## 10. Algebraical operations with real numbers.

We now proceed to define the meaning of the elementary algebraical operations such as addition, as applied to real numbers in general.

(i)  Addition.  In order to define the sum of two numbers $$\alpha$$ and $$\beta$$, we consider the following two classes: (i) the class $$(c)$$ formed by all sums $$c = a + b$$, (ii) the class $$(C)$$ formed by all sums $$C = A + B$$. Plainly $$c < C$$ in all cases.

Again, there cannot be more than one rational number which does not belong either to $$(c)$$ or to $$(C)$$. For suppose there were two, say $$r$$ and $$s$$, and let $$s$$ be the greater. Then both $$r$$ and $$s$$ must be greater than every $$c$$ and less than every $$C$$; and so $$C – c$$ cannot be less than $$s – r$$. But $C – c = (A – a) + (B – b);$ and we can choose $$a$$, $$b$$, $$A$$, $$B$$ so that both $$A – a$$ and $$B – b$$ are as small as we like; and this plainly contradicts our hypothesis.

If every rational number belongs to $$(c)$$ or to $$(C)$$, the classes $$(c)$$$$(C)$$ form a section of the rational numbers, that is to say, a number $$\gamma$$. If there is one which does not, we add it to $$(C)$$. We have now a section or real number $$\gamma$$, which must clearly be rational, since it corresponds to the least member of $$(C)$$. In any case we call $$\gamma$$ the sum of $$\alpha$$ and $$\beta$$, and write $\gamma = \alpha + \beta.$

If both $$\alpha$$ and $$\beta$$ are rational, they are the least members of the upper classes $$(A)$$ and $$(B)$$. In this case it is clear that $$\alpha + \beta$$ is the least member of $$(C)$$, so that our definition agrees with our previous ideas of addition.

(ii)  Subtraction. We define $$\alpha – \beta$$ by the equation $\alpha – \beta = \alpha + (-\beta).$ The idea of subtraction accordingly presents no fresh difficulties.

Example V

1. Prove that $$\alpha + (-\alpha) = 0$$.

2. Prove that $$\alpha + 0 = 0 + \alpha = \alpha$$.

3. Prove that $$\alpha + \beta = \beta + \alpha$$. [This follows at once from the fact that the classes $$(a + b)$$ and $$(b + a)$$, or $$(A + B)$$ and $$(B + A)$$, are the same, since, , $$a + b = b + a$$ when $$a$$ and $$b$$ are rational.]

4. Prove that $$\alpha + (\beta + \gamma) = (\alpha + \beta) + \gamma$$.

5. Prove that $$\alpha – \alpha = 0$$.

6. Prove that $$\alpha – \beta = -(\beta – \alpha)$$.

7. From the definition of subtraction, and Exs. 4, 1, and 2 above, it follows that $(\alpha – \beta) + \beta = \{\alpha + (-\beta)\} + \beta = \alpha + \{(-\beta) + \beta\} = \alpha + 0 = \alpha.$ We might therefore define the difference $$\alpha – \beta = \gamma$$ by the equation $$\gamma + \beta = \alpha$$.

8. Prove that $$\alpha – (\beta – \gamma) = \alpha – \beta + \gamma$$.

9. Give a definition of subtraction which does not depend upon a previous definition of addition. [To define $$\gamma = \alpha – \beta$$, form the classes $$(c)$$$$(C)$$ for which $$c = a – B$$, $$C = A – b$$. It is easy to show that this definition is equivalent to that which we adopted in the text.]

10. Prove that $\big||\alpha| – |\beta|\big| \leq |\alpha \pm \beta| \leq |\alpha| + |\beta|.$

## 11. Algebraical operations with real numbers (continued)

Multiplication. When we come to multiplication, it is most convenient to confine ourselves to positive numbers (among which we may include $$0$$) in the first instance, and to go back for a moment to the sections of positive rational numbers only which we considered in §§ 4-7. We may then follow practically the same road as in the case of addition, taking $$(c)$$ to be $$(ab)$$ and $$(C)$$ to be $$(AB)$$. The argument is the same, except when we are proving that all rational numbers with at most one exception must belong to $$(c)$$ or $$(C)$$. This depends, as in the case of addition, on showing that we can choose $$a$$$$A$$, $$b$$, and $$B$$ so that $$C – c$$ is as small as we please. Here we use the identity $C – c = AB – ab = (A – a)B + a(B – b).$

Finally we include negative numbers within the scope of our definition by agreeing that, if $$\alpha$$ and $$\beta$$ are positive, then $(-\alpha)\beta = -\alpha\beta,\quad \alpha(-\beta) = -\alpha\beta,\quad (-\alpha)(-\beta) = \alpha\beta.$

In order to define division, we begin by defining the reciprocal $$1/\alpha$$ of a number $$\alpha$$ (other than zero). Confining ourselves in the first instance to positive numbers and sections of positive rational numbers, we define the reciprocal of a positive number $$\alpha$$ by means of the lower class $$(1/A)$$ and the upper class $$(1/a)$$. We then define the reciprocal of a negative number $$-\alpha$$ by the equation $$1/(-\alpha) = -(1/\alpha)$$. Finally we define $$\alpha/\beta$$ by the equation $\alpha/\beta = \alpha \times (1/\beta).$

We are then in a position to apply to all real numbers, rational or irrational, the whole of the ideas and methods of elementary algebra. Naturally we do not propose to carry out this task in detail. It will be more profitable and more interesting to turn our attention to some special, but particularly important, classes of irrational numbers.

Example VI
Prove the theorems expressed by the following formulae:

 1. $$\alpha \times 0 = 0 \times \alpha = 0$$. 2. $$\alpha \times 1 = 1 \times \alpha = \alpha$$. 3. $$\alpha \times (1/\alpha) = 1$$. 4. $$\alpha\beta = \beta\alpha$$. 5. $$\alpha(\beta\gamma) = (\alpha\beta)\gamma$$. 6. $$\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$$. 7. $$(\alpha + \beta)\gamma = \alpha\gamma + \beta\gamma$$. 8. $$|\alpha\beta| = |\alpha|\, |\beta|$$.