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The word problem for an algebra is then to determine, given two expressions (words) involving the generators and operations, whether they represent the same element of the algebra modulo the identities. The word problems for groups and semigroups can be phrased as word problems for algebras. [1]
When an exponent is a positive integer, that exponent indicates how many copies of the base are multiplied together. For example, 3 5 = 3 · 3 · 3 · 3 · 3 = 243. The base 3 appears 5 times in the multiplication, because the exponent is 5. Here, 243 is the 5th power of 3, or 3 raised to the 5th power.
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 ...
Many easily stated number problems have solutions that require sophisticated methods, often from across mathematics. A prominent example is Fermat's Last Theorem . This conjecture was stated in 1637 by Pierre de Fermat, but it was proved only in 1994 by Andrew Wiles , who used tools including scheme theory from algebraic geometry , category ...
Unit fractions can also be expressed using negative exponents, as in 2 −1, which represents 1/2, and 2 −2, which represents 1/(2 2) or 1/4. A dyadic fraction is a common fraction in which the denominator is a power of two , e.g. 1 / 8 = 1 / 2 3 .
The definition of exponentiation can also be given by transfinite recursion on the exponent β. When the exponent β = 0, ordinary exponentiation gives α 0 = 1 for any α. For β > 0, the value of α β is the smallest ordinal greater than or equal to α δ · α for all δ < β. Writing the successor and limit ordinals cases separately: α 0 = 1.
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