Rational And Irrational Numbers Exercise Sheet Page 2
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1
2
3Exercise 0.4.
Show that
Between two rational numbers there is an irrational number.
Between two irrational numbers there is an rational number.
Proof.
The proof of the second part was already done in Extra Problems #3, Exercise 0.4
(in fact, we showed there were infinitely many rational numbers between any two numbers).
We will just do the first part.
Suppose p and q are rational numbers with p < q. We want to find an irrational number
between them. Let n be an integer large enough so that
n(q
p) > 2.
We can appeal to the decimal expansion of q
p to prove the existence of such an n. Simply
k+1
choose n = 10
where k = 1 if q
p
1 and otherwise choose k to be larger than the
number of zeros after the decimal point (before the first non-zero number).
Then we note between nq and np there must be at least two integers. Call them a and b
with a < b. Consider the number z =
3
1 + a. This number is certainly between a and
b, and thus also between nq and np. We wish to show that z/n is irrational.
Suppose it was rational, then z/n = ( 3
1 + a)/n = l/k for integers l, k. Then
3 =
nl+k ka
nl/k + 1
a =
so that
3 is a rational number. This is impossible since we know
k
3 is irrational by Extra Problems #3. Thus we know that z is irrational. But then since
np < z < nq, we have p < z/n < q, and we have found our irrational number, z/n.
This exercise can give a feeling that there are as many irrationals than rationals. Since
both sets have infinitely many numbers, this is true in some sense. However, sets with
infinitely many numbers can be quite different and we will argue in what follows that in
some sense there are more irrationals than rationals.
Theorem 0.5 (Cantor).
Show that there is no function f : N
[0, 1) that is onto.
Our next goal is to prove this theorem.
We will prove the theorem by contradiction. Suppose we have such a function f . Write
each f (n) as a decimal expansion and arrange them in a list like so,
f (1) = 0 . d
d
d
d
d
. . .
11
12
13
14
15
f (2) = 0 . d
d
d
d
d
. . .
21
22
23
24
25
f (3) = 0 . d
d
d
d
d
. . .
31
32
33
34
35
f (4) = 0 . d
d
d
d
d
. . .
41
42
43
44
45
. . .
We also assume that each decimal expansion does NOT have an infinite trail of 9s at the
end. (Remember 0.64999999999 . . . = 0.65). Recall that once we make this assumption, the
decimal expansion of a number is unique (that is, there is only one way to do it).
Exercise 0.6.
Prove Theorem 0.5.
H int: Draw a diagonal line down and to the right, starting at the digit d
. By considering
11
the digits on that diagonal, explain why you can construct a new decimal number that is
(a)
Not on the list.
(b)
Inside the interval (0, 1).
(c)
Does not have any 9s at all in it (let alone an infinite trail of 9s).
Proof.
Suppose that f : N
[0, 1) was onto. We define a new number e via decimal
expansion e = 0.e
e
e
e
e
. . . where each e
is chosen so that e
= d
and e
= 9. Then
1
2
3
4
5
n
n
nn
n
note that e cannot be on the list, because if it was, say e = f (m) for some m, we know
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