Rational And Irrational Numbers Exercise Sheet
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3EXTRA PROBLEMS #5
In this assignment, we consider again rational and irrational numbers.
We begin with a warm-up exercise (that will also be useful).
Exercise 0.1.
Suppose f : A
B is onto (here the domain of f is A and the codomain is
B). Also suppose that g : B
C is onto. Recall that g f (the composition of the functions
g and f ) is the function associated to the rule (g f )(x) = g(f (x)). Prove that g f : A
C
is also onto.
Proof.
Choose c
C arbitrary. We need to show there is an a
A such that g(f (a)) = c.
Since g is onto, there exists a b
B such that g(b) = c. Since f is onto, there exists an
a
A such that f (a) = b. Then
g(f (a)) = g(b) = c
as desired.
Exercise 0.2.
First show that there exists a function f : N
Q
that is onto. Here Q
>0
>0
is used to denote the set of positive rational numbers.
Hint: We are trying to organize the rational numbers into an infinite list. Consider arranging
the positive rational numbers like the diagram below.
1/1 1/2 1/3 1/4 . . .
2/1 2/2 2/3 2/4 . . .
3/1 3/2 3/3 3/4 . . .
4/1 4/2 4/3 4/4 . . .
. . .
. . .
. . .
. . .
. . .
Proof.
We simply need to arrange all the positive rational numbers into a list (where we
allow repeats). Once we have done this, we define f (n) to be the nth entry in the list. We
start with (1/1), (1/2), (2/1), (1/3), (2/2), (3/1), (1/4), . . .. We repeat this pattern by going
down the diagonal to the left, until we run out of numbers and then starting on the next
diagonal. Every positive rational number eventually appears in this list, so we are done.
Exercise 0.3.
Show that there is an onto function g : N
Q.
Hint: First construct an onto function from Z to Q, then use a previous extra math assign-
ment to find an onto function from N to Z. Conclude by applying the first exercise.
Proof.
We first define an onto function j : Z
Q by
f (n)
n > 0
j(n) =
0
n = 0
f ( n ) n < 0
By extra math assignment #3, there exists an onto function k : N
Z. Thus the compo-
sition j k : N
Q is surjective by the first exercise.
In Set #3 we showed that every interval contains infinitely many rational numbers. Use
the same kind of ideas to
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