
1936.] ON COMPUTABLE NUMBERS. 247
Let us suppose that there is such a process; that is to say, that we can
invent a machine
<D-
which, when supplied with the S.D of any computing
machine il will test this S.D and if il is circular will mark the S.D with the
symbol "u" and if it is circle-free will mark it with "
s
". By combining
the machines
<&
and U we could construct a machine
:l
I-
to compute the
sequence
j8'.
The machine
<O-
may require a tape. We may suppose that
it uses the jE'-squares beyond all symbols on .F-squares, and that when it
has reached its verdict all the rough work done by
l
0-
is erased.
The machine Ji has its motion divided into sections. In the first N—
1
sections, among other things, the integers 1,
2,...,
N—
1
have been written
down and tested by the machine
<
Q>-.
A certain number, say R(N— I), of
them have been found to be the D.N's of circle-free machines. In the N-th
section the machine
(
& tests the number N. If N is satisfactory, i.e., if it
is the D.N of a circle-free machine, then R(N) = l-\-R(N—l) and the first
R{N) figures of the sequence of which a $£N is N are calculated. The
R(N)-th figure of this sequence is written down as one of the figures of the
sequence/3'
computed by Ji. If N
is
not satisfactory, then R(N) = R(N—
1)
and the machine goes on to the (iV-(-l)-th section of its motion.
From the construction of
J
I-
we can see that .11- is circle-free. Each
section of the motion of Ji comes to an end after a finite number of steps.
For, by our assumption about Q, the decision as to whether N
is
satisfactor}'
is reached in a finite number of steps. If N is not satisfactory, then the
JV-th section is finished. If N is satisfactory, this means that the machine
il(JV) whose D.N is N is circle-free, and therefore its J?(iV)-th figure can be
calculated in a finite number of
steps.
When this figure has been calculated
and written down as the R(N)-th figure of
/3',
the iV-th section is finished.
Hence il is circle-free.
Now let K be the D.N of Ji. What does Ji do in the K-th. section of
its motion
1
It must test whether K is satisfactory, giving a verdict "
5
"
or "u". Since K is the D.N of JI- and since JI is circle-free, the verdict
cannot be "u". On the other hand the verdict cannot be "s". For if it
were, then in the K-th. section of its motion
J
I-
would be bound to compute
the first R(K—1) +
1
= R(K) figures of the sequence computed by the
machine with K as its D.N and to write down the R(K)-th as a figure of the
sequence computed by ill. The computation of the first R(K)
—
l figures
would be carried out all right, but the instructions for calculating the
R(K)-th. would amount to "calculate the first R(K) figures computed by
H and write down the R(K)-th". This R{K)-th figure would never be
found. I.e., 'i-l is circular, contrary both to what we have found in the last
paragraph and to the verdict "s". Thus both verdicts are impossible
and we conclude that there can be no machine '0-.