14-240/Tutorial-Sep30: Difference between revisions

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We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math>. By the addition and multplication tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).
We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math>. By the addition and multplication tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).



We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).
We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).

Revision as of 22:28, 4 October 2014

Boris

Problem

Find a set of two elements that satisfies the following:

(1) satisfies all the properties of the field except distributivity.

(2) .

Solution:

Let where is the additive identity and is the multiplicative identity and . After trial and error, we have the following addition and multiplication tables:


We verify that satisfies (1). By the addition and multiplication tables, . By the addition and multplication tables, then satisfies closure, commutativity, associativity and existence of identities and inverses. Since , then does not satisfy distributivity. Then satisfies (1).


We verify that satisfies (2). Since , then satisfies (2).

Nikita