Rational And Irrational Numbers Exercise Sheet Page 3
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1
2
3that e and f (m) have the same decimal expansion. But they don’t, since e
= f
. Thus
m
mm
we conclude that the function f cannot be onto because we found a number e
[0, 1) such
which f never “hits”.
Exercise 0.7.
Give an example of an onto function h : R
[0, 1).
Proof.
Define h to be the function which takes a number, views it as
(integer).(decimal expansion)
and outputs
0.(decimal expansion)
The function h is clearly onto [0, 1) since it is the identity function on [0, 1).
Exercise 0.8.
Prove that there is no onto function g : N
R.
H int: Suppose there was, then use the previous two exercises and exercise 0.1 to prove this
is impossible.
Proof.
Suppose that g was onto, we will aim for a contradiction. Compose with h to get a
function h g : N
[0, 1). Note that this function is onto by the first exercise of this set.
But this is impossible by Theorem 0.5.
We are now ready to show that there are “more” irrationals than rationals. But first a bit
of notation, R Q is just an expression that means all the real numbers that are not rational.
That is R Q is equal to the set of irrational numbers.
Exercise 0.9.
Show that there is no function f : Z
R Q that is onto.
H int: Use Exercise 0.5 together with Exercise 0.3 and also the last problem of Set 3.
Proof.
Again, suppose their was such a function, we will aim for a contradiction. Let l :
N
Z be an onto function which makes f
l : N
R Q an onto function. Construct a
new function k : Z
R.
f (l(n)) n > 0
65.12
n = 0
k(n) =
g( n )
n < 0
where g is the function from Exercise 0.3. Since f
l : N
R Q hits all the irrational
numbers, and g : N
Q hits all the rational numbers, our new function k : Z
R is also
onto. But then composing with l yet again, we get an onto function k l : N
R, which is
impossible by the previous exercise.
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