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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).


==Nikita==
==Nikita==

Revision as of 22:30, 4 October 2014

Boris

Problem

Find a set [math]\displaystyle{ S }[/math] of two elements that satisfies the following:

(1) [math]\displaystyle{ S }[/math] satisfies all the properties of the field except distributivity.

(2) [math]\displaystyle{ \exists x \in S, 0x \neq 0 }[/math].

Solution:

Let [math]\displaystyle{ S = \{ a, b \} }[/math] where [math]\displaystyle{ a }[/math] is the additive identity and [math]\displaystyle{ b }[/math] is the multiplicative identity and [math]\displaystyle{ a \neq b }[/math]. After trial and error, we have the following addition and multiplication tables:

[math]\displaystyle{ + }[/math] [math]\displaystyle{ a }[/math] [math]\displaystyle{ b }[/math]
[math]\displaystyle{ a }[/math] [math]\displaystyle{ a }[/math] [math]\displaystyle{ b }[/math]
[math]\displaystyle{ b }[/math] [math]\displaystyle{ b }[/math] [math]\displaystyle{ a }[/math]
[math]\displaystyle{ \times }[/math] [math]\displaystyle{ b }[/math] [math]\displaystyle{ a }[/math]
[math]\displaystyle{ b }[/math] [math]\displaystyle{ b }[/math] [math]\displaystyle{ a }[/math]
[math]\displaystyle{ a }[/math] [math]\displaystyle{ a }[/math] [math]\displaystyle{ b }[/math]


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


We verify that [math]\displaystyle{ S }[/math] satisfies (2). Since [math]\displaystyle{ aa = b \neq a }[/math], then [math]\displaystyle{ S }[/math] satisfies (2).

Nikita