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 Skip Navigation LinksMath Help > Sets, Set theory, Number systems > Sets > Rational and Irrational

In every closed interval with distinct endpoints, there is at least one rational number and one irrational number.

A Rational and an Irrational in Every Closed Interval

(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 * rational is rational

proof: (a/b)(c/d) = (ab)/(cd)

(7) The product of rational * irrational 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


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)*sqrt(2)/2 is
rational + rational*irrational, so it is irrational.

a < a + (c-a)*sqrt(2)/2 < a + (c-a) = c, so
a < b < c, and b is irrational, so statement (2) is proven


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*10n+1, and let c' = c*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 = floor(c')/10n+1, so b is a rational number, and

a  < b < c, so statement (3) is proven.


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).


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 f 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).

Related pages in this website

Definition of Interval

Subdividing an Interval

Calculus Theorems

 

The webmaster and author of this Math Help site is Graeme McRae.