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The following is a recursive definition of positive real numbers from book "Computer Theory" by I. Cohen.

  1. 1 is in positive R
  2. If x and y are in R, then so x+y, xy, and x/y

but the author said that

it does define some set, but it is not the set of positive real numbers

What does it mean as all the positive numbers are in the set defined by the above definition?

andand
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Irfan Ul Haq
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    This question appears to be off-topic because it is about math rather than programming. math.stackexchange.com might be a better place for it. – Jim Lewis Nov 03 '14 at 17:32
  • Okay, Jim I wasn't aware of it, I only come here to ask and answer question. – Irfan Ul Haq Nov 03 '14 at 17:43

1 Answers1

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Those are all rational operations, so that set is not the positive real numbers because it doesn't include any positive irrational numbers (e.g. sqrt(2)).

lmm
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  • well, but what about x/y, it can produce irrational too, aren't?. – Irfan Ul Haq Nov 03 '14 at 17:35
  • If x and y are rational then x/y is also rational. – lmm Nov 03 '14 at 17:37
  • if we want to make it R then another condition squrroot(x) is enough? – Irfan Ul Haq Nov 03 '14 at 17:39
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    No. Allowing `sqrt(x)` will give you an algebraic number field much smaller than R; it doesn't even include e.g. cube roots of rationals (thus the insoluble ancient geometric problem of "doubling the cube"). Even if you include the roots of any polynomial with rational coefficients, that only results in the algebraic numbers, and will not include e.g. pi. The usual way to define R would be to add the axiom that any subset of R which is bounded above has a least upper bound in R. (One constructive definition of a real number is a Dedekind cut). – lmm Nov 03 '14 at 20:10