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        riddles >> easy >> Between any two...
        
(Message started by: Sir Col on Aug 24^th, 2003, 7:44am)

Title: Between any two...
Post by Sir Col on Aug 24^th, 2003, 7:44am

Prove that between r and s, where r and s could be any two different,
(i) rational numbers, there exists a rational number.
(ii) rational numbers, there exists an irrational number.
(iii) irrational numbers, there exists an irrational number.
(iv) irrational numbers, there exists a rational number.

Title: Re: Between any two...
Post by towr on Aug 24^th, 2003, 11:09am

[hide]i) (r+s)/2
ii) r + (r-s)*sqrt(1/2) (unless r=-s, then take f.i. sqrt(1/2)*r)[/hide]

Title: Re: Between any two...
Post by Icarus on Aug 24^th, 2003, 12:20pm

A more general way of stating the problem is: show that between any two distinct real numbers, there is a rational and an irrational number.

This includes the other two cases Sir Col does not mention (rationals and irrationals between a rational and irrational).

Here is a nonconstructive argument for the existance of irrationals in the intervals, based on cardinality: [hide] Let r, s be any distinct real numbers with s > r. the function f(x) = r + (s-r)x is a one-to-one correspondance between the open intervals (0,1) and (r,s). Since (0,1) is uncountable, so is (r,s). Since the rationals are countable, (r,s) must contain some irrational values.[/hide]

This proof, while valid, is not as satisfying as actually constructing an example. There are fairly simple proofs for these, but I will leave them for others.

Title: Re: Between any two...
Post by Jamie on Aug 25^th, 2003, 6:16am

iii) [hide]Find rational numbers a and b s.t. ar+b < 1 and as+b > 2 (i.e. map the range onto (almost) (1,2)). Now, 1<[surd]2<2, so let x = ([surd]2-b)/a. Now x is irrational, and r < x < s.[/hide]

iv) [hide]Consider binary expansions, find the first position at which they differ, call it i. Now, wlog, r has a zero and s has a one at the i^th position. Let q be the common bits up to position i, with a final one. Clearly q > r, and q < s because s is irrational, and hence has an infinite binary expansion. Finally, q is rational because it's finite[/hide]