Racoon Posted June 9, 2006 Report Share Posted June 9, 2006 OK, got alittle thing w/ these...Can anyone help explain Irrational Numbers? I haven't done serious Math in over a year or so, and I am going to be taking some heavy stuff this Fall, and want to start brushing up Now! :) I did a search on this, and it seems CraigD had a pretty good thread going with Turtle a few months back.. so?: Pi = irrationalGolden Ratio = irrational? --> Craig had stated (5^.5 + 1) / 2Fibionachi Sequence = irrational? Can we get an Irrational discussion going on? thanks, and looking forward to comprehensible irrationality ;) Quote Link to comment Share on other sites More sharing options...

ronthepon Posted June 9, 2006 Report Share Posted June 9, 2006 Irrational numbers? Numbers which are infinite decimals which do not repeat. Example 1.101001000100001... You can never use these numbers in linear equations properly without using approximation methods. Quote Link to comment Share on other sites More sharing options...

Farsight Posted June 9, 2006 Report Share Posted June 9, 2006 You won't find me posting much on numbers. I find them somewhat artificial and struggle to maintain any interest. For example when it comes to irrational numbers, I think it's the measuring system that's irrational, not the number. Imagine, if I said measure the size of this third but only let you use a ruler divided into halves, and each half is divided into a half, etc. Sorry. Quote Link to comment Share on other sites More sharing options...

C1ay Posted June 9, 2006 Report Share Posted June 9, 2006 Irrational number Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted June 9, 2006 Report Share Posted June 9, 2006 The core of the matter: Rational numbers are a ratio, which means a ratio of two of the previously defined integer numbers. 2/3, 7/5, 6/35 Since Pythagoras, we've been aware of the possiblity of defining numbers that can't be so expressed. A very simple case is any root of any prime number. Popular, how would you change "the measuring system" in order to make the square root of 2 rational? Quote Link to comment Share on other sites More sharing options...

Farsight Posted June 9, 2006 Report Share Posted June 9, 2006 I don't know, Qfwfq. Use a saw-toothed ruler? ^^^^^^^^^^^^^^^ Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted June 9, 2006 Report Share Posted June 9, 2006 Or the [math]\norm L^1[/math] metric! ;) Actually, :doh: that would only mean the diagonal of the square ain't root two times the side... Find integers n and d such that [math]\norm(\frac{n}{d})^2 = 2[/math] and I'll swim non-stop from Bering Strait to Southern Australia, all alone and in the nude. Quote Link to comment Share on other sites More sharing options...

CraigD Posted June 9, 2006 Report Share Posted June 9, 2006 Can anyone help explain Irrational Numbers?Several members have done a pretty good job already:…Numbers which are infinite decimals which do not repeat.andRational numbers are a ratio, which means a ratio of two of the previously defined integer numbers.2/3, 7/5, 6/35Since any terminating decimal number can be written as a ratio (eg: 1.234 = 1234/1000), and, slightly less obviously, so can any infinitely repeating decimal number (eg: 0.333… = 1/3), these are in effect the same, correct definition of an irrational number.Pi = irrationalCorrectGolden Ratio = irrational? --> Craig had stated (5^.5 + 1) / 2Yes – [math]\sqrt{5}[/math] is irrational, so the golden ratio is tooFibionachi Sequence = irrational?No, at least not usually. The ratio of terms of the 2 (or any) Fibonachi sequence is rational, because the terms of any ”standard” Fibonachi sequence (eg: 1,1,2,3,5,8,…) are integers. You could create a weird Fibonachi sequence where the ration of terms is irrational so that this isn’t true, such as: 1,[math]\sqrt{2}[/math],[math]1+\sqrt{2}[/math],[math]\1+2\sqrt{2}[/math],[math]\2+3\sqrt{2}[/math],… To head off any potential confusion, let me expand a bit on something in Ron’s post:You can never use these [irrational] numbers in linear equations properly without using approximation methods.While it’s true that an irrational number can only be approximated by a rational number, irrational numbers can be used freely in equations without resorting to approximation. An expression involving Irrational numbers may even have a rational value, eg: [math]\sqrt{2} \times \sqrt{8} = 4[/math] It’s important, I think, to note that there’s nothing necessarily unknown or inexact about most irrational numbers, just that they can’t be written as fractions of integers. They’re less convient, but this is due to coincidental qualities - such as most electronic calculators and computers storing numbers in some way using integers, and that it’s not as easy in most text editing/displaying software to displaying [math]\sqrt{2}[/math] as it is 1.25 – not an inate quality of irrational numbers themselves. Quote Link to comment Share on other sites More sharing options...

Farsight Posted June 9, 2006 Report Share Posted June 9, 2006 LOL, Qfwfq. [math]\norm n^2 = 2d^2[/math] so [math]\norm n = sqrt2 d[/math] Quote Link to comment Share on other sites More sharing options...

ronthepon Posted June 9, 2006 Report Share Posted June 9, 2006 Aah, CriagD, surds are not linear. (I hope I'm not drastically mistaken) Quote Link to comment Share on other sites More sharing options...

ronthepon Posted June 9, 2006 Report Share Posted June 9, 2006 LOL, Qfwfq. [math]\norm n^2 = 2d^2[/math] so [math]\norm n = sqrt2 d[/math]You seem to want to have Qfwfq suffer. You will never find n and d, even if you refuse to give up until your mind goes into a enlightment. Quote Link to comment Share on other sites More sharing options...

Farsight Posted June 9, 2006 Report Share Posted June 9, 2006 I was offering a proof! If n² = d² x 9, n = d x 3, eg n = 9 and d = 3. If n² = d² x 4, n = d x 2, eg n = 8 and d = 4.If n² = d² x 2, n = d x [math]\bold sqrt2[/math], eg nuffin ..because root 2 is one of them irrational number jobbies. Quote Link to comment Share on other sites More sharing options...

CraigD Posted June 9, 2006 Report Share Posted June 9, 2006 Aah, CriagD, surds are not linear. (I hope I'm not drastically mistaken)Surds are a kind of number, so can’t appropriately be called linear or non-linear. "Linear equation" is usually a term for "linear polynomial equation", or "degree one equation", which is just any equation that has the form Y=aX + c, where Y and X are variable, a and c constant. Y, X, a, and c may be natural numbers, integers, rational numbers, real numbers (which include the irrationals), complex (which include the imaginary numbers), or numbers of any kind that are closed under addition and multiplication. “Linear” can also refer to a function [math]f[/math] where [math]f(x)+f(y) = f(x+y) [/math] and [math]f(ax) = af(x)[/math]. Many functions with irrational values aren’t linear – eg: for the square root function, [math]\sqrt{2}+\sqrt{3} \neq \sqrt{2+3}[/math]. However, a linear function (eg: [math]f(x)=2x[/math]) is still linear when irrational values of [math]x[/math] are used – eg: [math] (2\sqrt{2}+2\sqrt{3} = 2(\sqrt{2} + \sqrt{3})[/math]. The term “surd” is poorly defined, and archaic. As I’m most familiar with it, it’s a synonym for “constructable using compass and straigtedge”. Some old texts use it a synonym for irrational, either the mathematical or the common meaning. It’s likely best not to use the term, since it’s meaning is so poorly agreed on. Not all irrational numbers are surds (constructable with straightedge and compass). [math]\sqrt{2}[/math] and [math]\sqrt[4]{2}[/math] are, but [math]\sqrt[3]{2}[/math] is not. For computer-age folk who don’t like old-fashioned “radical” notation, [math]\sqrt[n]{m} = m^{\frac{1}{n}}[/math] = m^(1/n), while radicals without the n are assumed to be “square roots”, [math]\sqrt{m} = m^{\frac{1}{2}}[/math] = m^.5. Quote Link to comment Share on other sites More sharing options...

CraigD Posted June 9, 2006 Report Share Posted June 9, 2006 In an effort to reserve the Math and Physics thread for more speculative discussion, this thread has been moved to Science Projects and Homework. The idea is, if you want to discuss something you’re studying with the aim of learning its usual, academic meaning, SP&H is the place for it. If you want to discuss something new unconventional, or more advanced than would ever be considered “homework”, M&P is the place for it. Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted June 10, 2006 Report Share Posted June 10, 2006 More simply:[math]\norm n^2 = 2d^2[/math] so [math]\norm n = sqrt2 d[/math]if d is integer, will [math]\norm sqrt2 d[/math] also be integer? IOW will n also be integer? Quote Link to comment Share on other sites More sharing options...

ughaibu Posted August 2, 2006 Report Share Posted August 2, 2006 Irrational numbers are described along the lines of 'non-repeating, non-terminating decimals', doesn't this description suffer from a paradox? As there are only ten numerals, individual numerals will be repeated in sequences of more than ten terms, likewise, pairs of numerals will be repeated in sequences of more than one hundred terms, etc, as irrational numbers are infinite sequences this means that all occuring finite sequences, no matter how long, will repeat. So, the only non-repeating sequences will be infinite and as the only infinite sequence is the irrational number itself, the description "non-repeating" has no meaning. ? Quote Link to comment Share on other sites More sharing options...

CraigD Posted August 2, 2006 Report Share Posted August 2, 2006 Irrational numbers are described along the lines of 'non-repeating, non-terminating decimals', doesn't this description suffer from a paradox? As there are only ten numerals, individual numerals will be repeated in sequences of more than ten terms, likewise, pairs of numerals will be repeated in sequences of more than one hundred terms, etc, as irrational numbers are infinite sequences this means that all occuring finite sequences, no matter how long, will repeat. So, the only non-repeating sequences will be infinite and as the only infinite sequence is the irrational number itself, the description "non-repeating" has no meaning. ?I believe ughaibu is confusing the idea of ”occurring” with what is meant by “repeating” in many of the usual definition of an irrational number. It’s arguable - and, I’m fairly confident, provable - that and sequence of digits in the continuing decimal representation of an irrational number can be found in (“occurs”) another place in that expansion. For example, the first 4 digits of the decimal part of the irrational number [math]\pi[/math], “1415”, appears again in positions 7701 - 7704. This does not mean that [math]\pi[/math] has a 4 digit repeating decimal part – that is [math]\pi [/math] is not 3.1415 1415 1415 … “Repeating” in the context of describing the decimal (or any other numeral base) expansion of an irrational number means that the sequence repeats with no intervening space. For example, the rational number 7/13 = 0.538461 538461 538461 … You can continue expanding 7/13 forever, and the sequence “538461” will never fail to repeat with no intervening digits. You could find some number of consecutive (no intervening digits) occurrence of “538461” in an irrational number, but if you continue expanding it, eventually the next occurance will not be consecutive. Quote Link to comment Share on other sites More sharing options...

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