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A Very Nice Guide for Number theory
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Revised June 2, 2003
Copyleft 2002 by W. Edwin Clark
Copyleft means that unrestricted redistribution and modification are per- mitted, provided that all copies and derivatives retain the same permissions. Specifically no commerical use of these notes or any revisions thereof is per- mitted.
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iv PREFACE
Here are some examples of outstanding unsolved problems in number the- ory. Some of these will be discussed in this course. A solution to any one of these problems would make you quite famous (at least among mathemati- cians). Many of these problems concern prime numbers. A prime number is an integer greater than 1 whose only positive factors are 1 and the integer itself.
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Two quotations from G. H. Hardy: In the first quotation Hardy is speaking of the famous Indian mathe- matician Ramanujan. This is the source of the often made statement that Ramanujan knew each integer personally.
I remember once going to see him when he was lying ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. “No,” he replied, “it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways. ”
Pure mathematics is on the whole distinctly more useful than ap- plied. For what is useful above all is technique, and mathematical technique is taught mainly through pure mathematics.
Two quotations by Leopold Kronecker
God has made the integers, all the rest is the work of man.
The original quotation in German was Die ganze Zahl schuf der liebe Gott, alles Ubrige ist Menschenwerk.¨ More literally, the translation is “ The whole number, created the dear God, everything else is man’s work.” Note in particular that Zahl is German for number. This is the reason that today we use Z for the set of integers.
Number theorists are like lotus-eaters – having once tasted of this food they can never give it up.
A quotation by contemporary number theorist William Stein:
A computer is to a number theorist, like a telescope is to an astronomer. It would be a shame to teach an astronomy class without touching a telescope; likewise, it would be a shame to teach this class without telling you how to look at the integers through the lens of a computer.
(a) (an)m^ = anm (b) (ab)n^ = anbn (c) anam^ = an+m.
These rules hold for all n, m ∈ Z if a and b are not zero.
(a) (Transitivity) If a < b and b < c, then a < c. (b) If a < b then a + c < b + c. (c) If a < b and 0 < c then ac < bc. (d) If a < b and c < 0 then bc < ac. (e) (Trichotomy) Given a and b, one and only one of the following holds: a = b, a < b, b < a.
(a) P (n) is true if n = n 0. (b) Whenever P (n) is true for n 0 ≤ n ≤ k then P (n) is true for n = k + 1.
We use the usual conventions:
Important Convention. Since in this course we will be almost exclu- sively concerned with integers we shall assume from now on (unless otherwise stated) that all lower case roman letters a, b,... , z are integers.
In this section, I list a number of statements that can be proved by use of The Principle of Mathematical Induction. I will refer to this principle as PMI or, simply, induction. A sample proof is given below. The rest will be given in class hopefully by students.
A sample proof using induction: I will give two versions of this proof. In the first proof I explain in detail how one uses the PMI. The second proof is less pedagogical and is the type of proof I expect students to construct. I call the statement I want to prove a proposition. It might also be called a theorem, lemma or corollary depending on the situation.
Proposition 2.1. If n ≥ 5 then 2 n^ > 5 n.
Proof #1. Here we use The Principle of Mathematical Induction. Note that PMI has two parts which we denote by PMI (a) and PMI (b).
We let P (n) be the statement 2n^ > 5 n. For n 0 we take 5. We could write simply:
P (n) = 2n^ > 5 n and n 0 = 5.
Note that P (n) represents a statement, usually an inequality or an equation but sometimes a more complicated assertion. Now if n = 4 then P (n) be- comes the statement 2^4 > 5 · 4 which is false! But if n = 5, P (n) is the statement 2^5 > 5 · 5 or 32 > 25 which is true and we have established PMI (a).
Proof #2. We prove the proposition by induction on the variable n. If n = 5 we have 2^5 > 5 · 5 or 32 > 25 which is true. Assume
2 n^ > 5 n for 5 ≤ n ≤ k (the induction hypothesis).
Taking n = k we have 2 k^ > 5 k.
Multiplying both sides by 2 gives
2 k+1^ > 10 k.
Now 10k = 5k + 5k and k ≥ 5 so k ≥ 1 and therefore 5k ≥ 5. Hence
10 k = 5k + 5k ≥ 5 k + 5 = 5(k + 1).
It follows that 2 k+1^ > 10 k ≥ 5(k + 1)
and therefore 2 k+1^ > 5(k + 1).
Hence by PMI we conclude that 2n^ > 5 n for n ≥ 5.
The 8 major parts of a proof by induction:
Exercise 2.1. Prove that 2n^ > 6 n for n ≥ 5.
Exercise 2.2. Prove that 1 + 2 + · · · + n =
n(n + 1) 2
for n ≥ 1.
Exercise 2.3. Prove that if 0 < a < b then 0 < an^ < bn^ for all n ∈ N.
Exercise 2.4. Prove that n! < nn^ for n ≥ 2.
Exercise 2.5. Prove that if a and r are real numbers and r = 1, then for n ≥ 1
a + ar + ar^2 + · · · + arn^ =
a (rn+1^ − 1) r − 1
This can be written as follows
a(rn+1^ − 1) = (r − 1)(a + ar + ar^2 + · · · + arn).
And important special case of which is
(rn+1^ − 1) = (r − 1)(1 + r + r^2 + · · · + rn).
Exercise 2.6. Prove that 1 + 2 + 2^2 + · · · + 2n^ = 2n+1^ − 1 for n ≥ 1.
Exercise 2.7. Prove that 111︸ ︷︷ · · · 1 ︸
n 1 ’s
10 n^ − 1 9
for n ≥ 1.
Exercise 2.8. Prove that 1^2 + 2^2 + 3^2 + · · · + n^2 =
n(n + 1)(2n + 1) 6
if n ≥ 1.
Exercise 2.9. Prove that if n ≥ 12 then n can be written as a sum of 4’s and 5’s. For example, 23 = 5 + 5 + 5 + 4 + 4 = 3 · 5 + 2 · 4. [Hint. In this case it will help to do the cases n = 12, 13 , 14 , and 15 separately. Then use induction to handle n ≥ 16 .]
Definition 3.1. d | n means there is an integer k such that n = dk. d n means that d | n is false.
Note that a | b = a/b. Recall that a/b represents the fraction a b. The expression d | n may be read in any of the following ways:
Thus, the following five statements are equivalent, that is, they are all different ways of saying the same thing.
Theorem 3.1 (Divisibility Properties). If n, m, and d are integers then the following statements hold:
Exercise 3.1. Prove each of the properties 1 through 10 in Theorem 3.1.
Definition 3.5. If c = as + bt for some integers s and t we say that c is a linear combination of a and b.
Thus, statement 3 in Theorem 3.1 says that if d divides a and b, then d divides all linear combinations of a and b. In particular, d divides a + b and a − b. This will turn out to be a useful fact.
Exercise 3.2. Prove that if d | a and d | b then d | a − b.
Exercise 3.3. Prove that if a ∈ Z then the only positive divisor of both a and a + 1 is 1.
