In incomplete partition the quotient is 3, and the remainders 11 and 17 are in effect disregarded; if, after finding the quotient 3, we want to know what remainder would be produced by'a direct division, the simplest method is to multiply 3 by 2 4 0 and subtract the result from 935.
To compare them, or to add or subtract them, we must express them in terms of the same unit.
The process is the same as that of finding the sum or differ I ence of 3 sixpences and 5 fourpences; we cannot subtract 3 sixpenny-bits from 5 fourpenny-bits, but we can express each as an equivalent number of Ones.
To add or subtract fractional numbers, we must reduce them to a common denominator; and similarly, to multiply or divide surds, we must express them as power-numbers with the same index.
Thus to subtract 5s.
To subtract £3, 5s.
Into 12D., So That We Subtract £3, 5S.
On The Counting System It Will Be Found That, In Determining The Number Of Shillings In The Remainder, We Subtract 5S.
In actual practice, of course, we subtract large multiples at a time.
We first construct the multiple-table C, and then subtract successively zoo times, 30 times and I times; these numbers being the partial quotients.
Similarly we cannot subtract 8 from 15, if 15 means 1 ten + 5 ones; we must either write 15-815-8=(10+5)-8= (I o - 8)+5 = 2+5 = 7, or else resolve the 15 into an inexpressible number of ones, and then subtract 8 of them, leaving 7.
To find the square root of N, we first find some number a whose square is less than N, and subtract a 2 from N.
For example, 3.14 16= 3Xi g t o = 3x l g = 3X g fxz 9= 3(I + 1 1) (I - 2-50-6) Hence, to multiply by 3.1416, we can multiply by 34, and subtract 2 0 o (_ .0004) of the result; or, to divide by 3 .
1416, we can divide by 3, then subtract -h 2 of the result, and then add of the new result.
Dates expressed according to this era are reduced to the common era by subtracting 5502, up to the Alexandrian year 57 86 inclusive, and after that year by subtracting 5492; but if the date belongs to one of the four last months of the Christian year, we must subtract 5503 till the year 5786, and 5493 after that year.
- The era of Tyre is reckoned from the 19th of October, or the beginning of the Macedonian month Hyperberetaeus, in the year 126 B.C. In order, therefore, to reduce it to the common era, subtract 125; and when the date is B.C., subtract it from 126.
Multiply, therefore, the number of Armenian years elapsed by 365; add the number of days from the commencement of the current year to the given date; subtract 176 from the sum, and the remainder will be the number of days from the 1st of January 553 to the given date.
According to this law, Io+3+6-7= Io +3-7+6 = 3+6-7+10 = &c. But, if we write the expression as 3-7+6+10, this means that we must first subtract 7 from 3.
To subtract £3, 5s.
And to find the position, in time, of one event relatively to another, we have only to subtract the date of the second (taking account of its sign) from that of the first.
If the year is after Christ, and the event took place in one of the first six months of the Olympic year, that is to say, between July and January, we must subtract 776 from the number of the Olympic year to find the corresponding year of our era; but if it took place in one of the last six months of the Olympic year, or between January and July, we must deduct 777.
Thus To Deduce The New Moon Of Tisri, For The Year Immediately Following Any Given Year (Y), When Y Is Ordinary, Subtract (1 1 °) Days 15 Hours Ii Min.
If by pre-heating the blast we add to the sum of the heat available; or if by drying it we subtract from the work to be done by that heat the quantity needed for decomposing the atmospheric moisture; or if by removing part of its nitrogen we lessen the mass over which the heat developed has to be spread - if by any of these means we raise the temperature developed by the combustion of the coke, it is clear that we increase the proportion of the total heat which is available for this critical work in exactly the way in which we should increase the proportion of the water of a stream, initially too in.
1 And The Actual Forms For The First Three Weights Are 1 Aobzo, (Ao B 1 A 1 B O) Bo, (A O B 2 A 1 2 0 Bo, Ao(B2, 3 A1B2 A2B1 A O (B L B 2 3B O B 3) A I (B 2 1 2B 0 B 2); Amongst These Forms Are Included All The Asyzygetic Forms Of Degrees 1, 1, Multiplied By Bo, And Also All The Perpetuants Of The Second Binary Form Multiplied By Ao; Hence We Have To Subtract From The 2 Generating Function 1Z And 1 Z Z2, And Obtain The Generating Function Of Perpetuants Of Degrees I, 2.
For the retarded stream the only difference is that we must subtract R from at, and that the limits of x are o and +h.