d da db dc dd de df dg dh di dj dk dl dm dn do dp dr ds dt du dw dx dy dz

Перевод: divisor

[существительное]
делитель [мат.]

Тезаурус:

1. Assume the dividend and divisor are integral, in two's complement representation, For definiteness, assume that the remainder is zero or has the same sign as the dividend (see Stein and Munro 1971, Chapter 6 for more details).
2. Thus our assumption, namely that has led to the contradiction (absurdity) that numbers m, n with no common divisor are both even.
3. If the result is positive, it is the new partial remainder and the new quotient bit is one; otherwise the quotient bit is zero, and we restore the previous remainder by adding the divisor to the tentative remainder.
4. (e) For division, when being the dividend and D the divisor (see problem 2.4).
5. In division we have a dividend N and a divisor D, and require a quotient Q and remainder R such that The divisor, quotient, and remainder are n bits long, and the dividend is 2n bits long.
6. If, for instance, in 1.3.1 we had defined a to be a divisor of b if b = ac this would have left open the question of whether or not we are to call a divisor of b if no such c existed.
7. The division algorithm described in 2.1 is restricted to the case of a non-negative dividend, a positive divisor, and no overflow.
8. Proof We suppose so that we may write x = m/n where m, n Z. Clearly we may assume m, n have no common divisor such a common divisor can be eliminated before proceeding.
9. For each stage, the partial remainder is shifted left, and the divisor is subtracted from it to form a new tentative remainder.
10. Extend the algorithm to deal with a general dividend and divisor, and include a test for overflow (i.e. when the quotient cannot be represented in n bits).