14-240/Tutorial-Sep30: Difference between revisions

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==Boris==
==Boris==


====Challenge Problem====
====Problem====


Find a set <math>S</math> of two elements that satisfies the following:
Find a set <math>S</math> of two elements that satisfies the following:

* <math>S</math> satisfies all the properties of the field except distributivity.
(1) <math>S</math> satisfies all the properties of the field except distributivity.
*<math>\exists x \in S, 0x\neq 0</math>.

(2) <math>\exists x \in S, 0x \neq 0</math>.


Solution:
Solution:

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

{| class="wikitable"
|-
! <math>+</math>
! <math>a</math>
! <math>b</math>
|-
! <math>a</math>
| <math>a</math>
| <math>b</math>
|-
! <math>b</math>
| <math>b</math>
| <math>a</math>
|}

{| class="wikitable"
|-
! <math>\times</math>
! <math>b</math>
! <math>a</math>
|-
! <math>b</math>
| <math>b</math>
| <math>a</math>
|-
! <math>a</math>
| <math>a</math>
| <math>b</math>
|}


We verify that <math>S</math> satisfies (1). By the addition and multiplication 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).

====Elementary Errors in Homework====

(1) Prove <math>A \implies B</math>. Assume <math>A</math> and derive <math>B</math>. It is not the other way around.

(2) Prove <math>A \iff B</math>. Show that <math>A \implies B</math> and <math>B \implies A</math>.

(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.


==Nikita==
==Nikita==

Latest revision as of 23:13, 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, 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).

Elementary Errors in Homework

(1) Prove . Assume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} and derive . It is not the other way around.

(2) Prove . Show that and .

(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.

Nikita