# My Journal of Basic Concepts of Mathematics

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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.

- - -

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.

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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. :tongue2:

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.

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

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.

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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...

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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)

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

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.

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

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

Have fun. :D