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A negative base (or negative radix) may be used to construct a non-standard positional numeral system. Like other place-value systems, each position holds multiples of the appropriate power of the system's base; but that base is negative—that is to say, the base b is equal to −r for some natural number r ( r ≥ 2 ).
This is the minimum number of characters needed to encode a 32 bit number into 5 printable characters in a process similar to MIME-64 encoding, since 85 5 is only slightly bigger than 2 32. Such method is 6.7% more efficient than MIME-64 which encodes a 24 bit number into 4 printable characters.
If an unknown weight W is balanced with 3 (3 1) on its pan and 1 and 27 (3 0 and 3 3) on the other, then its weight in decimal is 25 or 10 1 1 in balanced base-3. 10 1 1 3 = 1 × 3 3 + 0 × 3 2 − 1 × 3 1 + 1 × 3 0 = 25.
8 + (−3) = 8 − 3 = 5 and (−2) + 7 = 7 − 2 = 5. In the first example, a credit of 8 is combined with a debt of 3 , which yields a total credit of 5 . If the negative number has greater magnitude, then the result is negative:
In decimal numbers greater than 1 (such as 3.75), the fractional part of the number is expressed by the digits to the right of the decimal (with a value of 0.75 in this case). 3.75 can be written either as an improper fraction, 375/100, or as a mixed number, 3 + 75 / 100 .
With base e the natural logarithm behaves like the common logarithm in base 10, as ln(1 e) = 0, ln(10 e) = 1, ln(100 e) = 2 and ln(1000 e) = 3 (or more precisely the representation in base e of 3, which is of course a non-terminating number).
A straightforward decimal rank system with a word for each order (10 十, 100 百, 1000 千, 10,000 万), and in which 11 is expressed as ten-one and 23 as two-ten-three, and 89,345 is expressed as 8 (ten thousands) 万 9 (thousand) 千 3 (hundred) 百 4 (tens) 十 5 is found in Chinese, and in Vietnamese with a few irregularities.
In general, if an increase of x percent is followed by a decrease of x percent, and the initial amount was p, the final amount is p (1 + 0.01 x)(1 − 0.01 x) = p (1 − (0.01 x) 2); hence the net change is an overall decrease by x percent of x percent (the square of the original percent change when expressed as a decimal number).