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I'm going through Apostol's Calculus Volume 1 and the first few chapters are more about real analysis than calculus. You can use all the properties of elementary algebra, inequalities and the least upper bound axiom(and fundamental properties of the supremum and infinum-should I elaborate?) I would like hints on how to prove that given two non equal reals there exists one and hence an infinite number of rational and irrational numbers between them. I've proven that there exist an infinite number of reals in between the two non equal arbitrary real but I want hints for proving the other statements. Hints instead of a complete solution will be appreciated.

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    $\begingroup$ How does Apostol define the real numbers? Are they the completion of Cauchy sequences of rationals? $\endgroup$ – davidlowryduda Nov 20 '14 at 2:14
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    $\begingroup$ If you know that between any two numbers there is a rational, you can conclude that there is also an irrational: If $a<b$, use that there is a rational $p$ with $(a-\sqrt 2)<p<(b-\sqrt2)$ to conclude that there is an irrational between $a$ and $b$. How to best argue that there are rationals between $a$ and $b$ depends on the specific way that reals are constructed For instance, if reals are given by Dedekind cuts, then that there are rationals in between is essentially by definition. $\endgroup$ – Andrés E. Caicedo Nov 20 '14 at 2:44
  • $\begingroup$ From the book: "The point of view we shall adopt is nonconstructive. We shall start rather far out in the process, taking the real numbers themselves as undefined objects satisfying a number of properties that we use as axioms. All the properties of real numbers can be deduced from these axioms." The axioms listed are the commutative and associative properties, the existence of identity elements, etc. $\endgroup$ – user42991 Nov 20 '14 at 4:28
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For the infinitude of rationals part try to find two rationals between the numbers.

then continuously average to show that there are an infinitude of rationals between them.

Have you proved that there are uncountably infinitely many reals between the two numbers or just infinitely many reals? The former makes the irrational part trivial to prove.

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  • $\begingroup$ Question: If x and y are arbitrary real numbers with x<y, prove that there is at least one real z satisfying x<z<y. My attempt: I showed that the real z = x+(y-x)/2 is between x and y. We can generalize with 2 being replaced by n, where n is 2,3,4,... Each real z thus generated is unique, if not, this implies that n1=n2(or any pair of non equal integers are equal). Also, since the set 2,3,4,... is unbounded above and each element corresponds to one real, unique z, there are an infinite number of reals between x and y(I don't know what you mean by uncountably infinite and just infinite). $\endgroup$ – user42991 Nov 20 '14 at 4:24

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