Numbers. Definition :Heading 4...1
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1 Numbers Table of Contents Definition :Heading NUMBERS We are all familiar with numbers, from an early age. We develop an intuitive grasp of counting, adding, multiplying and so on. Here we show how they can be based on some logical foundation. All we have are sets (and relations, and functions, which are both types of set) so we want a set-theoretic basis. For example, 1+1 = 2 Does it? What exactly are 1, +, = and 2, in terms of sets? Similarly, JavaScript already can handle numbers. But can we construct our own numbers in code, just starting from sets? THE NATURAL NUMBERS AND PEANO The natural numbers are the set 0,1,2,3,4... Some people miss out 0. They are natural numbers in the sense that they are used to count things. This set is usually called N. Note the difference between a number and its name. Three, 3, trois in French, III in Roman numerals, 11 base 2 are all ways of writing the same number. The most common way of providing a logical foundation is a set of axioms first set out by Giuseppe Peano (Italian, ): 1. There is a set named N which contains an element, named 0 It is tempting to say something we are trying to define. N contains at least one element. But we cannot, since 'one' is 2. There is a function succ (the successor function) with domain and co-domain N 3. succ is one-to-one 4. For all n N succ(n) = N. For every n N, succ(n) is also an element of successor of any natural number. 0 is false. In other words, 0 is not the
2 This asserts, as axioms, that N exists, that 0 is an element, and that succ is a function. Informally it says there is 0, and the number that follows that, and the numberthat follow sthat, and so on. Peano gives characteristics of this set N, without actually sayingwhat the elements of N are. But what exactly is 0, and what is the succ function? If we do not know what it is, how could we code it? One version is due to John von Neumann (American, ), as follows: 0 is the null set Φ succ(n) = n n That is it. so the successor of a number is a new set. It contains all the elements in the set which is the number, n, together with a new elements, which is n. ( Note the difference between a set A and the set A. A might contain any number of elements, but A contains just one - the set A ) We use the notation rather than 0, 1, 2, 3. This is because are already associated with many properties which do not have formal definitions, and we want to start with only properties which are stated explicitly. So 0 = 1 = 0 2 = 0, 1 3 = 0, 1, 2 and so on. This is convenient, since it means n is a set with n elements. But that does not mean that 5, say, is a set with 5 elements, since that would be a circular definition. It is also vague, since we have not yet defined what we mean by 'counting' - but we will later. This is not the only way to do this. Zermelo initially suggested: 0 = 1 = 0 = 2 = 1 = 3 = 2 =
3 In this representation, each n is a set with just one element - n-1. The set n contains another set, which contains another, containing another.. down to the the null set, with n 'inclusions'. We can relate von Neumann's version as iteration, one number after another, whilst Zermelo is recursive, one number inside another. We will code the von Neumann version, simply: function succ(n) return n.union(new Set([n])); so for example: var zero = new Set([]); setshow(zero); newline(); var one = succ(zero); setshow(one); newline(); var two = succ(one); setshow(two); newline(); var three = succ(two); setshow(three); newline(); which outputs zero one two and three: It is not easy to interpret these sets as numbers. But they show how the natural numbers can be treated as sets. Fortunately we can just write 0, 1, 2, and 3. For convenience, it is useful to set up the inverse function to succ, prev, the previous number: function prev(n) var zero = new Set([]); if (n.equals(zero)) return undefined; var len = n.elements.length-1; // copy the array of elements, missing out the last one var e = n.elements.slice(0, len); var result = new Set(e); // make a new set from this return result; This works because we can go 'backwards' by missing out the last element from the array. Real sets are not ordered, so there is no 'first' element. But this works for our implementation (a Zermolo implementation would need to be different). ADDITION AND MULTIPLICATION We define addition as a function, mapping pairs of natural numbers to natural numbers (like 3+4 = 7). The function is defined recursively like this: add(a, 0) = a add(a, succ(b)) = succ( add(a, b ) )
4 How does this work? So to add two numbers, the result is the successor of adding the previous number. As the recursion unwinds, we must eventually get to adding zero - and a+0 = a For example 1+1 = add(1,succ(0)) = succ(add (1,0)) = succ(1) = 2 and 1+2 = add(1, succ(1)) =succ(add(1,1)) = succ( add(1, succ(0)) ) = succ( succ(add(1,0))) = succ(succ(1)) = succ(2) = 3 Who knew 1+2=3 was so complicated? Similarly we can define multiplication recursively: mul(a, 0) = 0 mul(a, succ(b)) = add(a, mul(a,b)) In other words a.0=0 a.(b+1) = a + a.b add(a, succ( b )) = succ( add(a, b ) ) this this is the number before THE SIGNED INTEGERS So we have defined the natural numbers N in set-theoretic terms and using Peano's axioms, and have defined addition and multiplication which work as we expect (in the sense of agreeing with our intuitive grasp of adding and multiplying). But of course we cannot find an element of N such that 3 + n = 2 We want to define the signed integers, Z ( from the German Zahlen, number). We start with N 2, the Cartesian Product N X N. So this is the set of ordered pairs b (a,b), where a and b N We define a relation R on this such that (a,b)r(c,d) iff a+d = b+c (it will turn out that (a,b) corresponds to a-b, but we have not defined 4 3 T 2 T 1 T 0 T a subtraction yet). The signed integers
5 Here we show the set of elements (a,b) for all of which (1,0)R(a,b) is true. This relation therefore creates a set of equivalence classes. Z is this set of equivalence classes. So (5,3) = (4,2) = (3,1) = 2 in 'normal' notation, 2 is the name of the equivalence class (a+2,a) where a N By (5,3) = (4,2) we mean (5,3) and (4,2) belong to the same equivalence class. And the negative integers are, for example, (3,5)=(2,4) = -2 EXERCISE 2 Show that R is an equivalence relation. ADDITION AND MULTIPLICATION OF INTEGERS We can define addition on A=(a,b) and B = (c,d) as A B = (a + c, b + d) Here means the addition we are defining, while + means addition of natural numbers as already defined. so for example 2 2 = (5,3) (6,4) = (11,7) =11-7 = 4 Can we find an additive identity? In other words an element 0 such that A 0 = A for all A? Yes, the equivalence class (2,2) = (1,1) = (x,x) where x is any N since if A = (a,b), A 0 = (a+x, b+x) = (a,b) = A What is the additive inverse of A? If A = (a,b), it is (b,a), since A A -1 = (a+b, b+a) = (a,a) = 0 and multiplication as A B = (ac + bd, ad + bc) We define an order A < B iff a+d < b+c. An integer is positive iff 0 < A, and negative iff A < 0
6 THE RATIONAL NUMBERS Obviously we cannot find an integer a such that 2. a = 3 We need the rational numbers Q ( numbers like 3/2 ). We do something similar to the integers: Start with the set Z X ( Z-0) ( We do not allow division by 0 ) Define an equivalence relation R on this such that (a,b)r(c,d) iff a.d = b.c (so (a,b) is like a/b. The multiplication is as defined on Z). The rational numbers are the equivalence classes of R Addition on Q is defined as (a, b) (c, d) = (ad + bc, bd) and multiplication is (a, b) (c, d) = (ac, bd) Look at the website to see how this could be coded. ANSWERS 1. For example function mult(a, b) var zero = new Set([]); if (b.equals(zero)) return zero; return add(a, mult(a, prev(b))); which could be tested by: var zero=new Set([]); var one = succ(zero); var two = succ(one); setshow(two); newline(); var n = mul(two, two); setshow(n); newline(); 2. We need to show that R is reflexive, symmetric and transitive: a+b = a + b implies (a,b)r(a,b) so its reflexive a+d = c + d implies c+d = a+d so (a,b)r(c,d) implies (c,d)r(a,b) so its is symmetric if (a,b)r(c,d) and (c,d)r(e,f) then a+d = b+c and c+f = d+e so a + d + c + f = b + c + md + e so a+f = b + e so (a,b)r(e,f) so it is transitive. 3. For example: this.mul = mul; function mul(other) return new Integer(this.a*other.a+ this.b*other.b, this.a*other.b+this.b*other.a); used as var x=new Integer(6,3); // 3 var y=new Integer(2,0); // 2 var prod = x.mul(y); prod.show(); // 6
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