Procedure: 1. Divide the integers like you always do. 2. Follow these rules: (Note:+ is a positive integer, - is a negative integer and * is the multiplication symbol) a.+*+=+ b.+*-=- c.-*+=- d.-*-=+ Follow these rules and you will have your answer in no time.
One way to do it might be like this:Get / store the integer to be subtracted from, and the integer to subtract-we'll call them A and B.Decrement A and B by 1 each.If B = 0, stop. A now contains the result.If B ≠ 0, decrement A and B by one again.Repeat from step 3.
A rational number can be stated in the form a/b where and b are integers. Adding or multiplying such numbers always gives another number that can be expressed in this form also. So it is also rational.
No, they are not because fractions can be negative also. fractions aren't integers
The two integers are A and A+40 or, equivalently, B and B-40.
The algorithm is A/B * C/D = AB/CD.
Procedure: 1. Divide the integers like you always do. 2. Follow these rules: (Note:+ is a positive integer, - is a negative integer and * is the multiplication symbol) a.+*+=+ b.+*-=- c.-*+=- d.-*-=+ Follow these rules and you will have your answer in no time.
If a and b are integers, then a times b is an integer.
if you have the expression a + b*sqrt(c), the radical conjugate is a - b*sqrt(c). this is important because multiplying those two expressions together gives you an integer if a, b, and c are integers.
algorithm GCD (a, b) is:while (a b) doif a > b then a := a - b else b := b - aend whilereturn a
One way to do it might be like this:Get / store the integer to be subtracted from, and the integer to subtract-we'll call them A and B.Decrement A and B by 1 each.If B = 0, stop. A now contains the result.If B ≠ 0, decrement A and B by one again.Repeat from step 3.
Any negative integer can be factored to -1 times its positive value. Because negative one times itself is positive one, when multiplied by each other they cancel out. So if you're multiplying a negative integer A by a negative integer B. Replace A and B with -1*|A| and -1*|B| (You can do this because you know A and B are negative), and use the distributive property to rearrange them. Now you can see the -1*-1 term and equate it to 1, leaving only the |A| and |B| behind. Because two positive numbers multiplied together are always positive, the result will always be positive. Represented algebraically, as long as A and B are negative integers, the following is true: AB = -1|A|*-1|B| = -1*-1|AB| = |AB|.
A rational number can be stated in the form a/b where and b are integers. Adding or multiplying such numbers always gives another number that can be expressed in this form also. So it is also rational.
The algorithm can be easily stated as follows: if A is greater than B then return A, otherwise return B.
No, they are not because fractions can be negative also. fractions aren't integers
1.Declare three variables asint a,b,c;2.Get the Input of two numbers of Integers from the user:scanf("%d",&a);scanf("%d",&b);3.add a and b and store the result in c4. print c
A greedy algorithm will return as many results as possible. It depends on the algorithm what that means.An example would be in regular expressions. The regexp "/(a.+b)/" searches for a string that starts with "a" and ends with "b". So in the string "There's a bunny in the basket" a greedy algorithm would find "a bunny in the b", while a non-greedy search would find "a b".