VII

useful when we consider the meaning to be assigned to the operations of the addition and multiplication of ordered couples.

plus 0.75em minus 0.25em We will now go on to this question, and ask what meaning we shall find it convenient to assign to the addition of the two ordered couples $(x, y)$ and $(x^{'}, y^{'})$ . The interpretation must, (a) make the result of addition to be another ordered couple, (b) make the operation commutative so that $(x, y) + (x^{'}, y^{'}) = (x^{'}, y^{'}) + (x, y)$ , (c) make the operation associative so that ${(x, y) + (x^{'}, y^{'})} + (u, v) = (x, y) + {(x^{'}, y^{'}) + (u, v)},$ (d) make the result of subtraction unique, so that when we seek to determine the unknown ordered couple $(x, y)$ so as to satisfy the equation $(x, y) + (a, b) = (c, d),$ there is one and only one answer which we can represent by $(x, y) = (c, d) - (a, b) .$

All these requisites are satisfied by taking $(x, y) + (x^{'}, y^{'})$ to mean the ordered couple $(x + x^{'}, y + y^{'})$ . Accordingly by definition we put $(x, y) + (x^{'}, y^{'}) = (x + x^{'}, y + y^{'}) .$ Notice that here we have adopted the mathematical habit of using the same symbol $+$ in different senses. The $+$ on the left-hand side of the equation has the new meaning of $+$ which we are just defining; while the two $+$ 's on the right-hand side have the meaning of the addition of positive and negative numbers (operations) which was defined in the last chapter. No practical confusion arises from this double use.

As examples of addition we have

The meaning of subtraction is now settled for us. We find that $(x, y) - (u, v) = (x - u, y - v) .$ Thus $(+ 3, + 2) - (+ 1, + 1) = (+ 2, + 1),$ and $(+ 1, - 2) - (+ 2, - 4) = (- 1, + 2),$ and $(- 1, - 2) - (+ 2, + 3) = (- 3, - 5) .$

It is easy to see that $(x, y) - (u, v) = (x, y) + (- u, - v) .$ Also $(x, y) - (x, y) = (0, 0) .$ Hence $(0, 0)$ is to be looked on as the zero ordered couple. For example $(x, y) + (0, 0) = (x, y) .$

The pictorial representation of the addition of ordered couples is surprisingly easy. 10

plus 0.75em minus 0.25em Let $O P$ represent $(x, y)$ so that $O M = x$ and $P M = y$ ; let $O Q$ represent $(x_{1}, y_{1})$ so that $O M_{1} = x_{1}$ and $Q M_{1} = y_{1}$ . Complete the parallelogram $O P R Q$ by the dotted lines $P R$ and $Q R$ , then the diagonal $O R$ is the ordered couple $(x + x_{1}, y + y_{1})$ . For draw $P S$ parallel

to $O X$ ; then evidently the triangles $O Q M_{1}$ and $P R S$ are in all respects equal. Hence $M M^{'} = P S = x_{1}$ , and $R S = Q M_{1}$ and therefore

$O M^{'} = O M + M M^{'} = x + x_{1}, R M^{'} = S M^{'} + R S = y + y_{1} .$

Thus $O R$ represents the ordered couple as required. This figure can also be drawn with $O P$ and $O Q$ in other quadrants.

It is at once obvious that we have here come back to the parallelogram law, which