Simply Math

Spivak’s jk Analysis

I would say there are a few books that lowly undergraduate math majors (like myself!) should come in contact with before they hit the bucket (or just burn out). One of these books is Spivak’s Calculus. The title of this books is extremely deceiving, and if you handed this book to any calculus student they would probably not hesitate to work through this title on any given day.


Well unfortunately Spivak’s calculus is very deceiving. The cover of the book is very minimal (I like this), but to most it looks like “just another boring calculus text”. Well, let me tell you…Unless you are a super genius on crack and or are very smart you are going to have a difficult time plowing through this one. The book covers your standard calculus I and II material, but it only looks “standard” in the contents selection:

Part I (Prologue)
Part II (Foundations)
Part III (Derivatives and Integrals)
Part IV (Infinite Sequences and Infinite Series)
Part V (Epilogue)

So this is probably what you would expect to see in a standard calculus text. I can tell you that I’m still working through the “prologue” and I’ve already taken the entire “standard” calculus sequence. This book is really for the mathematician on steroids, and any of those who wish to challenge themselves. I think this book would be better called “Calculus (jk Introduction to Elementary Analysis)” Don’t let the title fool you!

So you might be asking yourself…

“Why is he even talking about this?”

Well of course I’m talking about it because I would like to share my terrible and  pathetic attempts at working through this book. To put it simply (as I like to on this blog!) the problems are hard, and I suck at them. 

Look around for more… (I hope)

~ J

Note: If you want to torture yourself, or you really have nothing better to do pick up a copy on amazon!

Induction (Problem Set 1)

In the next few days I will be posting all even numbered solutions to problems on “Mathematical Induction” (Pg. 6-8) in George Andrews book “Number Theory”. All the following problems were worked out by myself.



\displaystyle{\sum_{j=1}^{k} j^3 = \frac{k^2(k+1)^2}{4}}

Proceeding by mathematical induction we show the theorem holds for k=1

\displaystyle{\frac{1^2(1+1)^2}{4} = \frac{1(2)^2}{4} = \frac{4}{4} = 1}

We have shown the theorem holds for k=1.

Now we must show if the theorem holds k it must hold for (k+1).

\displaystyle{\sum_{j=1}^{k+1} j^3 = \sum_{j=1}^{k} j^3 + (k+1)^3}

\displaystyle{= \frac{k^2(k+1)^2}{4} + (k+1)^3}

\displaystyle{= \frac{k^2(k^2+2k+1)}{4} + k^3 + 3k^2 + 3k + 1}

\displaystyle{= \frac{k^2+2k^3+k^2}{4} + \frac{4k^3+12k^2+12k+4}{4}}

\displaystyle{= \frac{k^4+6k^3+13k^2+12k+4}{4}}

\displaystyle{= \frac{(k+1)^2 (k+2)^2}{4}}

We have shown the theorem holds for (k+1).




\displaystyle{\sum_{j=1}^{k} j(j+1) = \frac{k(k+1)(k+2)}{3}}

Proceeding by mathematical induction we show the theorem stands for k=1

\displaystyle{\frac{1(1+1)(1+2)}{3} = \frac{1(2)(3)}{3} = \frac{6}{3} = 2}

The theorem holds for k=1.

Now we show if the theorem holds for k, it must also hold for (k+1)

\displaystyle{\sum_{j=1}^{k+1} j(j+1) = \sum_{j=1}^{k} j(j+1) + (k+1)((k+1)+1)}

\displaystyle{= \frac{k(k+1)(k+2)}{3} + (k+1)((k+1)+1)}

\displaystyle{= \frac{k^3+2k^3 + k^2 + 2k}{3} + k^2 + 2k + k + 2}

\displaystyle{= \frac{k^3+2k^2+k^2 + 2k}{3}+\frac{3k^2+6k+3k+6}{3}}

\displaystyle{= \frac{k^3 + 6k^2 + 11k + 6}{3}}

\displaystyle{= \frac{(k+1)(k+2)(k+3)}{3}}

We have shown the theorem holds for (k+1).




\displaystyle{\sum_{j=1}^{n} \frac{1}{j(j+1)} = \frac{n}{n+1}}

Proceeding by mathematical induction we show the theorem holds for n=1

\displaystyle{\frac{1}{1+1} = \frac{1}{2}}

If the theorem holds for n, then it must hold for (n+1)

\displaystyle{\sum_{j=1}^{n+1} \frac{1}{j(j+1)} = \sum_{j=1}^{n} \frac{1}{j(j+1)} + \frac{1}{(n+1)((n+1) +1)}}

\displaystyle{= \frac{n}{n+1} + \frac{1}{(n+1)((n+1)+1)} = \frac{n^3 + 3n^2+3n+1}{(n+1)^2(n+2)}}

\displaystyle{= \frac{(n+1)^3}{(n+1)^2 (n+2)} = \frac{(n+1)}{(n+2)}}

We have shown the theorem holds for (n+1).


I will post more shortly.

~ J

New Theme, New Stuff….

As you can tell I’ve changed my theme. At heart, I’m a minimalist, and I love to keep things simple. I haven’t been posting in sometime, but I’ve been rather busy with some summer work. I’m still working through Dr. George Andrews “Number Theory”, but I’m going to change up the style a little. I’m going to focus much more on problems at the end of each chapter, and give short summaries of the chapters content. I feel this will help me learn the content of the book much better. This means I will be going back to Chapter 1 and rebooting my work. It may seem redundant, but I’d like to keep things consistant. I’ll mostly be focusing on problems though, so it should keep it fresh.

Stay tuned, and as always thanks for reading!

~ J

Euclid’s Division Lemma and The Basis Representation Theorem (Chapter 2: The Fundamental Theorem of Arithmetic )

Most of the following is going to be right from Dr. Andrews book. I decided to post the proof the basis representation theorem since it’s important for Euclid’s division lemma. I will include one or two problems at the end Euclid’s Lemma. (I can’t say I’ll have them solved, but I may skip over them for now.)

Anyway, Here we go!

First a few important things:

Corollary 1-1: If m and n are positive integers and


m > 1


n < m^n.

Theorem (1-2) 
\displaystyle{\sum_{j=0}^{n-1} x^j = 1 + x + x^2 + ... + x^{n-1} = \frac{x^n-1}{x-1}}

Proof: Let b_{k}(n) denote the number of representations of n to the base k. We must show that b_{k}(n) always equals 1.

It is possible that some of the coefficients a_{i} in a particular representation of n are equal to zero. Without affecting the representation, we may exclude terms that are zero. Thus suppose that

\displaystyle{n=a_{0}k^s+a_{1}k^{s-1} + ... + a_{s-t}k^{t}},

where now neither a_{0} nor a_{s-t} equals zero. Then

\displaystyle{n-1=a_{0}k^s + a_{1}k^{s-1} + ... + a_{s-t}k^t-1}

\displaystyle{= a_{0}k^s + a_{1}k^{s-1} + ... + (a_{s-t}-1)k^t + k^t-1}

\displaystyle{= a_{0}k^s + a_{1}k^{s-1} + ... + (a_{s-t}-1)k^t + \sum_{j=0}^{t-1} k(k-1)k^j}

by Theorem 1-2 with x=k. Thus we see that for each representation of n to the base k, we can find a representation of (n-1). If n has another representation to the base k, the same procedure will yield a new representation to the base k, the same procedure will yield a new representation of n-1. Consequently

\displaystyle{b_{k}(n) \leq b_{k}(n-1)}

It is important to note the inequality (1-2-2) holds even if n has no representation because b_{k}(n) = 0 \leq b_{k}(n-1) in that case. Inequality (1-2-2) implies the following inequalities:

\displaystyle{b_{k}(n+2) \leq b_{k}(n+1) \leq b_{k}(n)}

\displaystyle{b_{k}(n+3) \leq b_{k}(n+2) \leq b_{k}(n+1) \leq b_{k}(n)}

and, in general, if m \geq n+4,

\displaystyle{b_{k}(m) \leq b_{k}(m-1) \leq b_{k}(m-2) \leq .... \leq b_{k}(n+1) \leq b_{k}(n)}

since k^n > n by Corollary 1-1, and since k^n clearly has at least one representation (namely, itself), we see that

\displaystyle{ 1 \leq b_{k}(k^n) \leq b_{k}(n) \leq b_{k}(1) = 1}

The extreme entries in this set of inequalities are ones, so that all of the intermediate entries must be equal to 1. Thus b_{k}(n) = 1, and Theorem 1-3 is established.


Once a base k (k > 1) has been chosen, we can represent any positive integer n uniquely as a sume of multiples of powers of k:

\displaystyle{n=a_{s}k^s + a_{s-1}k^{s-1} + ... + a_{1}k + a_{0}}

Okay! Now for Euclid’s Division Lemma. The book will use the basis representation theorem to show proof of this lemma.

Euclid’s Division Lemma (Chapter 2)

In short Euclid’s Division Lemma is a rather basic concept. Whenever you perform division in mathematics you have a dividend, divisor, quotient, and remainder. It has a strong relationship to a certain properties about prime numbers, but that’s not of concern at the moment.

Theorem (2-1) (Euclid’s Division Lemma):
For any integers k (k>0) and j, there exists unique integers q and r such that 0 \leq r \leq k and


(Now again from the book!)

Proof: Note that we have simply rewritten a division problem in terms of multiplication and addition. In the notation used above, j is the dividend; k, the divisor; q, the quotient; and r, the remainder.

If k=1, r must be zero, so that q=j.

If k>1, suppose first that j>0. (We shall consider the cases in which j=0 and j<0 later.) By the basis representation theorem (see above!), j has a unique representation to the base k, say

\displaystyle{j = a_{s}k^s + a_{s-1}k^{s-1} + ... + a_{1}k + a_{0}}

\displaystyle{=k(a_{s}k^{s-1} + a_{s-1}k^{s-2} + ... + a_{1}) + a_{0}}


where 0 \leq r = a_{0} < k.

If a second pair q' and r' existed, we could find a representation for q' to the base k, say

\displaystyle{ q' = b_{t}k^t + ... + b_{1}k + b_{0}}

so that

\displaystyle{j=kq' + r'}

= \displaystyle{b_{t}k^{t+1} + ... + b_{1}k^2 + b_{0}k + r'}


\displaystyle{j=a_{s}k^s + a_{s-1}k^{s-1} + ... + a_{1}k + a_{0}}

By the uniqueness of the representation of j to the base k, we see that t = s-1, b_{i}=a_{i+1}, r' = a_{0} = r, and thus

\displaystyle{q' = b_{t}k^t + ... + b_{1}k + b_{0}}

\displaystyle{a_{s}k^{s-1} + ... + a_{2}k + a_{1}}


Consequently, the theorem is true for positive values of j.

If j=0, it is easy to verify that q=r=0 is the only possible solution of (2-1-1) with 0 \leq r < k.

If j < 0, then -j > 0, and there exist unique integers q'' and r'' such that.


If r''=0, then j=k(-q''); thus we may take q = -q'' and r=0. If r'' \neq 0, then

\displaystyle{j=-kq'' - r''}


and we may take q=-q''-1, and r=k-r''.

In either case, q and r satisfy equation (2-1-1). Uniqueness for negative j follows from uniqueness for -j, which is then positive.

Whew! Okay, I will try to avoid writing so much from the book in future posts, but in this case I felt I couldn’t sum it up any better than the book did. The proofs here are rather lengthy, but I’m sure they will be even longer as I progress through the book.

Here are some problems for thought:

(1) Without assuming Theorem (2-1), prove that for each pair of integers j and k(k>0), there exists some integer q for which j-qk is positive.

(2) The principle of mathematical induction is equivalent to the following statement called the least-integer principle: Every non-empty set of positive integers has a least element.

Using the least integer principle, define r to be the least integer for which j-qk is positive. (See (1)). Prove that
\displaystyle{0 < r \leq k}.

If anyone cares to gives these a shot leave a comment. For right now I think I’ll move on to the next part of Chapter 2.  “Divisibility” and “The Linear Diophantine Equation”. From there we will finish up with a proof of the “The Fundamental Theorem of Arithmetic”

If you would like to by Dr. Andrews book you can get it here

more on Euclid’s Lemma at Wikipedia

~ J

For a few proofs more…

This is a very short post.

While finishing up chapter 2,  I’ve managed to work on  two proofs from chapter 1. I’m putting them up on my blog just to keep track of my progress. I will finish up chapter 2 hopefully in the next few days.

(Exercise 9)

Let F_n be the nth Fibonacci number.


\displaystyle{ \sum_{j=1}^{n} F_{2n} = F_{2} + F_{4} + F_{6}+...+F_{2n} = F_{2n+1} - 1}

I. Proceeding by mathematical induction we start with the base case where n=1 (Showing the theorem holds for n=1).

\displaystyle{\sum_{j=1}^{1} F_{2n} = F_{2}}

The theorem holds for n=1

II. Now we assume the theorem holds for an arbitrary value k. That is

\displaystyle{\sum_{j=1}^{k} F_{2k} = F_{2} + F_{4} +F_{6} + ... + F_{2k} = F_{2k+1} - 1}

We now show it holds for the number (k+1)

\displaystyle{ \left(\sum_{j=1}^{k} F_{2k}\right) + F_{2(k+1)} = F_{2(k+1)+1}-1}

Note: \displaystyle{ \sum_{j=1}^{k} F_{2k} = F_{2k+1} - 1}

Now we can write

\displaystyle{F_{2k+1}-1 + F_{2(k+1)} = F_{2(k+1)+1} - 1}

\displaystyle{\left(F_{2k+1} + F_{2(k+1)}\right) - 1}

\displaystyle{\left(F_{2k+1} + F_{2k+2}\right) - 1}

\displaystyle{F_{2k+3} - 1 = F_{2(k+1) + 1} - 1}

We have shown the theorem holds for (k+1).

The theorem stands as is.


(Exercise 11) (Not fully completed)

Let F_{n} be the nth Fibonacci number.


\displaystyle{\sum_{j=1}^{n} \left(F_{2n-1}\right) \left(F_{2n}\right) = F_{1}F_{2} + F_{2}F_{3} + F_{3}F_{4} + ... + F_{2n-1}\left(F_{2n}\right) = F^{2}_{2n}}

I. Proceeding by mathematical induction we start with the base case where n=1 (Showing the theorem holds at n=1)

\displaystyle{\sum_{j=1}^{1} \left(F_{2n-1}\right) \left(F_{2n}\right) = F_{1}F_{2}}

Thus the theorem stands for n=1 (Our first value is F_{1}F_{2})

II. Now we assume the theorem holds for an arbitrary value m. That is

\displaystyle{\sum_{j=1}^{m} \left(F_{2m-1}\right) \left(F_{2m}\right) = F_{1}F_{2} + F_{2}F_{3} + F_{3}F_{4} + ... + F_{2m-1}\left(F_{2m}\right) = F^{2}_{2m}}

Now we will show it holds for the number (m+1)

\displaystyle{\left(\sum_{j=1}^{m} \left(F_{2m-1}\right) \left(F_{2m}\right)\right) + \left(F_{2(m+1)-1}F_{2(m+1)}\right)=F^{2}_{2(m+1)}}

Note: \displaystyle{\left(\sum_{j=1}^{m}\left(F_{2m-1}\right)\left(F_{2m}\right)\right)=F^{2}_{2m}}

So we can write

\displaystyle{F^{2}_{2m} + F_{2(m+1)-1}\left(F_{2(m+1)}\right) = F^{2}_{2(m+1)}}

\displaystyle{ F^{2}_{2m} + F_{(2m+2)-1}\left(F_{2(m+1)}\right) = F^{2}_{2(m+1)}}

\displaystyle{F^{2}_{2m} + F_{2m+1}\left(F_{2(m+1)}\right) = F^{2}_{2(m+1)}}

\displaystyle{F^{2}_{2m} + F_{2m+1}\left(F_{2m+2}\right) = F^{2}_{2(m+1)}}

At this point my algebra got very disorganized. I need to clean this up, but I think I’m on the right track. If  someone cares to correct me, please do. I’ll try to figure it out at some point later. Until then I need to finish up chapter 2 (Getting there!), and there should be quite a few posts coming related to that. My goal is to get to Chapter 3 by June 21. I’ve just been held back by a summer class, and other obligations.

~ J

The Basis Representation Theorem (Chapter 1: Basis Representation)

This is basically a continuation of my previous post, and the final post regarding the first chapter Dr. Andrews Dover book “Number Theory”. I’ve finished the proof that I said I was working on in the previous post. I will now show that, and then finish off talking about “The Basis Representation Theorem”.


Proof (Exercise 7)

Let F_{n} represent the nth Fibonacci number.

So that F_{1}=1, F_{2}=1, F_{3}=2, F_{4}=3, F_{5}=5

In general

F_{n} = F_{n-1} + F_{n-2} for n \ge 3


\displaystyle{\sum_{j=1}^{n} F_{n} = F_{n+2} -1}

I. Proceeding by induction we show that the theorem is true for n=1 (Basis Step)

\displaystyle{\sum_{j=1}^{1} F_{n} = 1 = F_{1+2} -1 = F_{3} - 1 = 2 - 1 = 1}

The theorem holds for n=1

II. We assume the theorem holds for an arbitrary value k. That is

\displaystyle{ F_{1} + F_{2} + F_{3} + ... + F_{k} = F_{k+2} - 1}

We now show it also holds for the the integer (k+1)

\displaystyle{(F_{1} + F_{2} + F_{3} + ... + F_{k}) + F_{k+1} = F_{(k+1)+2}-1}

Working from the left side

\displaystyle{[F_{k+2} - 1] + F_{k+1} \rightarrow (F_{k+2} + F_{k+1}) - 1 = F_{k+3} - 1}

\displaystyle{\therefore \quad F_{k+3}-1 = F_{(k+1) + 2}-1}

Condition (ii) has been satisfied. The theorem stands!



That ends the first section of chapter 1. There are many other proof exercises that become much more difficult as you go on, I may work these given some additional time, but for now I’d like to move on and finish the first chapter.


The Basis Representation Theorem

In Short this Theorem is basically saying that you can have a multitude of different bases in mathematics.

  • Base \quad 2
  • Base \quad 3
  • Base \quad k

Dr. Andrews gives an example in Base \quad 10

209 = 2 \cdot 10^2 + 0 \cdot 10^1 + 9 \cdot 10^0

and in Base \quad 2

23 = 10111 = 1 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0

If you’ve seen different number bases before this should be rather trivial; however, sometimes it’s wroth refreshing up on. The idea behind this is to really formalize representing unique numbers in different bases.

Dr. Andrews officially states the theorem as such

Theorem 1-3 (Basis Representation Theorem) 

Let k be any integer larger than 1. Then, for each positive integer n, there exists a representation

\displaystyle{n = a_{0}k^s+a_{1}k^{s-1} + ... + a_{s}}, where a_{0} \not= 0, and where each a_{i} is nonnegative and less than k. Furthermore, this representation of n is unique; it is called the representation of n to the base k.

Basically what’s going on here is that any natural number from the set of counting numbers, can be represented in any base k.

Proceeding this there is a proof about a page long. I’m going to leave that out of this post, since it plays on some tricky algebra. I’ll include a link to proof wiki at the end of this post. (I’m also not very fond of the books version). 

I haven’t had time to work out any problems in section (1-2), but I will leave a problem for the curious to try. 

(Exercise 4)

Prove that each nonzero integer may be uniquely represented in the form  

\displaystyle{ n = \sum_{j=0}^{s} c_{j}3^j},

where c_{s} \not= 0, and each c_{j} is equal to -1, 0 , or 1

I’ll leave that here for now, unless I quickly come up with a solution.

I’ll be updating with Chapter 2 where we will dive into “The Fundamental Theorem Of Arithmetic”

Proof Of Basis Representation Theorem

~ J


Principle Of Mathematical Induction (Chapter 1: Basis Representation)

The first stop on our journey through Dr. Andrews book “Number Theory” is an introduction to “The Principle Of Mathematical Induction” or to be less fancy…Proof by induction.

Dr. Andrews starts off with an absolute classic formula right on the front page. If you’re not familiar with the legend of mathematician Carl Gauss in which he adds all the numbers from 1 to 100 in the blink of an eye, then what you see before you is the formula for which he supposedly derived.

\displaystyle{S_{n} = \frac{n(n+1)}{2}}

This can all be written as…

\displaystyle{\sum_{j=1}^{n} j = \frac{n(n+1)}{2}}

Dr. Andrews also provides a table in the book to show you the various values that you get when you plug in n.

He also uses this as the first specimen on proof by induction. (I’ve omitted it, and left it up to you as an exercise.)

Now on to the proofs!!

When proving a theorem via induction there are two things we must take into account. Here it is as stated in the book:


Principle Of Mathematical Induction: A statement about integers is true for all integers greater than or equal to 1 if

(i) it is true for the integer 1, and

(ii) whenever it is true for all the integers 1,2,...,k then it is true for the integer (k+1)


This gives us the basic tool kit for proving various theorems by mathematical induction.

I will provide an example worked out by Dr. Andrews, then I will provide my own work from the exercise portion of the book.

Theorem  1-2: If x is any real number other than 1, then

\displaystyle{\sum_{j=0}^{n-1} x^j = 1 + x + x^2 +...+ x^{n-1} = \frac{x^n-1}{x-1}}

Proof: Again we proceed by mathematical induction.

If n=1, then \displaystyle{\sum_{j=0}^{1-1} x^j = x^0 = 1}  and  \displaystyle{\frac{(x-1)}{(x-1)}=1}

Thus the theorem is true for n=1. (Basis Case)

Assuming that \displaystyle{\sum_{j=0}^{k-1} x^j = \frac{(x^k-1)}{(x-1)}} we find that

\displaystyle{\sum_{j=0}^{(k+1)-1} x^j = \sum_{j=0}^{k-1} x^j + x^k = \frac{x^k -1}{x-1}+x^k}

\displaystyle{=\frac{x^k -1 + x^{k+1} - x^k}{x-1}=\frac{x^{k+1}-1}{x-1}}

Hence condition (ii) is fulfilled, and we have established the theorem.

This ends Dr. Andrews example, now I take it upon myself to write out my own proof.


1. Prove that

\displaystyle{1^2 + 2^2 + 3^2 +...+n^2 =\frac{n(n+1)(2n+1)}{6}}

Proceeding by induction we start by verifying that the theorem is valid for n=1 (Case (i))

\displaystyle{ 1 = \frac{1(1+1)(2(1)+1)}{6} \rightarrow 1 = \frac{1(2)(3)}{6} \rightarrow \frac{6}{6} \rightarrow 1=1}

We have shown the theorem is valid for n=1 (Basis Case)

We now proceed to the inductive step. (Case (ii))

Assume the formula holds for an arbitrary value k

\displaystyle{ 1^2 + 2^2 + 3^2 +...+ k^2 = \frac{k(k+1)(2k+1)}{6}}

Assuming this holds for k we will now show that this holds for the integer (k+1)

\displaystyle{(1^2 + 2^2 + 3^2 +...+ k^2) + (k+1)^2 = \frac{(k+1)[((k+1) + 1)(2(k+1)+1)]}{6}}

\displaystyle{\frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{(k+1)[((k+1) + 1)(2(k+1)+1)]}{6}}


\displaystyle{\frac{k^2 + k(2k+1)+6(k^2+2k+1)}{6}}

\displaystyle{\frac{2k^3 + k^2 +2k^2 + k + 6k^2 + 12k + 6}{6}}

\displaystyle{\frac{2k^3 + 9k^2 + 13k + 6}{6}}

We have now fully expanded the numerator to the polynomial \displaystyle{2k^3 + 9k^2 + 13k + 6}.

Factoring this we get..

\displaystyle{ (k+1)(k+2)(2k+3)} Notice that this numerator is now equal to the right hand side numerator.

\displaystyle{ (k+1)(k+2)(2k+3) = (k+1)[((k+1)+1)(2(k+1) + 1)]}


\displaystyle{ \therefore \quad \frac{(k+1)(k+2)(2k+3)}{6} = \frac{(k+1)[((k+1) + 1)(2(k+1)+1)]}{6}}

Case (ii) has been fulfilled. We have now established the theorem.



I’m still fairly new to proofs, so I’m not sure if I could have condensed this or done better, but that is my first proof from this chapter (Exercise 1). I’m working on another one with Fibonacci numbers (Exercise 7). If I finish in a timely manner I’ll publish it at the start of my next post. The next topic from the book will be “The Basis Representation Theorem” itself. The concept is rather simple, but it can get rather hairy later on. I’m just hoping that I don’t get stuck to fast in terms of my comprehension.

~ J

A Little Number Theory…

It’s summer, and that means I have a good amount of free time to do a lot of nothing, and a lot of free time to do a lot of productive work! Naturally I always end up choosing the path of least resistance, and I end up doing the nothing more than the productive work. Fortunately, I’ve been in a much more productive mood lately, and I think I should start some small scale projects. The first project I’ve decided to take on is working through a Number Theory book by Penn State professor Georege Andrews. Over time I’ve collected a lot of Dover books, but often these books are very hard to approach even when you have a strong mathematical background. Usually I decide to take on a topic such as “Abstract Algebra” and I’ll open the book to find that I’m rather lost or confused. Upon looking at “Number Theory” by Andrews, I find the text more approachable (Perhaps due to its combinatorial nature). The book has a strong focus not only on number theory, but also on combinatorics. I suppose you could say this is really a book about combinatorial number theory.


(Here’s a picture of it: courtesy my iPhone)

The book is broken into 15 chapters and 4 parts as follows:

Part I (Multiplicativity-Divisibility)  

Chapter 1 –> Basis Representation

Chapter 2 –> The Fundamental Theorem Of Arithmetic

Chapter 3 –> Combinatorial And Computational Number Theory

Chapter 4 –> Fundamentals Of Congruences

Chapter 5 –> Solving Congruences

Chapter 6 –> Arithmetic Functions

Chapter 7 –> Primitive Roots

Capter 8 –> Prime Numbers

Part II (Quadratic Congruences)

Chapter 9 –> Quadratic Residues

Chapter 10 –> Distribution Of Quadratic Residues

Part III (Additivity)

Chapter 11 –> Sums Of Squares

Chapter 12 –> Elementary Partition Theory

Chapter 13 –> Partition Generating Functions

Chapter 14 –> Partition Identities

Part IV (Geometric Number Theory)

Chapter 15 –> Lattice Points

I doubt I’ll work my way through this entire book over the summer, but I’m making it a goal to at least finish Part I and possibly start breaking into Part II. As I work through the book I’ll be sharing my results on this blog as part of a series of other posts. Many of the problems in this book are very interesting, and most of them are quite challenging. If I get stuck, I’ll probably be seeking out the help of the internet. I hope this blog, and this project will make concepts and ideas of mathematics more approachable to anyone (math major or not) who has a love for mathematics. I’ll follow up shortly.

~ J

Ronald Graham (Left)

“Why are numbers beautiful? It’s like asking why is Beethoven’s Ninth Symphony beautiful. If you don’t see why, someone can’t tell you. I know numbers are beautiful. If they aren’t beautiful, nothing is.”