cramya wrote:
That would be 5/6 and remainder =5
Thanks Logitech; dont know what I was smoking there...
You know the rule..puff puff and pass!
So I realized that I do not know some of the remainder concepts.
X = a Y + b
Lets go:
1) b < Y
The remainder for natural numbers
If a and d are natural numbers, with d non-zero, it can be proven that there exist unique integers q and r, such that a = qd + r and 0 ≤ r < d. The number q is called the quotient, while r is called the remainder. The division algorithm provides a proof of this result and also an algorithm describing how to calculate the remainder.
Examples
* When dividing 13 by 10, 1 is the quotient and 3 is the remainder, because 13=1×10+3.
* When dividing 26 by 4, 6 is the quotient and 2 is the remainder, because 26=6×4+2.
* When dividing 56 by 7, 8 is the quotient and 0 is the remainder, because 56=7×8+0.
* When dividing 3 by 10, 3 is the remainder as we always take the front number as the remainder when the second number is of higher value.
The case of general integers
If a and d are integers, with d non-zero, then a remainder is an integer r such that a = qd + r for some integer q, and with 0 ≤ |r| < |d|.
When defined this way, there are two possible remainders. For example, the division of −42 by −5 can be expressed as either
−42 = 9×(−5) + 3
as is usual for mathematicians, or
−42 = 8×(−5) + (−2).
So the remainder is then either 3 or −2.
This ambiguity in the value of the remainder can be quite serious computationally; for mission critical computing systems, the wrong choice can lead to dangerous consequences. In the case above, the negative remainder is obtained from the positive one just by subtracting 5, which is d. This holds in general. When dividing by d, if the positive remainder is r1, and the negative one is r2, then
r1 = r2 + d. :twisted:
The remainder for real numbers
When a and d are real numbers, with d non-zero, a can be divided by d without remainder, with the quotient being another real number. If the quotient is constrained to being an integer however, the concept of remainder is still necessary. It can be proved that there exists a unique integer quotient q and a unique real remainder r such that a=qd+r with 0≤r < |d|. As in the case of division of integers, the remainder could be required to be negative, that is, -|d| < r ≤ 0.
Extending the definition of remainder for real numbers as described above is not of theoretical importance in mathematics; however, many programming languages implement this definition — see modulo operation.
The inequality satisfied by the remainder
The way remainder was defined, in addition to the equality a=qd+r an inequality was also imposed, which was either 0≤ r < |d| or -|d| < r ≤ 0. Such an inequality is necessary in order for the remainder to be unique — that is, for it to be well-defined. The choice of such an inequality is somewhat arbitrary. Any condition of the form x < r ≤ x+|d| (or x ≤ r < x+|d|), where x is a constant, is enough to guarantee the uniqueness of the remainder.
what else ? Feel free to add.
You do not have the required permissions to view the files attached to this post.
