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For instance, if the one solving the math word problem has a limited understanding of the language (English, Spanish, etc.) they are more likely to not understand what the problem is even asking. In Example 1 (above), if one does not comprehend the definition of the word "spent," they will misunderstand the entire purpose of the word problem.
A simple fraction (also known as a common fraction or vulgar fraction, where vulgar is Latin for "common") is a rational number written as a/b or , where a and b are both integers. [9] As with other fractions, the denominator (b) cannot be zero. Examples include 1 / 2 , − 8 / 5 , −8 / 5 , and 8 / −5
The word problem on free lattices and more generally free bounded lattices has a decidable solution. Bounded lattices are algebraic structures with the two binary operations ∨ and ∧ and the two constants ( nullary operations ) 0 and 1.
An equivalent definition is sometimes useful: if a and b are integers, then the fraction a / b is irreducible if and only if there is no other equal fraction c / d such that | c | < | a | or | d | < | b |, where | a | means the absolute value of a. [4] (Two fractions a / b and c / d are equal or equivalent if and ...
In algebra, the partial fraction decomposition or partial fraction expansion of a rational fraction (that is, a fraction such that the numerator and the denominator are both polynomials) is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator. [1]
These include problems of bounding the length or maximum denominator in Egyptian fraction representations, finding expansions of certain special forms or in which the denominators are all of some special type, the termination of various methods for Egyptian fraction expansion, and showing that expansions exist for any sufficiently dense set of ...
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. Then
Then the word problem in is solvable: given two words , in the generators of , write them as words in and compare them using the solution to the word problem in . It is easy to think that this demonstrates a uniform solution of the word problem for the class K {\displaystyle K} (say) of finitely generated groups that can be embedded in G ...
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