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Sir Col
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Between any two...
« on: Aug 24th, 2003, 7:44am » |
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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.
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towr
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Re: Between any two...
« Reply #1 on: Aug 24th, 2003, 11:09am » |
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i) (r+s)/2
ii) r + (r-s)*sqrt(1/2) (unless r=-s, then take f.i. sqrt(1/2)*r)
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| « Last Edit: Aug 24th, 2003, 11:11am by towr » |
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Icarus
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Re: Between any two...
« Reply #2 on: Aug 24th, 2003, 12:20pm » |
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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: 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.
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.
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Jamie
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Re: Between any two...
« Reply #3 on: Aug 25th, 2003, 6:16am » |
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iii) 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.
iv) 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 ith 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
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