Rationals and irrationals in any
interval
By Patrick Aouad on Saturday, May 18, 2002
- 02:00 am:
How can you show that within any closed interval of length at
least 1 there exists at least one irrational number and at least
one rational number?
By Graeme Mcrae on Saturday, May 18, 2002
- 05:16 am:
(1) Every closed interval that contains more than one point
(that is, its endpoints are distinct) contains a rational number
and an irrational number.
To prove statement 1, I'll start with some simpler facts:
(2) Between every pair of rational numbers there is an irrational
number. And
(3) Between every pair of irrational numbers there is a rational
number.
The arguments for these facts depend, in turn, on some more
facts:
(4) The sum of rational + rational is rational
proof: a/b + c/d = (ad+bc)/(bd)
(5) The sum of rational + irrational is irrational
proof: if a is rational and b is irrational and a+b=c is
rational then c+(-a)=b is rational, a contradiction
(6) The product of rational xrational is rational
proof: (a/b)(c/d) = (ab)/(cd)
(7) The product of rational xirrational is irrational as long as
the rational is not zero
proof: if a is non-zero rational and b is irrational and ab=c
is rational then c(1/a) is rational, a contradiction
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Now, let's find an irrational number b that lies between any two rational
numbers a and c, where a < c:
Let b=a+(c-a)×Ö2/2 is rational + rational× irrational, so it
is irrational.
a < a+(c-a)×Ö2/2 < a+(c-a)=c, so a < b < c, and b is
irrational, so statement (2) is proven
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Now, let's find a rational number b that lies between any two
irrational numbers a and c, with a < c:
If c-a < 1, then let n = ceiling(-log10
(c-a))
If c-a > = 1, then let n = 0
Now let a' = a x 10n+1 , and let c' = c x
10n+1
c' - a' > 10, so
a' < ceiling(a') < floor(c') < c', so
a = a'/10n+1 < ceiling(a')/10n+1 <
floor(c')/10n+1 < c'/10n+1 = c
Let b = (a'+c')/10n+1 /2, so b is a rational number,
and
a < b < c, so statement (3) is proven.
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Now, let's tackle statement (1): that every closed interval with
distinct endpoints contains a rational number and an irrational
number.
Let the endpoints be a and c.
If a is rational and c is irrational, or vice-versa, then
statement (1) is true. Now let's assume they're either both
rational or both irrational.
If a and c are both rational, then statement (2) shows there is
also an irrational number between them, proving (1).
If a and c are both irrational, then statement (3) shows there is
also a rational number between them, proving (1).
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In fact, a statement stronger than (1) can be proven:
(8) Every interval, open or closed, that contains more than one
point (that is, its endpoints are distinct) contains an infinite
number of rationals and an infinite number of irrationals.
Proof: given two real numbers, a and f such that a < f, then
let b=(4a+f)/5, c=(3a+2f)/5, d=(2a+3f)/5, and e=(a+4f)/5
Then a < b < c < d < e < f, so the interval
between a and e contains two disjoint intervals [b,c] and
[d,e]
Since every interval contains two disjoint closed intervals,
every interval contains an infinite number of disjoint closed
intervals.
Since each closed interval contains a rational number and an
irrational number, every interval contains an infinite number of
rational numbers and an infinite number of irrational numbers,
proving statement (8).
By Patrick Aouad on Sunday, May 19, 2002 -
01:18 am:
That's excellent, thanks for all your time Graeme.