Basic material. We make use of the standard families of numbers. These now have standard names.

Transcription

1 (A) Basic material In this document we set down all the basic material you should already know, together with some notation and terminology. Of course, there are different terminologies and notations for the same thing. Here we fix what is used in these documents. You should already know the ideas behind these, but some of the notation may be new to you. [This document may change as the course progresses, because new material may be added. You should check this regularly. Once these remarks in this font disappears, that means it is the final version.] Contents 1 Various kinds of numbers Big problem: Numbers Logical symbols Exercises: Logical symbols Problems: Logical symbols Sets Exercises: Sets Problems: Sets Functions Exercises: Functions Problems: Functions Induction Exercises: Inductions Problems: Induction Various kinds of numbers We make use of the standard families of numbers. These now have standard names. N the natural numbers 0, 1, 2,... Z the integers 0, ±1, ±2,... Q the rational numbers those numbers of the form n/d where n, d Z R the real numbers C the complex numbers Each natural number and each integer has a standard name, as indicated. Each rational is formed by dividing one integer by a non-zero integer, so each such rational has the form n n/d or d a numerator over a denominator. Of course, different pairs of integers can give the same rational. The description is in canonical form if d > 0 and the pair a, d are relatively prime, that is the have no common divisor (except 1). As here, when dividing one thing by another two different notations are useful, depending on the complexity of the expression and where it occurs. For instance both the following (x 3 + xy + 4x 5)/7 1 x 3 + xy + 4x sin(x)

2 are readable and the first could be used in a line of text (not displayed), but the second is probably better in this given form. The real numbers R include the rationals Q and many more numbers as well, such as the algebraic numbers / where I have no idea how the last one ends. The reals also contains some non-algebraic numbers such as π, e, and some numbers such as e + π which no-one knows whether or not it is algebraic. Later in The rationals and reals as ordered sets we look at the relationship between Q and R as linearly ordered sets. The complex numbers C are mentioned here but will not play much of a part in these documents. The technique of induction over N (or final sections of Z) is important. You should look at the Exercises and Problems in Section Big problem: Numbers 1.1 Show that the value of the daft expression is actually 1. [This is a correction suggested by someone. I don t know the answer.] 2 Logical symbols In Mathematics we often have to deal with complicated statements, and then it is sometimes useful to have symbols for various logical notions. Here we use the following symbols for the informal readings. and or implies if and only if negation for all there exists Of these will cause no problem. Sometimes we will also use & for and. The technical name for this is conjunction. The use of is not quite the same as in everyday language. For two expressions θ, φ the compound θ φ means either θ or φ or both where down the pub it would probably mean either θ or φ but not both. The inclusive version is much better in Mathematics, but the exclusive version is more common in day-to-day language. The technical name for this is disjunction. 2

3 There is a story of how came to be used for this notion. In Latin there are two words vel for the inclusive or aut for the exclusive or and is just a version of the first letter of vel. Once this symbol was decided on it was obvious that should be used for and because of the way the two connectives interact across negation. See later for a further discussion of this. For two expressions θ, φ the compound θ φ can be phrased in several different ways. For instance θ implies φ if θ then φ are two standard readings. You may also see φ whenever θ which is a very good way of messing up the information being passed on. The compound θ φ says that θ and φ are equivalent, so (θ φ) (φ θ) is an expanded version of the assertion. You will probably see these symbols used in different lengths, such as, =, for instance. There is no intended difference in meaning. Usually the longer version helps to punctuate a more complicated expression. Or it may be that the wrong part of LaTeX has slipped in. You will also see single line arrows such as in different places. these have nothing to do with implication. They are used for different aspects of functions. The compound φ merely negates the expression φ. Whatever φ is saying, the compound φ says the opposite. Later we look at how interacts with the other formal symbols. The two quantifiers, are used very often, and can cause a lot of problems for some people (such as philosophers). Given some expression [ x ] which refers to some x, the compounds are read as ( x A)[ x ] ( x)[ x ] for all x A, we have x for all x, we have x where in the second version it is assumed that we know where x should live, that is in A as given by the first version. 3

4 The compounds are read as ( x A)[ x ] ( x)[ x ] there is some x A, such that x there is some x, such that x where, as before, in the second version it is assumed that we know where x should live. There are times when we want to sort out the negation of an assertion. To do this it is often helpful to write down the assertion in symbolic form and then hit it with, the negation symbol. We pass this symbol across all the others until it is deep inside the expression. Here is how passes across the connectives. (θ φ) becomes θ φ (θ φ) becomes θ φ (θ φ) becomes θ φ Notice how and match up in a dual fashion. To take across is is better to first rephrase the equivalence as a conjunction of two implications and then hit that. The two quantifiers are also dual gadgets. We have ( x A)φ becomes ( x A) φ ( x A)φ becomes ( x A) φ which you should think about. Finally two negations might meet up. This is not a problem. φ becomes φ If anyone tries to tell you something different, don t listen to him. He is from the other side. I should also tell you that although two wrongs don t make a right, three lefts do. 2.1 Exercises: Logical symbols 2.1 Consider an arbitrary function f : R R on the reals. Write down the formal ɛ-δ definition that f is (i) not continuous (ii) not uniformly continuous I suggest that before you do this you get the positive versions sorted out. 2.2 Problems: Logical symbols 2.2 Find a binary connective from which all the others,,,,,, can be obtained. 4

5 3 Sets A set is something that has members and subsets. x A X A x is a member of, or an element of, the set A x is contained in A X is a subset of, or a part of, the set A X is included in A These two notions should not be confused. Often in everyday language the two words contained and included are used to mean the same thing. That sometimes happens in Mathematics, but it is better to use them more carefully. Peano had a bit of trouble explaining the difference to some other mathematicians. When in doubt use the symbolic notations as above. Two sets are equal if the have the same members, or equivalently each is a subset of the other. A = B ( x)[x A x B] A B and B A Both of these characterizations are useful at different times. These two notions can be negated. Thus x / A means (x A) that is, x is not in A and this doesn t cause much trouble. In terms of inclusion we have X A means (X A) that is, X is not in a subset of A so this means there is some x X with x / A. We also write X A to mean X A with X A that is X A but X is not the whole of A. These notations are now fairly standard, but not universal. You will also find X A used in the literature. This usually means X A, but in the older literature it sometimes meant X A. We often display a finite set by listing its members. For instance {a, b, c, d, e, f} contains no more than 6 members (for we could have listed some twice). The order in which we list the elements is irrelevant. Thus {b, a, d, a, c, f, e, b} is the same set. If we want to have a list of elements where the order matters, then we use a different notation. We write (b, a, d, a, c, f, e, b) 5

6 for the list, and in this notation the order and the repetitions matter. There is a special symbol for the empty set, the unique set that has no members. Note that A for every set A. This symbol is the Scandinavian letter erf. Some people write it as φ or other wrong variants. 1 On many occasions we want to extract a subset from a given set A. We have a certain property φ( ) which an element of A may or may not satisfy. Then X = {a A φ(a)} is the extracted subset. Put slightly differently we have a X φ(a) for each element a A. Both notations are useful, but in proofs the second version is often easier to use. Sets can be combined in various ways. Here are the finitary boolean combinations. union x A B a A or x B intersection x A B a A and x B difference x A B a A and x / B Observe that the union and intersection operations are commutative and associative, that is A B = B A A B = B A and hold. However (A B) C = A (B C) (A B) C = A (B C) A B B A are very different (in more senses than one). Sometimes you will see the difference written as A\B, but not in these documents. There is also a more general version of the union and the intersection constructions. For each family A of sets we have the following. union x A ( A A)[a A] intersection x A ( A A)[a A] I said these two constructions exists for all families of sets A, but what do they mean when A is empty? Sometimes a family of sets is displayed as an indexed family, that is A = {A i i I} for I is the family of indexes. You may then find A written as A A written as A 1 Find the correct spelling i I i I 6

7 or something similar. This can cause a bit of a clutter. When all the sets we consider are subsets of one particular set S we also have the complement A = S A so that x A x / A for all x S. Notice that the notation ( ) depends on the parent set S. There are also other notations used for the complement. Every set A has a collection of subsets. This collection is the power set of A and is denoted as PA. We always have PA A PA but in general PA can have many more members. There is a useful way of dealing with certain aspects of subsets using characteristic functions. This is dealt with in Section 4. To conclude this section we look at ordered pairs and unordered pairs. We have seen that for any list a 1,..., a m of elements we write {a 1,..., a m } for the set whose elements are the ones listed. elements are listed is irrelevant. For example {a, b, c} = {a, c, b} = {c, a, b} = and so on. We may also list element elements more than once. Thus {a, b, c, a} = {b, a, b, a, c} = {a, b, c} = In particular, the order in which the and so on. In particular, we don t know that {a, b, c} has three elements, for we may not have noticed that c = a, for instance. All we know is that it has at least one element and no more than three. There are many times when the order of a listing is important. For that we use the notation (a, b, c) to indicate the list a, b, c in that order. Here repetitions and the order does matter. As a particular case we write (a, b) for the ordered pair. Thus (a, b) = (c, d) a = c and b = d for elements a, b, c, d. At this stage we should also point out that when we are looking at the reals R the notation (a, b) means the open interval between a and b, that is x (a, b) a < x < b 7

8 for each real x R. These two notations have nothing whatsoever to do with each other. It s just that the same symbols are being used in two entirely different ways. There are, of course, several kinds of reals intervals open closed left open, right closed left closed, right open (a, b) [a, b] (a, b] [a, b] and ± can also occur at one end or other. There used to be a convention that these four intervals were written as follows. ]a, b[ [a, b] ]a, b] [a, b[ This was a daft idea. What is perhaps a bit of a surprise is that the notation of an ordered pair can be produced using { and } only. 3.1 Exercises: Sets 3.1 (a) Verify the two distributive laws, that is (A B) C = (A C) (B C) (A B) C = (A C) (B C) for arbitrary sets A, B, C. (b) Verify the two De Morgan laws, that is C (A B) = (C A) (C B) C (A B) = (C A) (C B) for arbitrary sets A, B, C. (c) Suppose A, B are subsets of a parent set C. Write out the De Morgan laws in terms of ( ), the complement operation. 3.2 In this question A is an arbitrary non-empty family of sets. (a) Show that for each set B. (b) Show that B A = {B A A A} B A = {B A A A} B A = {B A A A} B A = {B A A A} for each set B. (c) Suppose each member of A is a subset of some parent set S. What are ( A ) ( A ) in terms of complements of members of A. 3.3 Show that A B = {A B A A, B B} A B = {A B A A, B B} for each pair A, B of non-empty families of sets. 8

9 3.4 Suppose all the sets we are interested in are subsets of S. What are ( ) ( ) in this context. 3.5 Find a family A of open intervals of R, each of finite length, such that Q A R. Can you find a pairwise disjoint family A with this property? 3.6 For arbitrary elements a, b let to produce a set. Show that a, b = {{a}, {a, b}} a, b = c, d a = c and b = d for all elements a, b, c, d. This is a formal way in which ordered pairs can be introduced into Axiomatic Set Theory. 3.2 Problems: Sets 3.7 In this problem we consider a doubly indexed family of sets. {A(i, j) i I, j J} We assume that both I and J are non-empty, and each indexed set A(i, j) is non-empty. We also consider the set [I J] of all functions f : I J. (See Section 4 for more information about this notation.) (a) Consider the following two statements concerning some element x. ( i I)( j J)[x A(i, j)] ( f [I J])( i I)[x A(i, f(i))] Show that each implies the other. (b) Can you see that one of the implications is not quite dodgy, but certainly requires further explanation. (c) Show that the two compounds { {A(i, j) j J} i I } { {A(i, f(i)) i I} f [I J] } are equal. (d) Are the two compounds { {A(i, j) j J} i I } { {A(i, f(i)) i I} f [I J] } equal? 9

10 3.8 Suppose we start of with a finite family A of sets, and we build up a compound by starting from the members of A and use and a finite number of times. Thus the use of these two operations could be highly nested. It turns out that each such compound can be put into one of two normal forms (called the disjunctive normal form and the conjunction normal form, but I can never remember which is which). (a) Show that each such compound E can be written as E = B 1 B n where each B = A 1 A n where each A A E = B 1 B n where each B = A 1 A n where each A A respectively. (b) Suppose we also use ( ) to build up the compound E. What does the normal form now look like. 3.9 In this problem we fix a set S and look at members of PS, that is subsets of S. (a) Let + be the operation on PS given by A + B = (A B) (A B) for sets A, B. (i) Show that A + B = (A B) (B A) for sets A, B. (ii) Show that the operation + is commutative, and A + A = A + S = A for each set A. (iii) Show that the operation + is associative. (iv) Show that (PS, +, ) is an abelian group. (b) Let be the operation on PS given by A B = A B for sets A, B. (i) Show that the operation is commutative and associative with A S = A for each set A. (ii) Show that for all sets A, B, C. (iii) Show that A (B + C) = A B + A C (PS, +,,, S) is a commutative ring of characteristic 2. 10

11 4 Functions A function f : S T S f T source target source target domain co-domain domain co-domain is something that has a source and a target, which are sometimes called the domain and co-domain. Each of these is a set, and the function has a behaviour, which converts each element x S of the source into a specified element f(x) of the target. These are sometimes called the input and output, respectively. Each of the two notations above is used when talking about a function. The version used depends on which is more convenient at the time. Informally, we also refer to the function f when the source and target are known, and need not be mentioned again. A function is often describe by giving a formula for the behaviour. Thus R R x x is an example of a polynomial function on the reals. The integer part function R Z can be defined by x = the unique n Z with n x < n + 1 for arbitrary x R. Observe that this function may not be doing quite what you expect to do on the negative reals. Two functions S 1 f 1 T 1 S 2 f 2 T 2 are equal, f 1 = f 2, if they have the same source and the same target and the same behaviour, that is S 1 = S 2 T 1 = T 2 f 1 (x) = f 2 (x) for each x S 1 = S 2. In practice we sometimes confuse (make equal informally) two functions that have the same behaviour on a sufficiently large set of interest. This often happens with functions of a real variable. Two functions S f F G g T are composible in the order shown if, and only if, F = G. Then the composite S g f T 11

12 is given by (g f)(x) = g(f(x)) for each x S. Sometimes this is written as gf, but that can lead to confusion at times. There are three important properties that a function may of may not have. This function f is A f B injective or an injection if ( x 1, x 2 A)[f(x 1 ) = f(x 2 ) = x 1 = x 2 ] surjective or a surjection if ( y B)( x A)[y = f(x)] bijective or a bijection if f is injective and surjective respectively. A bijection f with the same source and target, A = B is often called a permutation of that set A, even when the set is infinite. For each set A there is a certain function A id A A where the source and target are both A. This is the identity function given by id A (x) = x for each x A. This may not seem an important function, but it is useful in certain circumstances (a bit like the numbers 0 and 1). There are various other notations for this function that you may see, such as 1 A an id A. Sometimes when the set A is obvious it is omitted as a subscript, so id means id A in those circumstances. Given two sets A, B we often want to consider the set of all functions from A to B. Here we write [A, B] or [A B] for that set. Both notations are convenient at different times. You will also see B A and A B in certain places, but not here. As well as passing elements from its source to its target, a function can be used to pass subsets in both directions. Consider an arbitrary function. A f B For each subset X A we let f[x] = {f(x) x X} that is y f[x] ( x X)[y = f(x)] for each y B. This is the direct image of X across f, the set of outputs of f that arise from at least one input from X. The set f[a], the set of all outputs of f, is the range of f. Notice that f is surjective precisely when f[a] = B. For each Y B we extract f (Y ) A by x f (Y ) f(x) Y 12

13 for each x A. This is the inverse image of Y across f. There are variants of this notation. Sometimes f(x) is used for the direct image. The inverse image if often written as f 1 (Y ), but this can be confused with other related notions, such as the inverse function of a bijection. If you see ( ) 1 then be wary and think twice. Observe that the function f produces two functions PB f PA f[ ] between the power sets, and observe which way these go. There are certain properties of these two constructions which are dealt with in Exercise 4.3 and Problem 4.7. Let 2 = {0, 1} where here 0 and 1 are little more than tags. We could use and, or even false and true for the same purposes. Given a set A a characteristic function in A is merely a function χ : A 2 so that [A, 2] is the set of all such functions on A. Each subset X A gives a characteristic function χ X where { 1 if a X χ X (a) = 0 if a / X for each a A. Furthermore, each characteristic function χ gives a subset X A where a X χ(a) = 1 for each a A. Observe that this sets up a bijection between PA and [A, 2]. 4.1 Exercises: Functions 4.1 Consider a composible pair of functions A f B g C each of which may or may not be injective, surjective, or bijective. Sort out the 27 implications of the form f is a-jective and g is b-jective then g f is c-jective where a, b, c are the various possibilities. In each case give a proof of a true implication and a counter-example of a false implication. 4.2 Let f : A B be a function and consider the following statements. (i) The function f is injective. (ii) The function f is surjective. 13

14 (iii) There is a function g : B A with g f = id A. (iv) There is a function g : B A with f g = id B. Which implications hold between these four assertions? In each case give a proof of the implication. Can you see anything odd about (ii) (iv)? 4.3 Let f : A B be a function and consider the related direct and inverse image functions. For arbitrary X 1, X 2 A and Y 1, Y 2 B consider the following statements. (i) f[x 1 X 2 ] = f[x 1 ] f[x 2 ] (ii) f[x 1 X 2 ] = f[x 1 ] f[x 2 ] (iii) f (Y 1 Y 2 ] = f (X 1 ) f (Y 2 ) (iv) f (Y 1 Y 2 ] = f (X 1 ) f (Y 2 ) Which of these are true and which are false? equality, and a counter-example of a false equality. In each case give a proof of a true 4.4 Consider a commuting triangle of functions A h C f g B that is functions with h = g f. Show that the following induced triangles PA h[ ] PC PA h PC f[ ] PB g[ ] f PB g also commute. 4.5 Let A be an arbitrary set. (a) Show that the characteristic function assignment PA [A, 2] X χ X is a bijection. (b) For X A how are χ X and χ X related? (c) For X, Y A how can you calculate χ X Y and χ X Y in terms of χ X and χ Y? (d) Suppose f : B A is an arbitrary function. Each X A gives a subset Y = f (X) B. How are χ X and χ Y related? 14

15 4.2 Problems: Functions 4.6 We know that 2 = {0, 1} is a commutative ring using + 2, addition (mod 2), and standard multiplication Consider any set S together with the set of functions [S, 2], the characteristic functions on S. We know that the ring structure on 2 lifts to one on [S, 2]. (If you don t know this you should go away and prove it, and sort out a more general result.) The two operations are given pointwise, that is (χ + ξ)(x) = χ(x) + 2 ξ(x) (χ ξ)(x) = χ(x) ξ(x) for x S. The two constant functions give the zero and the one of [S, 2] In Problem 3.9 we produced a ring structure on PS, but the proof that addition is associative is a but messy. We can now do better. Consider the bijection PS [S, 2] A χ A which sends each subset A S to its characteristic function. (a) Show that for all A, B PS we have χ A + χ B = χ A+B χ A χ B = χ A B and hence the bijection is an isomorphism. 4.7 The notation in this problem may look a little eccentric, but it does have its uses. Let A f B be a fixed function. The inverse and direct image construction give two functions between the power sets. PA (f) = f[ ] I(f) = f PB Because f is fixed throughout we may write for (f) and I for I(f). (a) Show that each of these is monotone, that is X 1 X 2 = (X 1 ) (X 2 ) Y 1 Y 2 = I(Y 1 ) I(Y 2 ) for all X 1, X 2 PA and Y 1, Y 2 PB. (b) Show that (X) Y X I(Y ) for all X PA and Y PB. (This shows that and I form an adjunction, I, but we don t deal with that notion here.) (c) Now let (X) = (X ) 15

16 for each X PA. Show that this function is monotone. (d) Show that I(Y ) X Y (X) for all X PA and Y PB. (e) Find an example where the functions and are not the same. 4.8 A set A is transitive if A PA holds. Thus if A is transitive then each member of A is a set, and X A = X A holds. (a) Observe that is transitive, and produce an indexed family (n n N) of distinct transitive sets. (The sets you are looking for are called the von Neumann natural numbers.) (b) Show that if A is transitive, then so is PA. Use this to list a much larger family of transitive sets. (c) Show that if A is a family of transitive sets, then A is also transitive. Use this to produce an enormous family of transitive sets. The family you are looking for is called the Zermelo universe. It is inhabited by Klingons but not Vogons. To obtain Vogons you need a much more powerful construction. 5 Induction You should be aware of proof by induction over N, the natural numbers. There is also a related technique of definition by recursion which you will have seen but may not be so familiar with. A simple form of the recursion technique is often called definition by induction, which is nonsense. The following exercises and problems should get you going, in more senses than one. 5.1 Exercises: Inductions 5.1 In this exercise you may assume that each natural number n 2 is divisible by at least one prime. (a) Use induction to show that there are infinitely many primes. (b) Use induction to show that each natural number is a product of primes. (c) What is the difference between these two kinds of induction. 5.2 This question is concerned with iterated powers of 2, and illustrates how such exponentiations can explode. Let ℶ : N N N 2 : N N be the two functions given by ℶ(0, y) = y ℶ(1 + x, y) = 2 ℶ(x,y) 2(x) = ℶ(x, 1) 16

17 for each x, y N. (a) Show that ℶ(x, ℶ(u, y)) = ℶ(x + u, y) for all x, u, y N. (b) Write down explicit values of 2 x for 0 x 16, and suggest approximations for 2(10 n) for small n. (c) Write down explicit values of 2(x) for as many x as possible. (d) Write down explicit values of 2 r (1) for as many r as possible. 5.3 This exercise is concerned with the harmonic series and various approximations to its partial values. There is a couple of simple inductions involved, and you should locate these. Let P = N {0}, the set of strictly positive natural numbers. Let f : P Q be the function given by m f(m) = 1/r for each m P. (a) Show that for each n N. (b) Show that r=1 f(2 n+1 ) f(2 n ) 1/2 f(2 n ) 1 + n/2 for each n N. (c) Show that the harmonic series diverges. (d) Show that f(2 n 1) n for each n P. (e) Show that the harmonic series diverges very slowly. (f) For natural numbers l, r let us write l r to indicate that the two numbers are approximately equal with l slightly bigger. particular is an example. Show that 2 10 n 10 3 n for each n P (but where the discrepancy gets bigger with n). (g) Show that 2 42 is still quite a bit bigger than the Euro-debt. In 5.4 This exercise is concerned with jump functions, that is functions J : [N N] [N N] 17

18 which can be applied to a lower level function f : N N to produce another function J(f) : N N. The general idea is that the rate of growth of J(f) should be faster than f. We can then iterate the jump f J(f) J 2 (f) J 3 (f) to produce faster and faster functions. We look at two examples, the Robinson jump function and the Grzegorczyk jump function. It is interesting to see these developed in parallel. However, you may want to deal with one side before looking at the other. Consider the two jump functions given by R : [N N] [N N] G : [N N] [N N] R(f)(x) = f x+1 (1) G(f)(x) = f x (2) for each f : N N and each x N. We apply this to the successor function, that is S : N N on N (a) Prove the following (Ri) R(S)(x) = 2 + x (Gi) G(S)(x) = 2 + x (Rii) 3 + R 2 (S)(x) = 2(3 + x) (Gii) 2 + G 2 (S)(x) = 2(2 + x) (Riii) 3 + R 3 (S)(x) = 2 3+x (Giii) 2 + G 3 (S)(x) = 2 2+x (Riv) 3 + R 4 (S)(x) = 2(3 + x) (Giv) 2 + G 4 (S)(x) = 2(2 + x) (Rv) 3 + R 5 (S)(x) = 2 3+x (1) (Gv) 2 + G 5 (S)(x) = 2 2+x (1) each by an appropriate use of induction. (b) Why can t we handle 3 + R 6 (S) and 2 + G 6 (S)? 5.2 Problems: Induction 5.5 This series of problems is concerned with the Slow hierarchy, one of the very early attempts to produce an enumeration of R. (They all failed, and we now know why.) The various parts require (a) a 1-place induction (b) a 2-place induction (c) a 3-place induction (d) a 4-place induction and so on. Once you spot what is going on it is very easy to extend the idea to use 5-place induction, 6-placed induction, and so on. For each part we fix a function and then produce a function f : N F : N m+1 18 N N

19 where the arity m + 1 depends on the part. This is produced by an m-placed recursion. At each step we use ( ) to indicate the successor function on N. From part (b) onwards we use an arbitrary function ω : N in the construction. You may wonder why this function is called ω, and does it have anything to do with the ordinal ω. It does, and we may see what the connection is later. (a) Given the function f let F : N 2 N be recursively generated by N (0) F (0, x) = f(x) (1) F (r, x) = F (r, x) + 1 for each r, x N. Suggest an explicit value for F, and verify this by induction on r. (b) Given the function f let F : N 3 N be recursively generated by (0) F (0, 0, x) = f(x) (1) F (i, r, x) = F (i, r, x) + 1 (2) F (i, 0, x) = F (i, ω(x), x) for each i, r, x N. Suggest an explicit value for F, and verify this by induction on i, r. (c) Given the function f let F : N 4 N be recursively generated by (0) F (0, 0, 0, x) = f(x) (1) F (j, i, r, x) = F (j, i, r, x) + 1 (2) F (j, i, 0, x) = F (j, i, ω(x), x) (3) F (j, 0, 0, x) = F (j, ω(x), 0, x) for each j, i, r, x N. Suggest an explicit value for F, and verify this by induction on j, i, r. (d) Given the function f let F : N 4 N be recursively generated by (0) F (0, 0, 0, 0, x) = f(x) (1) F (k, j, i, r, x) = F (k, j, i, r, x) + 1 (2) F (k, j, i, 0, x) = F (k, j, i, ω(x), x) (3) F (k, j, 0, 0, x) = F (k, j, ω(x), 0, x) (4) F (k, 0, 0, 0, x) = F (k, ω(x), 0, 0, x) for each k, j, i, r, x N. Suggest an explicit value for F, and verify this by induction on k, j, i, r. (e) What has the function f got to do with all this? 19

20 5.6 There are two standard forms of induction over N. ( ) (bs) φ(0) ( n)[φ(n) φ(n + 1)] = ( n)φ(n) The Base-Step method (prog) ( n) [ ( m < n)ψ(m) ψ(n) ] = ( n)ψ(n) The Progressive method Here each of φ( ) and ψ( ) is an assertion that may contain other parameters, but these are fixed throughout. Show how each can be used to obtain the other. 20