In mathematics, a multiple of an integer is the product of that integer with another integer. In other words, a is a multiple of b if a = nb, where nis an integer. If b is not zero, this is equivalent to saying that a / b is an integer.0 is a multiple of every integer ().Source: http://en.wikipedia.org/wiki/Multiple_(mathematics)
In normalized scientific notation all numbers are written in the form a x 10^b (a times ten raised to the power of b) where a is a nonzero single-digit integer and b is an integer.
In normalized scientific notation all numbers are written in the form a x 10^b (a times ten raised to the power of b) where a is a nonzero single-digit integer and b is an integer.
He is pretty much responsible for the theory of congruences. Provided the conditions that a and b are integers and n is a positive integer, a and b, are said to be "congruent modulo n" if (a-b)/n is an integer. Written as
It is my first answer. Is the problem to solve A=B^X ? where A and B are positive integers and X the power exponent of B The given equation can be rewritten in a logarithm form. Log A = X * Log B solving for a unique X X = Log A / Log B The result: Any positive integer A can be rewritten as a positive integer B to the distinct power X. Where X is Log A divided by Log B A = B ^(Log A / Log B) I think, this is the solution. Roger Verbeeck
In mathematics, a multiple of an integer is the product of that integer with another integer. In other words, a is a multiple of b if a = nb, where nis an integer. If b is not zero, this is equivalent to saying that a / b is an integer.0 is a multiple of every integer ().Source: http://en.wikipedia.org/wiki/Multiple_(mathematics)
If a and b are integers, then a times b is an integer.
dim a as integer dim b as integer dim c as integer dim d as integer private sub command1_click () a=-1 b=1 d=1 while (d<=10) c=a+b print c a=b b=c next d end sub
It is a number that can be expressed as a fraction but is NOT an integer. For example. 3 is an integer and it is rational since we can write 3/1, but 1/3 is not an integer and it is rational since we wrote it as a fraction or a ratio. Remember that a rational number is one that can be written as A/B where A and B are integers. Now if B is 1, which is certainly an integer, A/1 is rational but since A is an integer, A/1 is an integer.
An even integer is a multiple of 2 so that if x is the even integer then there is some other integer a such that x = 2a. An odd integer is one which, when you divide it by 2, leaves a remainder of 1. That is, if y is the odd integer, then y = 2b + 1 for some other integer b. Now, the sum of the even and odd integer is x + y = 2a + 2b + 1 = 2(a+b) + 1 By the closure of integers under addition, a and b are integers implies that a+b is an integer. So 2(a+b) is even and so the sum is odd.
Because a is rational, there exist integers m and n such that a=m/n. Because b is rational, there exist integers p and q such that b=p/q. Consider a+b. a+b=(m/n)+(p/q)=(mq/nq)+(pn/mq)=(mq+pn)/(nq). (mq+pn) is an integer because the product of two integers is an integer, and the sum of two integers is an integer. nq is an integer since the product of two integers is an integer. Because a+b equals the quotient of two integers, a+b is rational.
A-b
If A and B are multiples of C, then A + B is also a multiple of C: If A is a multiple of C then A = mC for some integer m If B is a multiple of C, then B = nC for some integer n → A + B = mC + nC = (m + n)C = kC where k = m + n and is an integer → A + B is a multiple of C
A factor of a integer is an integer that divides the second integer into a third integer exactly; i.e. A is a factor of B if B/A is exactly C, where all of A, B and C are integers. A prime factor is a factor as above, but is also a prime number. This means that the only factors of that factor are one and the number itself; i.e. A is a prime factor of B if B/A is exactly C andthe only factors of A are 1 and A.
Subtraction a-b
a positive integer A that, if increased or decreased by the same positive integer B, yields 2 positive integers, A+B and A-B, that are both perfect squares" OK... i figured out kinda what it meant... i think the integer B is equal to A-1, like the rectangular number definition: n(n-1)
If a is even then there is an integer x such that a = 2x If b is even then there is an integer y such that b = 2y Then a+b = 2x+2y = 2(x+y) ie 2 is a factor of a+b, that is, a+b is even.