Graphics Reference
In-Depth Information
Infinity: countability
.1 23
.1 7 8
.0 5 4
.1 234
.1 7 89
.0 7 9 4
.0 3 6 0
.1 2345
.9 8 765
.2 0 6 84
.0 1 0 0 2
.01111
.1 23456
.2 2 2333
.2 7 1 809
.7 5 3 0 12
.77228 8
.000111
22 In the squares above, filled with terminating decimals between 0
and 1, think how you could fill in thelast row with a new
terminating decimal between 0 and 1, not in the earlier rows. Can
you think of an automatic procedure that would give you a new
row without further checking, for a 100
100 squareor biggr?
What about looking at the diagonal entries (marked bold here) and
making sure that the entries below them in your last line differed
from them?
23 If you made a sequence of infinite decimals between 0 and 1, could
you always construct a new infinite decimal between 0 and 1 that
was not in thelist you had mad?
One way of describing the result of qn 23 is to say that there are too
many infinite decimals between 0 and 1 to put them all in one
sequence. Any sequence is of the form
a
, . . .. So if thetrms
are all different, the suMx shows how to match the terms of the
sequence with the natural numbers,
,
a
,
a
,...,
a
N
.
24 By starting 0,
1,
1, . . . show how to put all the integers into a
sequence. This will match N to Z, one-to-one.
Another way to construct a sequence which includes every integer
is to match the positive integer
n
with 2
and the negative integer
n
with 2
· 3. Then running through the selected natural numbers
of theform 2
and 2
· 3 in order of size gives a way of building a
sequence which includes each of the
integers
exactly once.
What are the first seven integers when this second method is used
to construct a sequence running through
Z
?.
25 If
a
and
b
are positive integers, matching the rational number
a
/
b
(always taken in lowest terms) with the natural number 2
· 3
gives
a one-to-one matching of the positive rationals with a (rather
small!) subset of N. If we then run through the selected natural
numbers of the form 2
· 3
in order of size, we have a way of
