be obvious, but they are both formed in the same way in terms of recurrence relations. Whatever the sign ofn, the series of coefficientsnCk can be generated by starting withnC0= 1 and using the recurrence relation
nCk+1=n− k k + 1
nCk. (1.59)
The difference is that for positive integer n the series terminates when k = n, whereas for negative n there is no such termination – in line with the infinite series of terms in the corresponding expansion.
Finally we note that, in fact, equation (1.59) generates the appropriate coef-ficients for all values ofn, positive or negative, integer or non-integer, with the obvious exception of the case in whichx =−y and n is negative. For non-integer n the expansion does not terminate, even if n is positive.
1.7 SOME PARTICULAR METHODS OF PROOF
1.7.1 Proof by induction
The proof of the binomial expansion given in subsection 1.5.2 and the identity established in subsection 1.6.1 have already shown the way in which an inductive proof is carried through. They also indicated the main limitation of the method, namely that only an initially supposed result can be proved. Thus the method of induction is of no use for deducing a previously unknown result; a putative equation or result has to be arrived at by some other means, usually by noticing patterns or by trial and error using simple values of the variables involved. It will also be clear that propositions that can be proved by induction are limited to those containing a parameter that takes a range of integer values (usually infinite).
For a proposition involving a parameter n, the five steps in a proof using induction are as follows.
(i) Formulate the supposed result for generaln.
(ii) Suppose (i) to be true for n = N (or more generally for all values of n≤ N; see below), where N is restricted to lie in the stated range.
(iii) Show, using only proven results and supposition (ii), that proposition (i) is true forn = N + 1.
(iv) Demonstrate directly, and without any assumptions, that proposition (i) is true whenn takes the lowest value in its range.
(v) It then follows from (iii) and (iv) that the proposition is valid for all values ofn in the stated range.
(It should be noted that, although many proofs at stage (iii) require the validity of the proposition only forn = N, some require it for all n less than or equal to N – hence the form of inequality given in parentheses in the stage (ii) assumption.) To illustrate further the method of induction, we now apply it to two worked examples; the first concerns the sum of the squares of the firstn natural numbers.
Prove that the sum of the squares of the first n natural numbers is given by
n r=1
r2=16n(n + 1)(2n + 1). (1.60)
As previously we start by assuming the result is true forn = N. Then it follows that
N+1
r=1
r2=
N r=1
r2+ (N + 1)2
=16N(N + 1)(2N + 1) + (N + 1)2
=16(N + 1)[N(2N + 1) + 6N + 6]
=16(N + 1)[(2N + 3)(N + 2)]
=16(N + 1)[(N + 1) + 1][2(N + 1) + 1].
This is precisely the original assumption, but withN replaced by N + 1. To complete the proof we only have to verify (1.60) for n = 1. This is trivially done and establishes the result for all positiven. The same and related results are obtained by a different method in subsection 4.2.5.
Our second example is somewhat more complex and involves two nested proofs by induction: whilst trying to establish the main result by induction, we find that we are faced with a second proposition which itself requires an inductive proof.
Show that Q(n) = n4+ 2n3+ 2n2+n is divisible by 6 (without remainder) for all positive integer values ofn.
Again we start by assuming the result is true for some particular valueN of n, whilst noting that it is trivially true forn = 0. We next examine Q(N + 1), writing each of its terms as a binomial expansion:
Q(N + 1) = (N + 1)4+ 2(N + 1)3+ 2(N + 1)2+ (N + 1)
= (N4+ 4N3+ 6N2+ 4N + 1) + 2(N3+ 3N2+ 3N + 1) + 2(N2+ 2N + 1) + (N + 1)
= (N4+ 2N3+ 2N2+N) + (4N3+ 12N2+ 14N + 6).
Now, by our assumption, the group of terms within the first parentheses in the last line is divisible by 6 and clearly so are the terms 12N2and 6 within the second parentheses.
Thus it comes down to deciding whether 4N3+ 14N is divisible by 6 – or equivalently, whetherR(N) = 2N3+ 7N is divisible by 3.
To settle this latter question we try using a second inductive proof and assume that R(N) is divisible by 3 for N = M, whilst again noting that the proposition is trivially true forN = M = 0. This time we examine R(M + 1):
R(M + 1) = 2(M + 1)3+ 7(M + 1)
= 2(M3+ 3M2+ 3M + 1) + 7(M + 1)
= (2M3+ 7M) + 3(2M2+ 2M + 3)
By assumption, the first group of terms in the last line is divisible by 3 and the second group is patently so. We thus conclude thatR(N) is divisible by 3 for all N ≥ M, and takingM = 0 shows that it is divisible by 3 for all N.
We can now return to the main proposition and conclude that sinceR(N) = 2N3+ 7N is divisible by 3, 4N3+ 12N2+ 14N + 6 is divisible by 6. This in turn establishes that the divisibility ofQ(N + 1) by 6 follows from the assumption that Q(N) divides by 6. Since Q(0) clearly divides by 6, the proposition in the question is established for all values of n.
1.7.2 Proof by contradiction
The second general line of proof, but again one that is normally only useful when the result is already suspected, is proof by contradiction. The questions it can attempt to answer are only those that can be expressed in a proposition that is either true or false. Clearly, it could be argued that any mathematical result can be so expressed but, if the proposition is no more than a guess, the chances of success are negligible. Valid propositions containing even modest formulae are either the result of true inspiration or, much more normally, yet another reworking of an old chestnut!
1.7 SOME PARTICULAR METHODS OF PROOF
The essence of the method is to exploit the fact that mathematics is required to be self-consistent, so that, for example, two calculations of the same quantity, starting from the same given data but proceeding by different methods, must give the same answer. Equally, it must not be possible to follow a line of reasoning and draw a conclusion that contradicts either the input data or any other conclusion based upon the same data.
It is this requirement on which the method of proof by contradiction is based.
The crux of the method is to assume that the proposition to be proved is not true, and then use this incorrect assumption and ‘watertight’ reasoning to draw a conclusion that contradicts the assumption. The only way out of the self-contradiction is then to conclude that the assumption was indeed false and therefore that the proposition is true.
It must be emphasised that once a (false) contrary assumption has been made, every subsequent conclusion in the argumentmust follow of necessity. Proof by contradiction fails if at any stage we have to admit ‘this may or may not be the case’. That is, each step in the argument must be anecessary consequence of results that precede it (taken together with the assumption), rather than simply a possible consequence.
It should also be added that if no contradiction can be found using sound reasoning based on the assumption then no conclusion can be drawn about either the proposition or its negative and some other approach must be tried.
We illustrate the general method with an example in which the mathematical reasoning is straightforward, so that attention can be focussed on the structure of the proof.
A rational number r is a fraction r = p/q in which p and q are integers with q positive.
Further,r is expressed in its lowest terms, any integer common factor of p and q having been divided out.
Prove that the square root of an integerm cannot be a rational number, unless the square root itself is an integer.
We begin by supposing that the stated result isnot true and that we can write an equation
√m = r =p
q for integersm, p, q with q= 1.
It then follows thatp2=mq2. But, sincer is expressed in its lowest terms, p and q, and hencep2andq2, have no factors in common. However,m is an integer; this is only possible ifq = 1 and p2=m. This conclusion contradicts the requirement that q= 1 and so leads to the conclusion that it was wrong to suppose that√
m can be expressed as a non-integer rational number. This completes the proof of the statement in the question.
Our second worked example, also taken from elementary number theory, involves slightly more complicated mathematical reasoning but again exhibits the structure associated with this type of proof.
The prime integers piare labelled in ascending order, thusp1= 1, p2= 2, p5= 7, etc.
Show that there is no largest prime number.
Assume, on the contrary, that there is a largest prime and let it bepN. Consider now the numberq formed by multiplying together all the primes from p1topNand then adding one to the product, i.e.
q = p1p2· · · pN+ 1.
By our assumptionpN is the largest prime, and so no number can have a prime factor greater than this. However, for every primepi,i = 1, 2, . . . , N, the quotient q/pi has the formMi+ (1/pi) withMian integer and 1/pinon-integer. This means thatq/picannot be an integer and sopicannot be a divisor ofq.
Sinceq is not divisible by any of the (assumed) finite set of primes, it must be itself a prime. Asq is also clearly greater than pN, we have a contradiction. This shows that our assumption that there is a largest prime integer must be false, and so it follows that there is no largest prime integer.
It should be noted that the given construction forq does not generate all the primes that actually exist (e.g. forN = 3, q = 7 rather than the next actual prime value of 5, is found), but this does not matter for the purposes of our proof by contradiction.
1.7.3 Necessary and sufficient conditions
As the final topic in this introductory chapter, we consider briefly the notion of, and distinction between, necessary and sufficient conditions in the context of proving a mathematical proposition. In ordinary English the distinction is well defined, and that distinction is maintained in mathematics. However, in the authors’ experience students tend to overlook it and assume (wrongly) that, having proved that the validity of propositionA implies the truth of proposition B, it follows by ‘reversing the argument’ that the validity of B automatically implies that ofA.
As an example, let proposition A be that an integer N is divisible without remainder by 6, and propositionB be that N is divisible without remainder by 2. Clearly, ifA is true then it follows that B is true, i.e. A is a sufficient condition forB; it is not however a necessary condition, as is trivially shown by taking N as 8. Conversely, the same value of N shows that whilst the validity of B is a necessary condition forA to hold, it is not sufficient.
An alternative terminology to ‘necessary’ and ‘sufficient’ often employed by mathematicians is that of ‘if’ and ‘only if’, particularly in the combination ‘if and only if’ which is usually written as IFF or denoted by a double-headed arrow
⇐⇒ . The equivalent statements can be summarised by
A if B A is true if B is true or B =⇒ A, B is a sufficient condition for A B =⇒ A, A only if B A is true only if B is true or A =⇒ B, B is a necessary consequence of A A =⇒ B,
1.7 SOME PARTICULAR METHODS OF PROOF
A IFF B A is true if and only if B is true or B ⇐⇒ A, A and B necessarily imply each other B ⇐⇒ A.
Although at this stage in the book we are able to employ for illustrative purposes only simple and fairly obvious results, the following example is given as a model of how necessary and sufficient conditions should be proved. The essential point is that for the second part of the proof (whether it be the ‘necessary’ part or the
‘sufficient’ part) one needs to start again from scratch; more often than not, the lines of the second part of the proof willnot be simply those of the first written in reverse order.
Prove that (A) a function f(x) is a quadratic polynomial with zeros at x = 2 and x = 3 if and only if (B) the function f(x) has the form λ(x2− 5x + 6) with λ a non-zero constant.
(1) AssumeA, i.e. that f(x) is a quadratic polynomial with zeros at x = 2 and x = 3. Let its form beax2+bx + c with a= 0. Then we have
4a + 2b + c = 0, 9a + 3b + c = 0,
and subtraction shows that 5a + b = 0 and b =−5a. Substitution of this into the first of the above equations givesc =−4a − 2b = −4a + 10a = 6a. Thus, it follows that
f(x) = a(x2− 5x + 6) with a = 0, and establishes the ‘A only if B’ part of the stated result.
(2) Now assume thatf(x) has the form λ(x2− 5x + 6) with λ a non-zero constant. Firstly we note thatf(x) is a quadratic polynomial, and so it only remains to prove that its zeros occur atx = 2 and x = 3. Consider f(x) = 0, which, after dividing through by the non-zero constantλ, gives
x2− 5x + 6 = 0.
We proceed by using a technique known ascompleting the square, for the purposes of illustration, although the factorisation of the above equation should be clear to the reader.
Thus we write
x2− 5x + (52)2− (52)2+ 6 = 0, (x−52)2=14, x−52=±12.
The two roots off(x) = 0 are therefore x = 2 and x = 3; these x-values give the zeros of f(x). This establishes the second (‘A if B’) part of the result. Thus we have shown that the assumption of either condition implies the validity of the other and the proof is complete.
It should be noted that the propositions have to be carefully and precisely formulated. If, for example, the word ‘quadratic’ were omitted fromA, statement B would still be a sufficient condition for A but not a necessary one, since f(x) could then bex3− 4x2+x + 6 and A would not require B. Omitting the constant λ from the stated form of f(x) in B has the same effect. Conversely, if A were to state thatf(x) = 3(x− 2)(x − 3) then B would be a necessary condition for A but not a sufficient one.
1.8 Exercises