14-240/Tutorial-Sep30
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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, [math]\displaystyle{ S }[/math] is closed under addition and scalar multiplication. Since [math]\displaystyle{ a + b = b + a = b }[/math] and [math]\displaystyle{ ab = ba = a }[/math], then [math]\displaystyle{ S }[/math] is commutative. It can also be shown that [math]\displaystyle{ S }[/math] is associative. Observe that [math]\displaystyle{ a }[/math] is the additive and [math]\displaystyle{ b }[/math] is the multiplicative identity. Since [math]\displaystyle{ a(b + b) = a(a) = b \neq a = a + a = ab + ab }[/math], then [math]\displaystyle{ S }[/math] is not distributive. 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).
