14-240/Tutorial-Sep30

DBN: Classes: 2014-15: 14-240 / Navigation Welcome to Math 240! (additions to this web site no longer count towards good deed points) # Week of... Notes and Links 1 Sep 8 About This Class, What is this class about? (PDF, HTML), Monday, Wednesday 2 Sep 15 HW1, Monday, Wednesday, TheComplexField.pdf,HW1_solutions.pdf 3 Sep 22 HW2, Class Photo, Monday, Wednesday, HW2_solutions.pdf 4 Sep 29 HW3, Wednesday, Tutorial, HW3_solutions.pdf 5 Oct 6 HW4, Monday, Wednesday, Tutorial, HW4_solutions.pdf 6 Oct 13 No Monday class (Thanksgiving), Wednesday, Tutorial 7 Oct 20 HW5, Term Test at tutorials on Tuesday, Wednesday 8 Oct 27 HW6, Monday, Why LinAlg?, Wednesday, Tutorial 9 Nov 3 Monday is the last day to drop this class, HW7, Monday, Wednesday, Tutorial 10 Nov 10 HW8, Monday, Tutorial 11 Nov 17 Monday-Tuesday is UofT November break 12 Nov 24 HW9 13 Dec 1 Wednesday is a "makeup Monday"! End-of-Course Schedule, Tutorial F Dec 8 The Final Exam Register of Good Deeds Add your name / see who's in!

Boris

Problem

Find a set $S$ of two elements that satisfies the following:

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

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

Solution:

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

$+$	$a$	$b$
$a$	$a$	$b$
$b$	$b$	$a$

$\times$	$b$	$a$
$b$	$b$	$a$
$a$	$a$	$b$

We verify that $S$ satisfies (1). By the addition and multiplication tables, $S$ is closed under addition and scalar multiplication. Since $a+b=b+a=b$ and $ab=ba=a$, then $S$ is commutative. It can also be shown that $S$ is associative. Observe that $a$ is the additive and $b$ is the multiplicative identity. Since $a(b+b)=a(a)=b\neq a=a+a=ab+ab$, then $S$ is not distributive. Then $S$ does not satisfy distributivity. Then $S$ satisfies (1).

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