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Topic: Devlin adn Recursion
Replies: 17   Last Post: Dec 6, 2011 3:38 PM

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Dave L. Renfro

Posts: 4,792
Registered: 12/3/04
Re: Devlin adn Recursion
Posted: Nov 28, 2011 3:09 PM
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Dave L. Renfro wrote (in part):

> I'll look through some of my stuff at home (where I'm
> leaving for right after posting this) this weekend and
> try to come up a better understanding of whether this is
> something relevant for, say, a rigorous treatment of real
> analysis in which the Peano axioms are dealt with, or only
> relevant in hard-core mathematical logic and set theory work.

I looked at some things yesterday, but rather than answer the
question I raised, I'll just toss out some of what I found.


Abraham A. Fraenkel, "Abstract Set Theory",
2nd edition, 1961.

>From a footnote on p. 180:

The definitions of addition and of other operations were
for the first time strictly based [for the first time
rigorously established] in Dedekind 1888 [...] Peano's
procedure (from 1889 on, cf. Peano 1895) is not sufficient
as long as it is not supplemented by a new primitive
symbol for addition or by an appropriate rule of inference;
othewise one has, as here in Theorem 7, to prove that
_a_function_exists_ which satisfies Peano's recursive

Among the references that Franekel then provides are
the following two: A. Church, [review of a book],
Journal of Symbolic Logic 17 (1952), 199-200; Beth,
"The Foundations of Mathematics", 1959, chapter 6.


Ivor Grattan-Guinness, "The Search for Mathematical
Roots. 1870-1940", 2000.

>From p. 106:

To found arithmetic suitably Dedekind treated mathematical
induction with a new level of sophistication. [...]
It legitimated inductive _proofs_ in arithmetic by
providing a justification for inductive _definitions_;
for it guaranteed the existence of a means by which [...]
Despite all the concern with the foundations of arithmetic
in the succeeding decades, the profundity of this section
of Dedekind's booklet [1888] was not appreciated.
[footnote] The classic location for an appreciation of
Dedekind's treatment of inductive definitions is the
textbooks on foundations of analysis by Edmund Landau
(1930a, preface and ch. 1), where he acknowledged an
idea by Laszlo Kalmar. However, he [Landau] made the
point already in his obituary of Dedekind, in a version
for which he thanked Zermelo (Landau 1917a, 56-57).
[Nachrichten der Königlichen Gesellschaft der
Wissenschaften zu Göttingen, Geschäftliche Mitteilungen,
pp. 50-70.]


Leon Henkin, "On mathematical induction", American Mathematical
Monthly 67 #4 (April 1960), 323-338.

Near top of p. 328:

We may phrase this observation by saying that the _axiom_
of mathematical induction does not itself justify
_definitions_ by mathematical induction.
[footnote] This fact was clearly brought out by Dedekind
in his famous book: Was sind und was sollen die Zahlen?
(See "Bemerkung," paragraph 130, section 9.)

>From p. 338 (last paragraph of the paper):

While the method of presenting the material in this
paper may have some claim to originality, several of
the proofs which were given are well known to
mathematicians. In particular, this is true of the
two proofs (one given in detail, the other outlined)
of Theorem I. According to Professor Alonzo Church,
the origin of the first proof goes back to the
Hungarian mathematician L. Kalmar, while the idea
of the second proof should be credited to P. Lorenzen,
and D. Hilbert and P. Bernays, who discovered the
proof independently and published their work nearly
simultaneously. The proof of Theorem II is given by
E. Landau who credits it to L. Kalmar. However, Landau
fails to note the significance of the fact that the
proof does not use Axioms P1 and P2.


Two of the most detailed _mathematical_references_ I was
able to find are Enderton's book (his set theory book, not
his logic book I mentioned previously in this thread)
and Hrbacek/Jech's book. Each gives a careful discussion
of the existence proof.

Herbert B. Enderton, "Elements of Set Theory", 1977.

See pp. 70-78. A section titled "Digression" on p. 76
begins with the following sentence: "There is a classic
erroneous proof of the recursion theorem that people
have sometimes tried (even in print!) to give."

Karel Hrbacek and Thomas Jech, "Introduction to Set
Theory", 3rd edition, 1999.

See Sections 3.3-3.4 (pp. 46-55).

>From p. 47 of Hrbacek/Jech. (Recall that a sequence
in a set X is, mathematically, a function from the
set of natural numbers into the set X.)

(a) The sequence s: N --> N is defined by

s_0 = 1;
s_{n+1} = n^2 for all n [belonging to] N.

(b) The sequence f: N --> N is defined by

f_0 = 1;
f_{n+1} = f_n x (n+1) for all n [belonging to] N.

The two definitions, in spite of superficial similarity,
exhibit a crucial difference. The definition of s gives
_explicit_ instructions how to compute s_x for any
x [belonging to] N. More precisely, it enables us to
formulate a property P such that

s_x = y if and only if P(x,y);

namely, let P be "either x = 0 and y = 1 or, for some
n [belonging to] N, x = n + 1 and y = n^2." The
existence and uniqueness of a sequence s satisfying
(a) then immediately follows from our axioms:

s = {(x,y) [belonging to] N x N | P(x,y)}.

In contrast, the instructions supplied by the definition
of f tell us only how to compute f_x _provided_ that
the value of f for some smaller number (namely, x - 1)
was already computed. It is not immediately obvious how
to formulate a property P, not involving the function f
being defined, such that

f_x = y if and only if P(x,y).

We might view the definition (b) as giving _conditions_
the sequence f ought to satisfy: "f is a function on N
to N which satisfies the 'initial condition': f_0 = 1,
and the 'recursive condition': for all n [belonging to] N,
f_{n+1} = f_n x (n + 1)." Such definitions are widely used
in mathematics; the reader might wish to draw a parallel,
e.g., with the implicit definitions of functions in
calculus. However, a definition of this kind is justified
only if it is possible to show that there exists some
function satisfying the required conditions, and that
there do not exist two or more such functions. In calculus,
this is provided for by the Implicit Function Theorem.
We now state and prove an analogous result for our
situation. [They go on to state and prove "The Recursion


Each of the following book reviews by Reuben Louis
Goodstein (1912-1985) gives some worthwhile comments
about these issues, for those who want to keep digging
into this topic.

Review of A. A. Fraenkel "Abstract Set Theory", Mathematical
Gazette 38 #323 (February 1954), 54-55.

Review of J. B. Rosser "Logic for Mathematicians",
Mathematical Gazette 38 #325 (September 1954), 224-227.

Review of H. A. Thurston "The Number System", Mathematical
Gazette 41 #336 (May 1957), 158-159.

Review of B. K. Youse "The Number System", Mathematical
Gazette 50 #374 (December 1966), 436-437.


Here's a proof of existence. It's not the general recursion
theorem result proved in Enderton and Hrbacek/Jech (cited
above), but rather it's specifically written for the special
case of addition in PA.

Let S be the successor function and N be the set of positive
integers. For better readability, instead of S(2) and
S(S(g(3))) I'll write S2 and SSg(3). Let E be the set of
positive integers n such that there exists (no claim of
uniqueness being made) a function f_n: N --> N such that

(a) f_n(1) = Sn

(b) f_n(Sk) = Sf_n(k) for each k = 1, 2, 3, ...

We prove that E = N by induction. To this end we will prove
that 1 belongs to E and, if n belongs to E then Sn belongs
to E.

[[ proof that 1 belongs to E ]]

Define f_1 = S. Then (a) and (b) are readily verified.
(The verifications are easy, but there is something to be
done, namely prove the above equations hold when n = 1
and f_1 = S.)

[[ proof that n belongs to E implies Sn belongs to E ]]

Since n belongs to E (induction hypothesis), there exists
(at least one) function f_n that satisfies (a) and (b) above.
Let f_{Sn} = Sf_n (composition of S and f_n, with f_n being
taken first).

proof that (a) holds for f_{Sn}:

f_{Sn}(1) = Sf_n(1) [definition of f_{Sn}]

= SSn [induction hypothesis: (a) holds for f_n]

Note that we've shown f_j(1) = Sj for j = Sn, so we've
shown that (a) holds for f_{Sn}.

proof that (b) holds for f_{Sn}:

Let k be any (fixed choice of a) positive integer. Then

f_{Sn}(Sk) = Sf_n(Sk) [definition of f_{Sn}]

= SSf_n(k) [induction hypothesis: (b) holds for f_n, so f_nS = Sf_n]

= Sf_{Sn}(k) [definition of f_{Sn}]

Note that we've shown f_j(Sk) = Sf_j(k) for j = Sn, so we've
shown that (b) holds for f_{Sn}.

Now define m + n to be the value of f_n(m).

Now check (easy) that + is defined on all pairs of
natural numbers and we have the property (m,n) = (m',n')
implies m + n = m' + n'.


Dave L. Renfro

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