My Journal of Basic Concepts of Mathematics

[JasonRox]

This journal is the journal that will record my progress through the text:

Basic Concepts of Mathematics by Elias Zakon

It can be downloaded off the internet with a simple search on Google.

I spent the entire summer doing no mathematics, and I lost my drive to doing higher mathematics. In short, the purpose of this journal is to ensure that I continue to take small steps to higher mathematics.

Even though I have a course in Basic Analysis in January, it will probably not be as throughout as this. Also, this will give me the opportunity to score in my class, and to move on further into Mathematical Analysis with greater understanding.

Journal Rules

1. Post a short summary of the section just read.
2. Post atleast one full solution a week.
3. Post the answers to every question. A short summary to how the answer shall be reached will suffice.

Readers may correct solutions or assist in finding a solution. It would be appreciated if the readers did not post solutions to future questions.

Also, to make the journal a little bit more exciting. I will post an inspirational quote from time to time.

Note: I love mathematics, but I had plans to do some independent study during the summer. I was held quite busy for the summer, and I don't punish myself for that. Again, the journal is to get these plans rolling.

 

[quasar987]

Sounds like a plan! Good luck to you and most importantly, have fun!

Looking foward to those inspirational quotes too.

 

[JasonRox]

Thanks, quasar.

I will sure have some good inspirational quotes. I'm currently reading Fermat's Last Theorem by Amir D. Aczel.

My first summary should be posted tomorrow.

 

[JasonRox]

A short summary on Sections 1 and 2.

Section 1 - Introduction. Sets and Their Elements.

This section was to describe sets, and other fundamentals, like reflexitivity, transitivity and anti-symmetry.

Section 2 - Operations on Sets.

Just like the title, operations on sets were discussed. We have union, intersection, and difference. Theorem 1 went through the Idempotent, Commutative, Associative and Distributive Laws. Theorem 2 was de Morgan's duality laws. Theorem 3 was the Generalized Distributive Laws.

Kept short and sweet.

Some solutions shall come later on during the day.

 

[JasonRox]

Summary of Solutions

Section 2 - Problems in Set Theory

1.

c) , d), and f) can be answered using a simple style they used on Page 5. Very similiar.

g) Also very straight forward.

2.

Prove that -(-A)=A

Let S be the space that A is derived from.

S-(S-A)=A

Let xeS, but not in A. Therefore, xe(S-A), but x is not an element of S-(S-A) using the definition of difference.

Let yeA, then y is not an element of (S-A). But then y is an element of S-(S-A) because yeS, but not (S-A), which again is the definition of difference.

We have just shown using two cases that when something is not in A, it is also not in S-(S-A), and when something is in A, then it is also in S-(S-A).

We now conclude that S-(S-A)=A.

I'll post some more tomorrow. If I can't get to it, I'll be sure to post an inspirational quote at the very least.

 

[JasonRox]

I did not have the time today to do any work.

I worked 10 hours, then my gf came over for the night. I like spending time with her, but now she's leaving for school tomorrow afternoon. I'll plenty of time for myself now... sort of.

Anyways, here is the inspirational quote...

Mathematics is an independent world created out of pure intelligence.

- Wordsworth, William

 

[quasar987]

I like spending with her

Sounds like you got yourself a golddigger fella.

 

[JasonRox]

I have all day off.

I just have to run a couple errands and that is it. Also, I have to drop off my gf at the bus station.

I shall make this a productive day.

New Journal Rules

4. I may read the next section when I reach Question 11 on the current section.
5. I can not complete more than 10 questions if all the questions of the previous section are not yet completed.

These rules are to avoid me from working on one section too long, and it also avoids me from reading too far ahead.

I will return with some solutions.

Note to Readers: You may correct my solutions.

 

[JasonRox]

Summary of Solutions

Section 2 - Problems in Set Theory

3.

a) Show that $A \cap (B - C) = (A \cap B ) - (B \cap C)$

$$x\ \epsilon \ A \cap ( \ B - C \ ) \Rightarrow x \epsilon A, B \ \mbox{while x is not an element of C}$$

$$x\ \epsilon \ ( \ A \cap B \ ) \ \mbox{while x is not an element in} \ ( \ A \cap C \ )$$

$$x\ \epsilon \ ( \ A \cap B \ ) - ( \ B \cap C \ )$$

b) Similiar as above.

Note: I really hope you appreciate the time I put in Latex. Just started using it and it took awhile to build that so it looks neat. If anyone has any tips, it'd be appreciated... I need to know how to make that "not an element of" epsilon.

4.

I want to write the solution for this. It's pretty simple. I just want a good and short way of writing it.

Anyways, enjoy your day.

 

[rachmaninoff]

I need to know how to make that "not an element of" epsilon.

\notin

$$\notin$$

see
http://amath.colorado.edu/documentation/LaTeX/Symbols.pdf

 

[JasonRox]

\notin

$$\notin$$

see
http://amath.colorado.edu/documentation/LaTeX/Symbols.pdf

Thanks!

That was really quick.

 

[quasar987]

I need to know how to make that "not an element of" epsilon.

Also, it is not an epsilon. It is really a symbol of its own with the meaning "is an element of". It is done in latex like so (click on the symbol):

[tex]\in[/itex]

 

[JasonRox]

Also, it is not an epsilon. It is really a symbol of its own with the meaning "is an element of". It is done in latex like so (click on the symbol):

[tex]\in[/itex]

Even better. It doesn't get any easier than that.

 

[Eratosthenes]

I think a journal is a good idea. It will reinforce what you learn. Is there a post that discusses how to create mathematical symbols like that? I ran a search and could not find one.

 

[z-component]

Is there a post that discusses how to create mathematical symbols like that?

https://www.physicsforums.com/showthread.php?t=8997

 

[JasonRox]

I'm posting an inspirational as I haven't had much time this weekend, but I'll get back to it tomorrow.

Since the mathematicians have invaded the theory of relativity, I do not understand it myself any more.

Albert Einstein

 

[JasonRox]

Summary of Solutions

Section 2 - Problems in Set Theory

4.

We have two conditions...

$$( \ A \cup C \ ) \ \subset \ ( \ A \cup \ B \ )$$

...and...

$$( \ A \cap C \ ) \ \subset \ ( \ A \cap \ B \ )$$

We must show that if these two conditions are met, then...

$$C \ \subset \ B$$

If C and A have a common element that is not in B, then that violates the second condition as a subset.

If C has an element outside of B and A, then that violates the first condition as a subset.

Clearly, C does not have any elements outside of B, but the first condition implies that B has elements outside of C.

This reasoning is proof that $C \subset B$.

The converse is not true because B and C can both be subsets of A, which then violates the first condition.

 

[JasonRox]

Summary of Solutions

Section 2 - Problems in Set Theory

5.

This question is straight forward.

6.

I barely understand what they are asking, so I'll just re-read later tonight. Nice and slow.

7.

A has n! subsets, and it has (n-1)! proper subsets.

 

[JasonRox]

Summary of Solutions

Section 2 - Problems in Set Theory

8.

Show that...

$( A \cup B ) \cap ( B \cup C ) \cap ( C \cup A ) = ( A \cap B ) \cup ( B \cap C ) \cup (C \cap A)$

Left Side...

$( A \cup B ) \cap ( B \cup C ) \cap ( C \cup A ) = ( A \cap C ) \cup B \cap ( C \cup A )$

*Theorem 1 - f)

$( A \cap C ) \cup B \cap ( C \cup A ) = ( A \cap C ) \cup ( A \cap B ) \cup ( B \cap C )$

*Theorem 1 - e)

$( A \cap C ) \cup ( A \cap B ) \cup ( B \cap C ) = ( A \cap B ) \cup ( B \cap C ) \cup (C \cap A)$

I have just shown that the left side equals the right side.

 

[JasonRox]

Summary of Solutions

Section 2 - Problems in Set Theory

9.

Show that the following relations hold iff $A \subseteq E$...

i) $( E - A ) \cup A = E$

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) $E - ( E - A ) = A$

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) $A \cup E = E$

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

Similiar approaches for iv) and v).

 

[JasonRox]

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]

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]

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]

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)

$(xy)^{-1} = x^{-1}y^{-1}$

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

$(xy)^{-1} = x^{-1}y^{-1}$

$(xy)(xy)^{-1} = (xy)x^{-1}y^{-1} \ \mbox{...multiply both sides by xy}$

$1 = xx^{-1}yy^{-1} \ \mbox{Axiom V and II}$

$1 = 1 * 1 \ \mbox{Axiom V}$

$1 = 1 \ \mbox{Axiom V}$

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

Similiar solution as previous question. I have no idea why this is being asked again.

 

[JasonRox]

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]

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]

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]

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...

 

[JasonRox]

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)

$a^{m}a^{n} = a^{m+n}$

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.

$a^{m}a^{k+1} = a^{m}a^{k}a^{1} = a^{m+k}a^{1}$

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

$a^{mk}a^{1} = a^{m+k+1} = a^{m+n}$

ii)

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

$(a^{m})^{k+1} = (a^{m})^{k}(a^{m})^{1} = a^{mk}a^{m}$

$a^{mk+m} = a^{m(k+1)} = a^{mn}$

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]

I've just started working on this text as well. I take it you are having some luck doing chapter 2 first?

 

[JasonRox]

Yeah, it's recommend that you do Chapter 2. It explains all that in the Introduction/Preface.

Have fun. :D