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An irreducible fraction (or fraction in lowest terms, simplest form or reduced fraction) is a fraction in which the numerator and denominator are integers that have no other common divisors than 1 (and −1, when negative numbers are considered). [1]
For example, is not in lowest terms because both 3 and 9 can be exactly divided by 3. In contrast, is in lowest terms—the only positive integer that goes into both 3 and 8 evenly is 1. Using these rules, we can show that 5 / 10 = 1 / 2 = 10 / 20 = 50 / 100 , for example.
In number theory the standard unqualified use of the term continued fraction refers to the special case where all numerators are 1, and is treated in the article Simple continued fraction. The present article treats the case where numerators and denominators are sequences { a i } , { b i } {\displaystyle \{a_{i}\},\{b_{i}\}} of constants or ...
Using this form for f(n), it can be shown that the parity sequences for two numbers m and n will agree in the first k terms if and only if m and n are equivalent modulo 2 k. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Hailstone cycles, then their corresponding parity ...
The numbers or the objects to be added in general addition are collectively referred to as the terms, [6] the addends [7] [8] [9] or the summands; [10] this terminology carries over to the summation of multiple terms. This is to be distinguished from factors, which are multiplied.
Arithmetic is closely related to number theory and some authors use the terms as synonyms. [8] However, in a more specific sense, number theory is restricted to the study of integers and focuses on their properties and relationships such as divisibility, factorization, and primality. [9] Traditionally, it is known as higher arithmetic. [10]
The lowest common denominator of a set of fractions is the lowest number that is a multiple of all the denominators: their lowest common multiple. The product of the denominators is always a common denominator, as in: + = + =
As () is a repeated factor, we now need to find two numbers, as so we need an additional relation in order to solve for both. To write the relation of numerators the second fraction needs another factor of ( 1 − 2 x ) {\displaystyle (1-2x)} to convert it to the LCD, giving us 3 x + 5 = A + B ( 1 − 2 x ) {\displaystyle 3x+5=A+B(1-2x)} .