89. To determine the sign of the product, observe the following rule : If the multiplier and multiplicand have the same sign, the product is positive; if they have different signs, the product is negative. 1st +ax+b=+ab; because in this case a is to be taken positively b times; therefore the product ab must be positive. 2d. -ax+b=-ab; because a is to be taken b times; that is, we must take gd. - ab. +ax-b=−ab; for a quantity is said to be multiplied by a negative number b, if it be subtracted b times; and a subtracted b times is ab. This also appears from Art. 92. Ex. 2. 4th -ax-b=+ab. Here a is to be subtracted b times, that is, -ab is to be subtracted; but subtracting - ab is the same as adding +ab (Art. 86); therefore we have to add +ab. The 2d, and 4th cases may be thus proved; a-a=0, multiply these equals by b, then ab together with axb must be equal to bx0, or nothing; therefore a multiplied by b must give -ab, a quantity which when added to ab makes the sum nothing. Again, a-a=0; multiply these equals by -b, then - ab together with ax-b must be equal to 0; therefore - ax-b=+ab. 90. If the quantities to be multiplied have coefficients, these must be multiplied together as in common arithmetic; the sign and the literal product being determined by the preceding rules. Thus 3ax5b=15ab; because 3×a×5×b = 3×5×a×b=15ab (Art. 88). · 6d × 4m = 24md. In 91. The powers of the same quantity are multiplied together by adding the indices; thus axa3a", for aax aaaaaaaa. the same manner axala16; and -3a2x3× 5axy2 = −15a3x1Ÿ3. It is a common mistake of beginners to say that an algebraical expression which appears under the form ax0 is equal to a, by supposing it to signify a not multiplied at all; whereas, since axb signifies a taken 6 times, in the same manner ax0 signifies a taken 0 times, and is therefore equal to 0.-ED. To prove generally that a×a"="+", m and n being any positive integers. By Def. Art. 63. a"=axaxaxax &c. continued to m factors; also a"=a×a×a×a× &c. n ............... ........; ‚. a”×a"=a×axa...to m factorsxaxaxa...to n factors, =axaxa...to m+n factors, =am+, by Def. Art. 63. It will be proved hereafter that the same rule holds when m and n are either fractional or negative. (Arts. 132, 162.) Also ab"×ab=aTM.a2.b".ba=aTM+pbn+s ̧ 92. If the multiplier or multiplicand consist of several terms, each term of the latter must be multiplied by every term of the former, and the sum of all the products taken for the whole product of the two quantities. Here a+b is to be taken c+d times, that is, c times and d times, or (a + b)c + (a + b)d. Here a+b is to be taken ed times, that is, c times wanting d times; or c times positively and d times negatively; that is, (a + b)c − (a+b)d, or ac + bc − (ad+bd), or ac+bc - ad - bd, Art. 87. Prod. =x3-(p − a)x2 + (q−ap),r+aq Here the coefficients of 2 and a are bracketed, since - (p-a)x3 ➡−px2 + ax2; and (q − ap)x = qx — apx. Ex. 12. Mult. ma+na" by nam+ma" 2m mna2+n3aTM+" It may be useful to exhibit the Rules for Multiplication as follows:- p(a-b)-pa-pb........ .(1), ..(2), Assuming (1) and (2), which are too obvious to need a proof, to prove (3), let a+b=m, then, N.B. The Rules for the management of Brackets, given in Art. 87, apply only to the Addition and Subtraction of quantities so enclosed. If a collection of quantities within brackets is to be multiplied or divided by any quantity or collection of quantities, the brackets must not be struck out until the multiplication or division is actually performed. Thus (a+b)x (c+d) signifies that a+b is to be taken c+d times, and is obviously not the same as either a+b(c+d), or (a+b)c+d. Again, (a+b)÷(c+d) is not equivalent to either a+b÷(c+d), or (a+b)÷c+d; but it may be written a+b the line which separates the numerator and denominator serving as a vinculum to both. c+d' The learner would do well to practise multiplication of quantities by means of brackets as early as possible. Thus, Ex. 1. (a-b)(c-d)=(a-b)c-(a-b)d, =ac-bc-(ad-bd), =ac-bc-ad+bd. Ex. 2. (x+a)(x+b)=(x+a)x+(x+a)b, = x2+ax+bx+ab, = x2+(a+b)x+ab. Ex. 3. (x+1)(x+2)(x+3)=(x2+2+1.x+2)(x+3). (Ex. 2.) =(x2+3x+2)x+(x2+3x+2)3 It is also useful to commit to memory at an early stage the three following results; as will appear from the subjoined Examples :— ......... (A+B).(A+B) or (A+B)2=A2+B2+2AB..... .(i) .(ii) .(iii) whatever quantities A and B may represent, simple or compound. Ex. 1. (ax+by)(ax+by)=(ax)2+(by)2+2.ax.by, by (i), =a2x2+b3y2+2abxy. Ex. 2. (ax-by)(ax—by)=(ax)2+(by)3–2.ax.by, by (ii), =a2x2+b3y'-2abxy. Ex. 3. (ax+by)(ax—by)=(ax)3— (by)3, by (iii), =a3x3—b3y3. Ex. 4. (ax+b+cy+d)2= (ax+b+cy+d)3, =(ax+b)2+(cy+d)2+2(ax+b)(cy+d), +2acxy+2adx+2bcy+2bd. Ex. 5. (a+b+c)(a+b−c)=(a+b+c)(a+b−c), =(a+b)3-c2, by (iii), =a2+b2+2ab-c2. Ex. 6. (b+c-a)(c+a−b)=(c−a−b)(c+a−b), =c(a-b), by (iii), =c2-(a+b2-2ab), =c-a-b2+2ab, SCHOLIUM. The method of determining the sign of a product from the consideration of abstract quantities has been found fault with by some algebraical writers, who contend that - a, without reference to other quantities, is imaginary, and consequently not the object of reason or demonstration. In answer to this objection we may observe, that whenever we make use of the notation a, and say it signifies a quantity to be subtracted, we make a tacit reference to other quantities. Thus, in numbers, a represents a number to be subtracted from those with which it is connected; and when we suppose - a to be taken b times, we must understand that a is to be taken b times from some other numbers. In estimating lines, or distances, -a represents a line, or distance, in a particular direction. The negative sign does not render quantities imaginary or impossible, but points out the relation of real quantities to others with which they are concerned. |