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The first step is to determine a common denominator D of these fractions – preferably the least common denominator, which is the least common multiple of the Q i. This means that each Q i is a factor of D , so D = R i Q i for some expression R i that is not a fraction.
In elementary algebra, root rationalisation (or rationalization) is a process by which radicals in the denominator of an algebraic fraction are eliminated.. If the denominator is a monomial in some radical, say , with k < n, rationalisation consists of multiplying the numerator and the denominator by , and replacing by x (this is allowed, as, by definition, a n th root of x is a number that ...
As (+) = and (+) + =, the sum and the product of conjugate expressions do not involve the square root anymore. This property is used for removing a square root from a denominator , by multiplying the numerator and the denominator of a fraction by the conjugate of the denominator (see Rationalisation ).
To begin solving, we multiply each side of the equation by the least common denominator of all the fractions contained in the equation. In this case, the least common denominator is () (+). After performing these operations, the fractions are eliminated, and the equation becomes:
Generalization to fractions is by multiplying the numerators and denominators, respectively: = (). This gives the area of a rectangle A B {\displaystyle {\frac {A}{B}}} high and C D {\displaystyle {\frac {C}{D}}} wide, and is the same as the number of things in an array when the rational numbers happen to be whole numbers.
Any equation containing fractions or rational expressions can be simplified by multiplying both sides by the least common denominator. This step is called clearing fractions . Rule of three
In other words, multiply the remainder by 100 and add the two digits. This will be the current value c. Find p, y and x, as follows: Let p be the part of the root found so far, ignoring any decimal point. (For the first step, p = 0.) Determine the greatest digit x such that (+).
Karatsuba's basic step works for any base B and any m, but the recursive algorithm is most efficient when m is equal to n/2, rounded up. In particular, if n is 2 k , for some integer k , and the recursion stops only when n is 1, then the number of single-digit multiplications is 3 k , which is n c where c = log 2 3.