# 14-240/Tutorial-Sep30

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## Boris

#### Problem

Find a set ${\displaystyle S}$ of two elements that satisfies the following:

(1) ${\displaystyle S}$ satisfies all the properties of the field except distributivity.

(2) ${\displaystyle \exists x\in S,0x\neq 0}$.

Solution:

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

${\displaystyle +}$ ${\displaystyle a}$ ${\displaystyle b}$
${\displaystyle a}$ ${\displaystyle a}$ ${\displaystyle b}$
${\displaystyle b}$ ${\displaystyle b}$ ${\displaystyle a}$
${\displaystyle \times }$ ${\displaystyle b}$ ${\displaystyle a}$
${\displaystyle b}$ ${\displaystyle b}$ ${\displaystyle a}$
${\displaystyle a}$ ${\displaystyle a}$ ${\displaystyle b}$

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

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

#### Elementary Errors in Homework

(1) Prove ${\displaystyle A\implies B}$. Assume ${\displaystyle A}$ and derive ${\displaystyle B}$. It is not the other way around.

(2) Prove ${\displaystyle A\iff B}$. Show that ${\displaystyle A\implies B}$ and ${\displaystyle B\implies A}$.

(3) This is for Boris's section only. When a proof requires a previous result, there are two possibilities:

(a) The result is already proved in class or in a previous homework. Then state the result and use it without proof.
(b) The result is neither proved in class nor in a previous homework. Then reference it in the textbook or prove it yourself.