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My Journal of Basic Concepts of Mathematics

by JasonRox
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JasonRox
#19
Sep11-05, 09:08 PM
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Summary of Solutions

Section 2 - Problems in Set Theory

8.

Show that...

[itex] ( A \cup B ) \cap ( B \cup C ) \cap ( C \cup A ) = ( A \cap B ) \cup ( B \cap C ) \cup (C \cap A) [/itex]

Left Side...

[itex] ( A \cup B ) \cap ( B \cup C ) \cap ( C \cup A ) = ( A \cap C ) \cup B \cap ( C \cup A )[/itex]

*Theorem 1 - f)

[itex]( A \cap C ) \cup B \cap ( C \cup A ) = ( A \cap C ) \cup ( A \cap B ) \cup ( B \cap C ) [/itex]

*Theorem 1 - e)

[itex]( A \cap C ) \cup ( A \cap B ) \cup ( B \cap C ) = ( A \cap B ) \cup ( B \cap C ) \cup (C \cap A)[/itex]

I have just shown that the left side equals the right side.
JasonRox
#20
Sep11-05, 09:25 PM
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Summary of Solutions

Section 2 - Problems in Set Theory

9.

Show that the following relations hold iff [itex]A \subseteq E[/itex]...

i) [itex]( E - A ) \cup A = E [/itex]

If A is not a subset of E, and was subject to elements outside of E, then the left side would have also have elements outside of E, which makes the statement false.

ii) [itex]E - ( E - A ) = A[/itex]

The left side are elements that belong in E, only in E, so all the elements in A must also be in E for this to hold. Therefore, the above relation must hold.

iii) [itex] A \cup E = E [/itex]

If the above relation did not hold, then A has elements outside of E. This would make the above statement false, since [itex]x \in A , x \notin E \ \Rightarrow \ x \in ( A \cup E ) , x \notin E[/itex].

Similiar approaches for iv) and v).
JasonRox
#21
Sep13-05, 08:23 PM
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There has been a change of plans. After reading the introduction, I've decided to take their advice. It says to start with Chapter 2, then 3, and then 1. This is recommended for Freshmen's. Although I am not a Freshmen, I am still relatively new to this, and doing this independently.

I'm leaving what I've done in Chapter 1, and I will return later to finish it off.

I'm moving on to Chapter 2.
JasonRox
#22
Sep13-05, 08:56 PM
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Reading Summary

Chapter 2

Section 1 - Introduction


A real basic discussion of how things will work.

Section 2 - Axioms of an Ordered Field

I will write the following axioms of addition and multiplication.

Axioms of Additiona and Multiplication

I - Closure Laws
II - Commutative Laws
III - Associative Laws
IV - Existence of Neutral Elements
V - Existence of Inverses
VI - Distributive Law

Axioms of Order

VII - Trichotomy
VIII - Transitivity
IX - Monotonicity of Addition and Multiplication

A field satisfies Axioms I-VI.

An ordered field satisfies Axioms I-IX.
JasonRox
#23
Sep14-05, 07:40 PM
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Reading Summary

Chapter 2

Section 3 - Arithmetic Operations in a Field


This section just went through some basic Corollaries to show some examples of deductive reasoning.

Section 4 - Inequalities in an Ordered Field, and Absolute Values.

Again, just went through some basic Corollaries that apply to Ordered Fields.

The one I liked best was the density of an ordered field.
JasonRox
#24
Sep14-05, 08:11 PM
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Summary of Solutions

Section 4 - Problems on Arithmetic Operations and Inequalities in a Field.

3.


i)

abcd = cbad

Let ab = m.

By Axiom II, m = ba.

We have mcd = bacd.

Now, it's just a matter of doing this over and over again until you get the particular order you desire. Axiom II allows you to do this.

Note: Not sure if you can do this for the general case.

I'm not doing this for addition since it is very similiar.

ii)

The hint gave it away. You can also do this by using transitivity, which may me a little longer.

iii)

[itex](xy)^{-1} = x^{-1}y^{-1}[/itex]

We can just show that they are equivalent with the following steps.

[itex](xy)^{-1} = x^{-1}y^{-1}[/itex]

[itex](xy)(xy)^{-1} = (xy)x^{-1}y^{-1} \ \mbox{...multiply both sides by xy}[/itex]

[itex]1 = xx^{-1}yy^{-1} \ \mbox{Axiom V and II}[/itex]

[itex]1 = 1 * 1 \ \mbox{Axiom V}[/itex]

[itex]1 = 1 \ \mbox{Axiom V}[/itex]

Done.

The reason why neither, x or y, can be equal zero is because the product xy will be equal to zero. Axiom V states that zero does not have a multiplicative inverse.

Note: I have to idea was is meant by the hint.

iv)

Note: <> means not equal to.

If [itex]x <> 0, y <> 0 / /mbox{and} / z <> 0[/itex], then [itex](xyz)^{-1} = x^{-1}y^{-1}z^{-1}[/itex].

Similiar solution as previous question. I have no idea why this is being asked again.
JasonRox
#25
Sep15-05, 09:04 PM
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Although I'm at Section 6, of Chapter 2, I am going to post an inspirational quote because I do not have the time to write it out. I will be sure to write it out tomorrow.

Here it is:

I do not believe in the gifted. If they (the students) have ganas
(Spanish for desire), I can make them do it.


Jaime Escalante
JasonRox
#26
Sep16-05, 10:10 PM
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Summary of Solutions

Section 4 - Problems on Arithmetic Operations and Inequalities in a Field.

The questions here are very basic, so I'm not going to bother going through them. I will in fact go through the proof of Corollary 7, and Question 5, (iv) and (v). Unfortunately not today though.

- - -

Reading Summary

Chapter 2

Section 5 - Natural Numbers and Induction.


A discussion of Natural Numbers was done. The First and Second Induction Laws were introduced, which we call Weak and Strong at school. Examples using induction were done, along with some Theorems.

Section 6 = Induction.

Some more definitions, mostly recursive. Some basic stuff that can help you with proofs that are to come.

- - -

That's all for today. Have a good weekend.
JasonRox
#27
Sep18-05, 08:42 PM
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Alright, this is just a quick remark. I will be posting many solutions in the morning as I have written many of them on my whiteboard. I'll even post a picture of them although they are very short and not detailed.

So, I wouldn't mind if people got involved in the journal. I know I do lack some entertainment, but that can't come until I get really stuck and frustrated. I wish that never happens.

Just to let you know, you can download the text for free at...

http://www.trillia.com/zakon1.html

Follow along if you like.

Note: My whiteboard is quite large, and I show it off to people that visit. It's a mathematicians dream, but it's also a nightmare for everyone else.
JasonRox
#28
Sep19-05, 10:33 PM
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For most of the solutions, Section 6 of Chapter 2, I've just written a short solution. I've only included the second step of the proofs by induction.

I've excluded some questions because they just seemed too obvious. I will show the solutions to Questions 9, and 15-20 another day because some of them are related to sets and problems in the first chapter.

Here they are as they were on my whiteboard...
Attached Thumbnails
math1.JPG   math2.JPG   math3.JPG  
JasonRox
#29
Sep20-05, 07:53 PM
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Summary of Solutions

Chapter 2

Section 6 - Problems on Natural Numbers and Induction.


11.

I'll show a few, and the rest is very similiar.

i)

[itex]a^{m}a^{n} = a^{m+n}[/itex]

This is true for n=1, now let's assume the induction hypothesis that it is true for n=k. Let's now show it is true for n=k+1.

[itex]a^{m}a^{k+1} = a^{m}a^{k}a^{1} = a^{m+k}a^{1}[/itex]

We know we can do this by the induction hypothesis, now we also know it's true when n=1, so it follows...

[itex]a^{mk}a^{1} = a^{m+k+1} = a^{m+n}[/itex]

ii)

Let n = k + 1, after checking n = 1 and assuming the induction hypothesis for n = k.

[itex](a^{m})^{k+1} = (a^{m})^{k}(a^{m})^{1} = a^{mk}a^{m}[/itex]

[itex]a^{mk+m} = a^{m(k+1)} = a^{mn}[/itex]

It's a simple process like the above for the rest of the solutions in Question 11.

12.

The solution can be found online using google. The trick is in the properties of factorials, and it's pretty straightforward from there.

Note: I will have learn more Latex to write the other solutions because I don't know how to use the Summation and Product.
Jeff Ford
#30
Sep22-05, 09:34 AM
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I've just started working on this text as well. I take it you are having some luck doing chapter 2 first?
JasonRox
#31
Sep22-05, 05:03 PM
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Yeah, it's recommend that you do Chapter 2. It explains all that in the Introduction/Preface.

Have fun. :D


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