from the mathematical analysis that concluded that if a number cannot be expressed as a ratio of integers, then that number falls "outside of logical reasoning". The irony, of course, is that numbers with non-repeating decimals are perfectly logical, and in fact dominate the field of "real" numbers. (Which leads to the other point that "imaginary" numbers are quite real themselves.)

I consider it important to state this up front in order to reduce the obvious potential for confusion.

ChrisnHouston 22:24, 1 November 2007 (UTC)

But the part about non-repeating decimals should not be treated as if it were the essence of the idea. Michael Hardy (talk) 03:33, 17 March 2008 (UTC)

maybe i'm just insane, as this has no useful integration to real life.

i had a thought one day, about using decimals as a form of formulaic production (it's more of a sum really.) anyhow, i created an idea I originally called Diades (said as though plural) and using this i ran into a stump. Diades work through the use of multiple decimals, two specifically, and by using these decimals a number can be shared, so i guess it's more of a notation. it works by placing three numbers side by side like so: 2.4.8 or a.b.c Diades are performed by using this sequence (ac.(b/c)) so in words it is: a times b with a decimal value of b over c. making the aforementioned Diade equal to 8.8 the problems i encountered were pi, and remainders. using pi would render the following п.x and remainders could render a 4.9.5. as i thought i decided that pi shall count as three for a and 1415926... for b. and the remainder issue was solved as a x10 shift to allow the decimals to line up. so 4.9.5 would be 21.8. that just left pi's issue to be resolved as the problem became 3.1415926....x would equal infinite.infinite over 3 (I don't know how to write that in a proper manner. i'll work on that.) and as that became, on my paper, i wondered, is this number irrational? imaginary? or something else? —Preceding unsigned comment added by 24.187.112.51 (talk) 08:18, 15 February 2008 (UTC)

Proof that there exists a pair of irrational numbers that a^b is rational

The article proves that either sqrt(2)^sqrt(2) or (sqrt(2)^sqrt(2))^sqrt(2) is such a pair; is it known whether or not sqrt(2)^sqrt(2) is rational? Or is it like the open questions from the following paragraph, where numbers like pi+e are strongly suspected to be irrational, but never conclusively proven? - Mike Rosoft (talk) 18:52, 15 September 2008 (UTC)

${\sqrt {2}}^{\sqrt {2}}$ is transcendental by the Gelfond-Schneider Theorem. --Zundark (talk) 19:09, 15 September 2008 (UTC)

The square roots of other numbers proof wrong

The proof at the start of this section is wrong. It can be fixed either to something a little shorter using the Fundamental theorem of arithmetic or else using Richard Dedekind's proof in [1]. I prefer the latter as it is more self contained and assumes less, the fundamental theorem wasn't properly proved till Gauss came along. I'll fix it in the next day or so if no-one else does and there's no objection. Dmcq (talk) 17:34, 16 September 2008 (UTC)

Assuming that you were referring to second proof I added and not the first, can you explain the flaw? --Bowlhover (talk) 19:42, 20 September 2008 (UTC)

The proof falls down on 'where p and q are integers and ${\frac {p^{2}}{q^{2}}}$ can be assumed to be in lowest terms'. It can't be assumed to be of this form when they are in lowest terms, this has to be proved. This can probably be done using Euclid's first theorem which grounds the thing a bit further back than the Fundamental theorem of arithmetic. In fact the whole business of lowest terms is iffy without rather a lot more proof at this level but wouldn't affect the proof too much I think. Dmcq (talk) 23:38, 20 September 2008 (UTC)

The square root of 10 proof could be fixed by showing that (10n+m)² does not end in 0 for any m between 1 and 9 which can be done by enumerating the cases. It can't just be stated for any radix in general though without some proof otherwise it is assuming in effect what it is trying to prove.Dmcq (talk) 23:53, 20 September 2008 (UTC)

In that case, assume that p/q is in lowest terms instead. If p/q is not, one can always choose another p and another q so that p/q is in lowest terms. --Bowlhover (talk) 04:28, 21 September 2008 (UTC)

A couple of counterexamples to the reasoning are perhaps in order.

If we declare the number 3 unclean then our acceptable natural numbers are 1,2,4,5,6,7,8,9,10... and our prime numbers are 1,2,5,6,7,9,11,13,15 etc. Multiplying acceptable numbers together always gives another acceptable number. Then we can see that 9 is not the square of an acceable number, however its square root can be represented as the acceptable rational number 6/2.

Addition and subtraction have to be brought in to make the proof work, this is done in the proof of Euclid's first theorem. Dmcq (talk) 06:47, 21 September 2008 (UTC)

As to the proof using radixes try using the radix 8. Then 4*4=20 has an odd number of zeros at the end but is still a square.

The reasoning can be tricky. There was probably a very good reason why Theodorus of Cyrene stopped at 17 when proving the surds of non-square numbers are irrational. See What is a number? and Square root of 2 is irrational from cut-the-knot for some good proofs. In general one should follow Wikipedia:No original research - Wikipedia does not publish original research or original thought. You need to get proofs like this from published sources unless they are blindingly obvious which this isn't. Dmcq (talk) 06:47, 21 September 2008 (UTC)

Under the sub title "General roots", you have stated (although the proof is not clearly demonstrated), that "if an integer is not an exact k^th power of another integer then its k^th root is irrational" . What follows below is a proof that shows that this is generally true even for the more difficult case of fractions, a fact that was not apparent until a proof for Fermat’s Last Theorem was recently found.

Theorem

For natural numbers n and integers a, b, the nth Root of $[(b/a)^{n}+1]$ is irrational for n > 2. . Hence this formula can be used to generate an infinite number of irrational numbers.

Proof

Assume that the nth Root of [$(b/a)^{n}+1]$ is rational, then so is nth Root $[(b/a)^{n}+1]*a$

Hence, nth Root [$(b/a)^{n}+1$] = c/q . . . . . . for some integers c and q

So, $a^{n}+b^{n}=(c/q)^{n}$

And $q^{n}*a^{n}+q^{n}*b^{n}=c^{n}$ . . . . . let d = q * a and e = q * b,

Thus, $d^{n}+e^{n}=c^{n}$, which since d and e are integers, contradicts “Fermat’s Last Theorem” which has recently been proved by Andrew Wiles. Hence nth Root $[(b/a)^{n}+1]$ must be irrational, for n > 2.

NB this result was already known for the case where b/a is actually a whole number (due to the fundamental theorem of arithmetic and the fact that the nth root of primes are irrational), and in this respect provides an alternative proof.. However this was not previously known to be true for fractions, as demonstrated above. For example if we take 16, which is a square of 4 and add 1, the square root is irrational. However if we take a = 4 and b =3, the fraction 3/4 when squared and added to one, does not yield an irrational number when square rooted. This result can only occur when n = 2 but the process will always produce an irrational number for n > 2. Indeed, if the above theorem could be shown using an alternative method, it would supply a rather quick proof of Fermat’s Last Theorem. --Pgb23 (talk) 19:29, 7 November 2008 (UTC)