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What is the positive real number (say less than one) that is not a rational nor an irrational number?

I have encountered a mathematical problem that confused me about the definition of real numbers, so I thought there may be some other set of numbers that is not well defined !

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closed as off-topic by James Kingsbery, Keelan♦, jobermark, iphigenie, virmaior Jun 9 '15 at 0:28

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  • "Questions on the definitions or semantics of words or phrases are off-topic here as they are already well-answered elsewhere. There are many fine dictionaries available on The Internet, and Wikipedia offers good introductions to most common schools of philosophy." – James Kingsbery, Keelan, iphigenie
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    This question may have metaphysical implications, but they are not apparent. Please, consider editing the question to reflect those concerns, or remove the tag. – André Souza Lemos Jun 8 '15 at 14:14
  • I don't know why my answer is not visible normally, please note that math-exchang is not taking the question into consideration, so can you help in this regard the way you or somebody else finds it suitable? – bassam karzeddin Jun 8 '15 at 14:25
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    See Irrational number : "In mathematics, an irrational number is any real number that cannot be expressed as a ratio of integers." Thus, the real numbers are made of two "disjoint" subsets : that containing the rationals and its complement including the irrationals. No way out from this : irrational are defined as those that are not rational. – Mauro ALLEGRANZA Jun 8 '15 at 14:33
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    Philosophy.SE is not configured to accept MathJax, so you'll have to go without it. Additionally, keep in mind that your question will be treated here for its philosophical content only. – André Souza Lemos Jun 8 '15 at 14:42
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    The math site is dealing with this appropriately, despite what the questioner imagines, he would get no better reading here. – jobermark Jun 8 '15 at 20:27
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I'll give you the standard, schematic answer, but be aware that this question opens a Pandora's box.

Irrational numbers are those that cannot be indicated by a fraction of two integers. Because it is a negative definition, it can be a bit misleading, and unhelpful. See, all numbers that are not rational have to be... irrational.

There is, however, an "outer circle", a set of numbers that aren't rational, but that can be positively defined as those that are the solution of some algebraic equation. These are called the algebraic numbers.

The next circle outside is the set of computable numbers. These are, roughly speaking, the numbers that can be approximated to an arbitrary precision by a finite calculation process.

Numbers that are not algebraic are called transcendental. Most of these numbers are also not computable. Another negative definition, as you can see. There are (and probably will always be) more things that we do not know about these numbers than things that we do know, so if you are perplexed by them, you are in good company.

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  • Can I exclude the transcendental numbers, since all real positive transcendental numbers that are less than one are also irrational numbers that I have already excluded? – bassam karzeddin Jun 8 '15 at 12:52
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    If you exclude the irrationals, all that you are left with are the rationals. Simple logic here. – André Souza Lemos Jun 8 '15 at 12:53
  • I shall give an argument that may break this simple logic here! – bassam karzeddin Jun 8 '15 at 12:58
  • Go ahead! It will be interesting. – André Souza Lemos Jun 8 '15 at 13:09
  • If math breaks logic, we're doomed... – Quentin Ruyant Jun 8 '15 at 20:14
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If $x$ is positive rational number less than $\frac{1}{2}$, can the following logarithmic expression be equivalent to any real number , say $r$?

$$\frac{\log(1-x)}{\log x} = r$$, Where $r$ is positive real number less than one,

Assume first $r$ is a rational number say $n/m$, then you would get this equation, $x^n=(1-x)^m$, where solution for $x$ will be clearly algebraic - irrational number contradicting the assumption of rationality of $x$, hence $r$ cannot be rational number & must be irrational number, if so, you get this following equation, $x^r = 1-x$, the right hand side is a rational number, but the left hand side is transcendental number according to - Gel-schneider theorem, that contradicts again the assumption of irrationality of $r$, so what is that number suppose to be?

I guess someone will simply say it is a transcential number given by definition, if so assume $x$ an integer power of a rational number, say $x=a^p$, where $p$ is odd prime number, then by substitution you get this simple equation which is a reduced form of FLT for rational numbers , $a^p + (a^r)^p = 1$, but we already know that $a^r = b$, where $b$ is a rational number that doesn't exist from the proof of Andrew Wiles & Taylor to Fermat's Last Theorem, thus $log(b) / log(a)$ is not Transcendental number, it is a kind of numbers that have to be redefined again & precisely

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  • You may kindly view it clearly here: math.stackexchange.com/questions/1311922/… – bassam karzeddin Jun 8 '15 at 14:51
  • Your question was well accepted there. The jury is still out on your answer, so if your patient enough, maybe what they have to say will satisfy your curiosity. As I said, it can only be of interest to us here for its philosophical content, i.e. as a matter of conceptual clarification. – André Souza Lemos Jun 8 '15 at 15:27
  • Is it, that by FLT you mean Fermat's Last Theorem, and you think the above might be a short sweet proof? Then just think about it: he would surely have found place to write that in the margin. It's too short! :) – Cheers and hth. - Alf Jun 8 '15 at 16:40
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    If the answer you're looking for is purely mathematical, why is this question here on Philo.SE? – Keelan♦ Jun 8 '15 at 19:38
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    You cannot use FLT because a is not an integer (nor a^r). – Quentin Ruyant Jun 9 '15 at 10:53

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