This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1879 Excerpt: ... ay. For ax--ay=a (x--y); but x--y is positive only, and a negative only; hence a (x--y) is negative only. (Art. 77.) Note.--In what follows, unless the contrary is stated, the fundamental symbols employed are supposed to be single and positive. XIII. DIVISION. 148. To find the primary parts into which a universe is ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1879 Excerpt: ... ay. For ax--ay=a (x--y); but x--y is positive only, and a negative only; hence a (x--y) is negative only. (Art. 77.) Note.--In what follows, unless the contrary is stated, the fundamental symbols employed are supposed to be single and positive. XIII. DIVISION. 148. To find the primary parts into which a universe is divided by any number of independent selective operations x, y, z, etc. Suppose we have two symbols x and y. Now X+(i--x)--i, and y+(i-y) = i, multiply these two equations together, xy+x( i--y)+(i--x)y + (i-x)( i--y) = i Since x and y are each positive, the Fig. 5. parts xy, x(i-y), (i-x)y, (i-x)(i-y) are each positive (Art. 120). Also their sum is equivalent to the whole. Since each of the terms is positive, and their sum single, they must be exclusive of one another (Art. 145); that is, each term, each sum of two, each sum of three is single. These then are the primary parts of i, when we have two independent operations x and y. Fig. 5. 149. There are other sets of positive parts, the sum of which is equivalent to the whole; but they each consist of fewer terms than the primary set. For example: since xy+x(i--y)=x; x+(i--x)y+(i--x)(i--y)=i, where we have three positive parts. 150. Suppose that we have three symbols, x, y, z. Multiply xy+x(i--y)+(i--x)y+(i--x)(i--y)=i by z+(i--z)=i; then xyz+xy(i--z)+x(i--y)z+x(i--y)(i--z) + (i--x)yz + (i-x)y(i-z)+(i-x)(i-y)z+(i-x)(i-y)(i-z)=i. It may be shown, as before, that xyz, xy i--z), etc., are each positive; and that they are exclusive of one another. Hence they are the primary parts, into which the universe is divided by the three independent operations x, y, z. And so for any number of independent symbols. 151. The mode, in which three independent symbols divide a universe, can be well illustrated by ...
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