# My Journal of Basic Concepts of Mathematics

1. Sep 4, 2005

### JasonRox

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

Basic Concepts of Mathematics by Elias Zakon

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.

Last edited: Sep 4, 2005
2. Sep 4, 2005

### quasar987

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

Looking foward to those inspirational quotes too. :tongue2:

3. Sep 4, 2005

### 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.

4. Sep 5, 2005

### 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.

5. Sep 5, 2005

### 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.

6. Sep 7, 2005

### 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

Last edited: Sep 7, 2005
7. Sep 7, 2005

### quasar987

Sounds like you got yourself a golddigger fella. :tongue:

8. Sep 7, 2005

### 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.

9. Sep 7, 2005

### 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.

Last edited: Sep 7, 2005
10. Sep 7, 2005

### rachmaninoff

11. Sep 7, 2005

### JasonRox

12. Sep 7, 2005

### quasar987

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):

$$\in[/itex] 13. Sep 8, 2005 ### JasonRox Even better. It doesn't get any easier than that. 14. Sep 8, 2005 ### Eratosthenes I think a journal is a good idea. It will reinforce what you learn. Is there a post that discusses how to create mathematical symbols like that? I ran a search and could not find one. 15. Sep 8, 2005 ### z-component 16. Sep 10, 2005 ### JasonRox I'm posting an inspirational as I haven't had much time this weekend, but I'll get back to it tomorrow. Since the mathematicians have invaded the theory of relativity, I do not understand it myself any more. Albert Einstein 17. Sep 11, 2005 ### JasonRox Summary of Solutions Section 2 - Problems in Set Theory 4. We have two conditions... [tex]( \ A \cup C \ ) \ \subset \ ( \ A \cup \ B \ )$$

...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.

Last edited: Sep 11, 2005
18. Sep 11, 2005

### 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.

19. Sep 11, 2005

### 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.

20. Sep 11, 2005

### 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).