HOMEWORK 1 SOLUTIONS
JOEL FRIEDMAN
Copyright: Copyright Joel Friedman 2015. Not to be copied, used, or revised
without explicit written permission from the copyright owner.
Exercise 7.1.1
(1) Every positive integer n > 0 can be written as 1+1+· · ·+1 (with 1 appearing
n times), and hence is the meaning of
one plus one plus . . . plus one
(2) No. Every string of alternating numbers (one and two) and operations (plus
and times) that begins and ends in a number has a meaning (which might depend
on the precedence of operations); any other string is meaningless. Alternitively, the
paradox cannot be build since none of the words of W are self-referential, i.e., none
refer to strings in W .
Exercise 7.1.2
(1) The one-word sentences that are not meaningless are one, whose meaning is
1, and two, whose meaning is 2. Hence the sentence “moo one” evaluates to be 3.
(2) The two-word sentences that are not meaningless are supposed to include
the two-word sentence “moo two.” But this sentence is supposed to be assigned an
integer whose values do not include its own value. Hence “moo two” is not supposed
to be equal to itself.
(3) This is the same type of paradox as Paradox 3, in that the self-referential
term moo tries to claim that we should be able to ascribe a meaning to a sentence
whose meaning is not supposed to be equal to its own meaning.
(4) This a paradox where the term moo is not sufficiently precise, in that the
sentence moo moo can’t be evaluated. If you make the term moo more precise,
this “paradox” could be resolved. There are many reasonable ways to resolve this
paradox.
One way is to define any self-referential sentence to be meaningless. Since the
sentene “one” is the only sentence that evaluates to 1, this effectively means that
any sentence beginning with a moo, excepting the sentence “moo one”, is meaningless.
Another way to make the word moo more precise is to define “moo followed by
a subsentance that evaluates to k to be the smallest positive integer not described
by a sentence by k words or fewer that can be explicitly evaluated.” Then moo moo
would evaluate to 4 (i.e., not equal to the meanings of one, two, and moo one.
Research supported in part by an NSERC grant.
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JOEL FRIEDMAN
Exercise 7.2.1
Since each Ci is countable, there is for each i a sequence
ci1 , ci2 , ci3 , . . .
such that each element of Ci occurs somewhere in the sequence. Then the sequence
c11 , c21 , c12 , c31 , c22 , c13 , . . .
which lists all cij in order of the value of i + j (starting with j = 1, and increasing
in j) contains all elements of C1 ∪ C2 ∪ · · ·
Exercise 7.2.3
If Σ is countable, then Exercise 7.2.2 implies that Σ2 = Σ × Σ is countable;
another application of this exercise shows that Σ3 = Σ × Σ2 is countable; by
induction we show that for any positive integer i we have Σi is countable.
Since Σ∗ = Σ0 ∪ Σ1 ∪ · · · , and each Σi is countable, Exercise 7.2.1 implies that
∗
Σ is countable.
Exercise 7.2.7
If C is countable, then C ∪ C is countable, using Exercise 7.2.1. Hence R would
be countable, which is impossible.
Department of Computer Science, University of British Columbia, Vancouver, BC
V6T 1Z4, CANADA, and Department of Mathematics, University of British Columbia,
Vancouver, BC V6T 1Z2, CANADA.
E-mail address: jf@cs.ubc.ca or jf@math.ubc.ca
URL: http://www.math.ubc.ca/~jf