Logic and Deduction

2.1 Introduction

We said in Chapter 1 that if we start with the equation

x+3=√1+x−√1−x

then an apparently correct argument leads us to show that =35 or x = –1, yet neither of these numbers satisfies the equation. The process produces spurious solutions. We also saw the converse phenomenon with the cubic equation: that process produces only correct solutions, but it does not find all of them. In both cases we have to do additional work to modify the list of solutions we first obtain. What on earth is happening here?

Let us begin with the equation

x+3=√1+x−√1−x.

We square both sides of this, to produce

+3=1+x−2√1+x√1−x+1−x

and carry on from there. Now what we really mean at this step is that if √ (x + 3) = √ (1 + x) − √ (1 − x) then the square of the left hand side equals the square of the right hand side. From that second step we notice that if it is true then

+1=−2√1+x√1−x

and, omitting some detail, then =35 or x = − 1. That is, what we have actually shown is that if x satisfies the original equation then =35 or x = − 1. Nobody has yet said that x does satisfy the original equation, but we have obtained some useful information. If x satisfies the equation then it has to be one of two numbers, so we have cut down the range of potential solutions from the whole collection of real numbers to just two, and we can check individually whether or not these two potential solutions do satisfy the equation. The checking process establishes whether or not the statements “=35 is a solution” or “x = − 1 is a solution” are true. This could be put slightly more formally by saying that we check whether or not it is true to say that “if x = 5 then √ (x + 3) = √ (1 + x) − √ (1 − x)”.

The checking process may be tedious and it is obviously important to know when we need to use it. Moreover, if we consider the example of the cubic equation, we see that we also need to know what to check. For the cubic equation y3 + py + q = 0, we find s and t satisfying s – t = –q and st = (p/3)3 and then set =s13−t13. The logic here is that if we can find s and t and construct y, then that y satisfies the equation y3 + py + q = 0. That is, the y we find will be a solution of the cubic, but the process is tantamount to saying that a number of a certain kind is a solution; it does not tell us whether some numbers of a different kind are also solutions.

The point to notice here is that in both cases what we have actually shown to be true is a statement of the form “If A then B” where A and B are statements themselves. This is actually the heart of mathematics, in that the subject proceeds by noticing that if one thing holds then necessarily some other statement has to be true. The statement “if A then B” is often expressed as “A implies B” and put in symbols as A ⇒ B. What this tells us is that the two statements A and B are related in such a way that should A happen to be true then B must also be true. It does not indicate whether or not A and B themselves arc true, nor does it give us any useful information about B in the case where A happens to be false. From the two statements A ⇒ B and B ⇒ C we can deduce that A ⇒ C.

Let us return to the equation √ (x + 3) = √ (1 + x) − √ (1 − x). Then, using implication signs, we have

+3=1+x−1−x⇒x+3=1+x−21+x1−x+1−x⇒x+1=−21+x1−x⇒x2+2x+1=4−4x2⇒5x2+2x−3=0⇒5x−3(x+1)=0⇒5x−3=0orx+1=0

where we mean here that each statement implies the one immediately following it, so we conclude that

x+3=√1+x−√1−x⇒x=35orx=−1.

This is not exactly what we want, but it is a great step in the right direction, since we have now shown that at most two numbers are candidates for solutions of the equation √ (x + 3) = √ (1 + x) − √ (1 − x). We now need to check the truth of the opposite implications, e.g. x = − 1 ⇒ √ (x + 3) = √ (1 + x) − √ (1 − x), which happens to be false. Also =35 does not imply that √ (x + 3) = √ (1 + x) − √ (1 − x) so that neither of the candidates is a solution. We conclude that the equation has no real solutions.

2.2 Implication

There are, then, two possible implications connecting the statements A and B: “A ⇒ B” and “B ⇒ A”. For a given pair of statements A and B it may happen that one of these implications is true but not the other, or they may both be true, or neither may be true. The most satisfactory case, which occurs frequently, is when both are true. For example, if A is “x = y + 5” and B is “x – y = 5” then A ⇒ B (if A is true, subtract y from both sides and we see that B is true), and B ⇒ A (if B is true, add y to both sides and we see that A is true). Since we can write B ⇒ A just as neatly as A ⇐ B (and we take that to be the meaning of ⇐) we shall write A ⇔ B to mean “A ⇒ B and B ⇒ A”. This means that if A is true then B is true, and if B is true then A is true, so that either A and B are both true or they are both false. This is the most useful logical connection, although in practice it is usually easier to establish “A ⇒ B” and “B ⇒ A” separately. “A ⇔ B” is usually pronounced “A (is true) if and only if B (is true)”.

It is important to distinguish between the three logical connectives ⇔, ⇒ and ⇐, particularly since everyday speech often blurs the distinction. If, in ordinary life, I say “I will give you this book if you pay me a pound”, then in addition to the statement just made there is an unspoken presumption that I will not give you the book if you do not give me the pound. Everyday human behaviour may lead us to that conclusion, and it may be correct, but it is not what was actually said.

In our arguments above we used the statement

+3=1+x−1−x⇒x+3=1+x−21+x1−x+1−x.

This is as much as we can say immediately: if two numbers are equal then their squares are equal. The converse is false in general since the squares of two numbers may well be equal without the two numbers being equal; for example 22 = (–2)2. We cannot say that the right hand side above implies the left hand side without additional information.

Consider a simpler...