Fall 20 Independent Study Selberg s proof of the prime number theorem Sana Jaber
List of study units 000 0002 The notion of finite limit of a function at infinity. A first discussion of prime numbers. 0003 Every natural number greater than is either a prime or a product of prime numbers. 0004 If a number that divides the product of two numbers is coprime with the first, it must divide the second. 0005 A prime that divides a product of primes must be equal to one of them. 0006 0007 0008 0009 000 00 002 003 004 005 The squeeze theorem for limits at infinity. An exercise on limits. The limit of any negative power of x at infinity is zero. Addition and finite limits at infinity. The Fundamental theorem of Arithmetic. The Möbius function. Euler s function. Multiplicative and fully multiplicative functions. Definition of infinite limit at infinity. Functions that are bounded at infinity. 006 The sum of two functions that tend to infinity, also tends to infinity. 007 The product of two functions that tend to infinity, also tends to infinity. 008 009 0020 Introduction to congruences. Elementary properties of congruences. The Chinese remainder theorem.
002 0022 0023 0024 0025 0026 Definition of limit at infinity being negative infinity. A divisibility proposition that we have often used. The notion of group. An equivalence for some finite limits. The multiplicativity of the Möbious function. A very important property of the Möbious function. 0027 The sum of two function that have limit negative infinity also has limit negative infinity. 0028 The set of congruence classes mod n. 0029 The product of a function that tends to + and one that tends to tends to. 0030 003 0032 0033 Upper bounds, lower bounds. Definition of sequence. If f has limit + and g has limit l > 0 then fg has limit +. If f has limit and g has limit l > 0 then fg has limit. 0034 Notational conventions for functions - Formal definition for a function. 0035 The Möbius invesion formula. The function π(x) and the statement of the prime number the- The von Mangoldt function. The Chebyshev functions. Introduction to permutations. The statement of the mean value theorem. The number of permutations of n objects chosen k at a time. 0036 orem. 0037 0038 0039 0040 004 0042 The integer part of a real number.
0043 0044 0045 0046 0047 0048 0049 The definition of finite limit of a sequence. An exercise on limits. The largest power of a prime p that divides n!. The number of combinations of n objects chosen k at a time. Newton binomial formula. The supremum of a set. Characterization of the least upper bound of a set. 0050 Convention about the least upper bound of unbounded sets and of the empty set. 005 0052 An increasing function has a limit at infinity. Characterization of the greatest lower bound of a set. 0053 The greatest lower bound of a nonempty, bounded below set always exists. 0054 0055 0056 The limit at infinity of a decreasing function always exists. The big O and the small o notation. ϑ(x) = O(x) 0057 The limit of bounded function times one that tends to 0 is 0. 0058 The product of two function with finite limits at infinity. 0059 The limit of the reciprocal of a function with nonzero limit is the reciprocal of its limit. 0060 006 Definition of limit + for a sequence. Definition of limit for a sequence. 0062 Inferring limits of functions from limits sequences and conversely. 0063 0064 An additive group structure for Z/nZ. The operation of multiplication on Z/nZ.
0065 Using the mean value theorem to get estimates for differences between the values of a function. 0066 0067 0068 0069 0070 007 The limit of the sum of two sequences with finite limits. The Bernoulli inequality. Definition of monotonic sequences. Every monotonic sequence has a limit. Definition and characterization of bounded sequence. Any convergent sequence is bounded. 0072 The product of a bounded sequence and of one that tends to zero, tends to zero. 0073 0074 0075 Definition of series and a few simple examples. Some terminology relating to series. A nonnegative series with bounded partial sums is convergent. 0076 The multiplicative group (Z/nZ). 0077 Changing the first few terms of a sequence does not change its limit properties. 0078 Changing the first few summands of a series does not change its convergence properties. 0079 0080 008 0082 0083 The integral test. Definition of Cauchy sequence. The notion of subsequence. Any Cauchy sequence is bounded. The number e revisited. 0084 A sequence which is not bounded above has a subsequence that tends to +. 0085 An inequality involving factorials.
0086 0087 0088 limit. log(m!) = m log m + O(m) The upper and lower limit of a sequence. Comparing limits of subsequences with the upper and the lower 0089 The upper and the lower limit do occur as limits of subsequences of a given sequence. 0090 009 0092 0093 Why we need upper and lower limits. Limit of a function at a point. The upper and lower limits for a function. Upper and lower limits for functions and sequences. 0094 Expressing upper limits of functions through upper limits of sequences. Same for lower limits. 0095 0096 The rising sun lemma and the Bolzano-Weierstrass theorem. Any Cauchy sequence is convergent. 0097 An inequality between functions implies an inequality between their upper and lower limits. 0098 0099 000 00 002 003 004 A slightly different representation for the function ψ(x). Restating the prime number theorem. A general asymptotic formula. The Cauchy criterion for series. Absolute convergence of a series implies convergence. Mertens formula: p x log p p = log x + O() Limits of sequences characterize limits of functions. 005 Partial summation and the Abel-Pringsheim criterion for convergence. 006 Continuity
007 008 009 Preliminary work needed for Dirichlet s estimate. Every continuous function on a closed interval is bounded. The Dirichlet estimate. 00 Every continuous function on a closed interval has a minimum and a maximum. 0 The estimate p x log p log(x/p) = O(x). A continuous function on a closed interval is uniformly contin- 02 03 04 05 06 07 08 09 020 02 022 023 uous. The Intermediate Value Theorem The estimate log p log q = O(x). p α q β x α> The derivative and Fermat s theorem. The estimate ( log 2 x ) = O(x). n n x The second Möbius inversion formula. Rolle s theorem. The estimate p α x α> log 2 x = O(x). The proof of the mean value theorem and its consequences. The estimate d x µ(d) d = O(). The notion of uniform continuity. Selberg s function σ n (x) = ( µ(d) log 2 x ). d d n 024 Computation of σ n (x) for the case that n has three distinct prime divisors.
025 Upper and lower sums for a bounded function over a closed interval. 026 The estimate τ(n) n = ( 2 log2 x + c log x + c 2 + O 4 ). x n x 027 028 029 Inequalities for left and right sums as we refine the partition. Adjusting a sum. Proving that log 2 ([x] ) = log 2 x + O ( 4 x ). 030 If θ n = O( n), then 03 032 n [x] n [x] ( ) log n n(n + ) = constant + O 4 x The estimate n x σ n (x) = x d x ( θ n n(n + ) = constant + O 4 ). x µ(d) d ( x ) log2 + O(x) d 033 The definition of a Riemann integrable function and the Riemann integral. 034 The estimate σ n (x) = log 2 p + log p log q + O(x) n x p x pq x 035 036 037 038 The Riemann integrability criterion. The estimate µ(d) ( x ) d log2 = 2 log x + O(). d d x Every continuous function is integrable. Selberg s formula. 039 The Fundamental Theorem of Calculus (st form - Newton- Leibniz formula). 040 The estimate p x log p log q = O(x log x). 04 The inequality log k log(k + ) < k log 2 k.
042 The estimate 2 n x ( ) x log n = O. log x 043 The estimate pq x log p log q log(pq) = O(x). 044 The second form of the Fundamental Theorem of Calculus (existence of antiderivatives). 045 046 The estimate pq x The estimate p x log p log q = log x + pq x log p + pq x log p log q log(pq) log p log q log(pq) + O(x) ( ) x = 2x + O log x 047 048 The sum of two integrable functions is integrable. The inequality log(n + ) log n n < n 2. 049 The estimate p x log 2 p p = 2 log2 x + O(log x). 050 05 052 053 054 055 The estimate log p log q = pq 2 log2 x + O(log x). pq x log n log(n + ) n log 2 n < n 2 log 2 n + 2 n 2 log 3 n The estimate pq x log p log q = log x + O(log log x). pq log(pq) The estimate log p = O(log log x). p( + log(x/p)) p x ( ) ( x Every quantity which is O is also O log x The estimate p x log p + pq x log p log q log(pq) x + log x ( = 2x + O ). x + log x ).
056 057 pq x The estimate p x log p = 2x pq x log p log q = 2x log x pq x 058 The estimate ϑ(x) log x = pq x log p log q log(pq) ( ) x + O + log x ( ) log p log q x log(pq) ϑ +O(x log log x) pq ( ) log p log q x log(pq) ϑ +O(x log log x) pq 059 The function R(x) and an equivalent formulation of the Prime Number Theorem 060 06 062 R(x) log x + ( ) x log pr = O(x) p p x R(x) log x = ( ) log p log q x log(pq) R + O(x log log x) pq pq x The inequality R(x) log x ( ) log p x R + O(x) p p x 063 R(x) log x ( ) log p log q x log(pq) R + O(x log log x) pq pq x 064 2 R(x) log x ( ) log p x R + ( ) log p log q x p log(pq) R + pq p x pq x O(x log log x) 065 Rewriting ( ) log p x R. p p x 066 Rewriting pq x log p log q log(pq) ( ) x R. pq 067 The inequality R(a) R(b) ϑ(a) ϑ(b) + a b. 068 The estimate ( n x + log n n x ) = O(x log log x). n + n x
069 070 07 The inequality n + log n n + log(n ) + log(n ) ( n ( ( )) x x ϑ ϑ = O(x log log x) + log n n) n + n x n x 072 n x 2n n + log n 073 R(x) log x ( R ( x ) ( ) ) x R = O(x log log x) n n + ( R ( x ) ( ) ) x R n n + ( x R n n x = 2 n x ) ( ) x log log x + O log x ( x ) R + O(x) n 074 ϑ(n) The adjustment n(n + ) = n x n x ϑ(n) n 2 + O() 075 076 The estimate n x The estimate n x ϑ(n) n 2 R(n) n 2 = log x + O() = O() 077 There exists K > 0 such that for every two positive numbers x > x, we have n 2 < K. x n x R(n) 078 If x > x are two positive numbers with x > and x > x + 3, then n > 2 (log x log x) x n x 079 There exists a constant C > 0 such that log x log x < C for evey x, x 4 with x x x + 3. 080 There exists a constant K > 0 such that for every x, x 4 with x x x + 3, there exists a number y [x, x ] with R(y) y < K log(x /x)
08 There exists a constant K 2 > 0 so that if R(n) does not change sign for n [x, x ] where x, x 4 are two numbers with x > x + 3, then there exists y [x, x ] such that: R(y) y < K 2 log(x /x) 082 If n 4 is a natural number with R(n)R(n + ) < 0, then R(n) < log n. 083 There exists a constant K 3 such that if x > x 4 and R(n) changes sign in the interval [x, x ] then there exists y [x, x ] with: R(y) y < K 3 log x log x 084 There exists a constant K so that if x > x 4, then there exists y [x, x ] such that: R(y) y < K log x log x 085 There exists K > 0 such that for every δ > 0 and every x > 4, there exists y in the interval [ x, e K/δ x ] with: 086 087 R(y) < δy ( ) log y If y > y > 0, then R(y ) R(y) y y + O If y > 4 and y 2 y 2y, then: ( ) y R(y ) R(y) + y y + O log y y 088 The inequality e x/2 < x for 0 < x <. 089 If 0 < δ <, K > x > 4 and y ( x, xe k/δ), then the interval [ e δ/2 y, e δ/2 y ] contains a closed subinterval of the form [ z, e δ/2 z ] that is fully contained in ( x, xe k/δ).
090 There exist constants K > 0 and L > 0 such that for every δ with 0 < δ < and every x > e L/δ, the interval ( x, e K/δ x ) contains a subinterval [ z, e δ/2 z ] such that for every t in that subinterval. n x R(t) < 4δt 09 The inequality R(x) < (4 log 2 + )x for x > 0. 092 R(x) ( x ) ( ) x R + O log x n log x 093 Our assumptions for the final argument. 094 R(x) < x log x n (x/x o) n ( x ) ( ) x n x R + O n log x 095 R(x) < ax aδ ( ) x 4 log ρ + O log x ( ) 096 The sequence a n with a = b and a n+ = a n a2 n 300K 097 Finally! The Prime Number Theorem
STUDY UNITS
000 The purpose of this unit is to define and discuss the limit of a function f(x) as x tends to infinity. Intuitively speaking, we say that f(x) = l if as lim x + x gets larger and larger, f(x) approaches the value l. For example: lim x + x = 0 This is intuitively clear since as x gets larger and larger x gets smaller and smaller. The formal definition of limit ensures that if f(x) = l, then lim x + f(x) approaches l as close as one might wish. Before we present the formal definition, it would be a good idea to look at the classical approach and try to understand what is missing. The first inclination as far as the statement lim f(x) = l is concerned, x + would be to start pluging in large values of x and observe what f(x) does. This has the disadvantage that, in general, it is unpredictable how long one has to wait before the values of f(x) start approaching l to a satisfactory extent. Even the term satisfactory may be quite subjective. What is close for one situation might not be close enough for another. The formal definition of limit addresses both of these questions. It does so by demanding to check that any possible closeness between f(x) and l is eventually achieved. Here is the formal definition: Definition: Suppose that the domain of definition of the function f includes a set of the form (a, + ). We will then say that f(x) approaches the real number l as x tends to infinity and write f(x) = l, if for every ɛ > 0 lim x + there exists an x o > 0 such that if x > x o then f(x) l < ɛ. Let s try to understand how this definition addresses the points made above. First of all, ɛ expresses the desired amount of proximity between f(x) and l that one may wish to achieve. In other words, the definition is based on a challenge: Is it possible to achieve the given amount of proximity? The response to that is the x o that is provided so that for all x s greater than x o that proximity is achieved.
Let s see how that works in practice. We will prove that lim x + x = 0. We fix ɛ > 0. We wish to find x o > 0 so that if x > x o, then x 0 < ɛ, i.e. x < ɛ. If we solve this inequality for x, then we get x >. So if we set ɛ x o = ɛ, then for x > x o we have < ɛ which is what we wanted. x Homework. Use the formal definition of limit to show that lim x + x 5 = 0 Solution: We fix ɛ > 0. We wish to find x o > 0 so that if x > x o, then x 5 0 < ɛ,i.e. x 5 < ɛ. If we solve this inequality for x, we get ɛ < x 5 or x > ɛ + 5. We can set x o = ɛ + 5, then for x > x o we have x > ɛ + 5 which is equivalent to < ɛ as desired. x 5 2. Use the formal definition of limit to show that lim x + x 2 = 0 Solution:We fix ɛ > 0. We wish to find x o > 0 so that if x > x o, then x 2 0 < ɛ, i.e. x 2 < ɛ. If we solve this inequality for x2, we obtain x 2 >. Let s solve for x by taking the sqaure root of both sides to get ɛ x >. We can set x o =, then for x > x o we have x >. Square ɛ ɛ ɛ both sides of the inequality to obtain x 2 > ɛ which is equivalent to x 2 < ɛ as desired. 3. Use the formal definition of limit to show that lim x + x 2 + x + = 0 Solution: We fix ɛ > 0. We wish to find x o > 0 so that if x > x o, then
x 2 + x + 0 < ɛ. We know x 2 + x + for x > 0. So it is enough x to require that x < ɛ. But then this is equivalent to x > ɛ. So if we set x o = ɛ, then x > x o implies x < ɛ and since x 2 + x + x, we also have x 2 < ɛ, as desired. + x + x 4. Use the formal definition of limit to show that lim x + x + = 0 Solution: We fix ɛ > 0. We aim to show x o > 0 so that if x > x o then x x + < ɛ. By arithmetic this is equivalent to x + < ɛ,i.e. x + < ɛ. If we solve this inequality for x, then we get x >. Let ɛ x o = ɛ, then for x > x o we have < ɛ which is what we wanted. x +
0002 Let us first recall that the set of natural numbers N consists of all positive integer numbers:,2,3,... A natural number is called prime, if its only divisors are and itself. So the first few prime numbers are: 2, 5, 7,, 3, 7, 9, 23, 29, 3, 37, 4, 43, 47, 53,... People were fascinated by prime numbers from the very beginning. First of all they form the building stones for all other natural numbers. According to the Fundamental Theorem of Arithmetic every natural number greater than can be written in a unique way as a product of prime powers. Also, there are many fundamental questions about prime numbers that, despite the enormous progress of the last few centuries, still remain open. For example: Two prime numbers are called twin if they differ by 2. It is conjectured that there are infinitely many pairs of twin primes. It is conjectured (the Goldbach conjecture) that every even natural number greater than 2 can be written as the sum of two prime numbers. It is conjectured that there exist infinitely many prime numbers of the form n 2 + where n is a natural number. More important are the questions about the distribution of prime numbers. The prime number theorem roughly states that the number of primes n between and n is. Considerations about the actual error that one log n n makes by assuming that there are lead to very deep theories that touch log n upon the single most famous conjecture in mathematics: the Riemann hypothesis. The purpose of this study will be to understand the mathematics involved in proving the prime number theorem with methods that do not go beyond traditional Calculus.
0003 Proposition: Every natural number greater than is either a prime or a product of prime numbers. Proof: We show this by induction. In this case induction begins at 2 and not at since the statement is given for all numbers greater than. Induction can always be used with any set that satisfies the well ordering principle (i.e. every nonempty subset of that set has a least element) and certainly the natural numbers that are greater than form such a set. For n = 2 this is obvious because 2 is a prime number. Suppose that the statement is true for all k n. We will show that it is true for n +. Indeed, if n + is a prime number, then there is nothing to prove further. If n + is composite, then it can be written as the product of two natural numbers n and n 2 where < n, n 2 < n +. Thus each of the numbers n and n 2 is less than or equal to n. According to the inductive hypothesis each of them is either prime or otherwise a product of prime numbers. In all possible cases n n 2 is a product of prime numbers, i.e. the statement is correct for n +. This completes the proof.
0004 The purpose of this unit is to show that: Proposition: If n, a, b are natural numbers with n ab and gcd(n, a) =, then n b. Proof: Since n ab, there exists a natural number k such that ab = nk. Since gcd(n, a) = there exist integer numbers x and y such that nx + ay =. But if we multiply both sides by b, we get: bnx + bay = b = bnx + nky = b = n(bx + ky) = b The above relation implies that n b.
0005 The purpose of this unit is to show that: Proposition: Let p, q,..., q k be prime numbers such that p q q k. Then p is equal to q i for some i between and k. Proof: Suppose that the statement of the proposition is not always true. Then there exist prime numbers p, q,..., q k, so that p q q k and p q i for i =,..., k. We choose numbers p, q,..., q k, so that k (i.e. the number of primes q) is the least possible. Since p q, we have that (p, q ) =. By what was proven in unit 0004, we have that p q 2 q k. But this is impossible because we had assumed that k was the least possible amount of prime factors that can occur in such a situation and we just came across a similar situation with k factors. This contradiction shows that the statement of our proposition is true and our proof is complete. Is it a contradictin since we said k is the least possible and now we are working with k factors whcih is not much different (referring to the last sentence of the second paragraph)?
0006 Proposition: Suppose that three functions f, g, h are defined on (a, + ), for some a R, and are such that: f(x) g(x) h(x) for every x (a, + ). If lim g(x) = l. x + lim f(x) = lim h(x) = l for some l R, then x + x + Proof: We fix ɛ > 0. We need to find x o so that if x > x o, then: g(x) l < ɛ Since lim f(x) = l, there exists x R such that if x > x then f(x) l < x + ɛ. Also, since lim h(x) = l, there exists x 2 R such that if x > x 2 then x + h(x) l < ɛ. We set x o = max(x, x 2 ). We will show that this does the job. Indeed, it is enough to show that if x > x o, then: l ɛ < g(x) < l + ɛ But we know that for x > x o we have: l ɛ < f(x) < l + ɛ l ɛ < h(x) < l + ɛ Thus g(x) f(x) > l ɛ and g(x) h(x) < l +ɛ which completes the proof.
0007 Exercise: Suppose that a > 0 is a fixed real number. Show that lim ( x + a x) = 0 x + Solution: We multiply and divide by the conjugate quantity and we have: x + a x) a x + a + x x + a + x x + a + x = So it suffices to show that lim x + 0 x + a x x + a + x = a x + a + x = 0. We have that: a x + a + x a x By the squeeze theorem, it is enough to show that lim x + a x = 0. For this, we will work directly with the definition of limit. We fix an ɛ > 0. We wish to find an x o > 0 so that x > x o then a 0 x < a ɛ,i.e. < ɛ. If we solve the inequality for x, we ge x > a2. If we take x ɛ2 x o = a2 ɛ 2 then for x > x o we have x > a2 a which is equivalent to < ɛ, as ɛ2 x desired. Since lim x + a x = lim 0 = 0 the squeeze theorem implies that: x + lim x + a x + a + x = 0
0008 A general property of limits: Proposition: If a > 0 is a real number then, lim x + x a = 0. Proof: We fix ɛ > 0. We will show that there exists x o > 0 such that if x > x o then x a 0 < ɛ. This can be rewritten as x a < ɛ which is equivalent to x > ɛ /a. Thus if we take, x o = ɛ /a, then for x > x o we have < ɛ and this shows xa that lim = 0, as desired. x + xa
0009 How addition relates to limits at infinity. Proposition: Suppose that lim f(x) = l and lim g(x) = m. Then x + x + lim (f(x) + g(x)) = l + m x + Proof: We fix ɛ > 0. We need to find x o so that if x > x o, then: (f(x) + g(x)) (l + m) < ɛ Since lim f(x) = l, there exists x R such that if x > x then f(x) l < x + ɛ/2. Also, since lim g(x) = l, there exists x 2 R such that if x > x 2 then x + g(x) l < ɛ/2. We set x o = max(x, x 2 ). We will show that this does the job. Indeed, if x > x o, then: (f(x)+g(x)) (l+m) = (f(x) l)+(g(x) m) f(x) l + g(x) m < ɛ 2 + ɛ 2 = ɛ as needed.
000 We are now ready to prove the Fundamental Theorem of Arithmetic: Theorem: Let n be a natural number greater than. Then there exist unique prime numbers: p <... < p k where l and natural numbers m,..., m k such that; n = p m p m k k Proof: Let n be a natural number greater than. First of all, we have already shown that any natural number greater than is a prime or can be written as a product of prime numbers. In either case this means that there are prime numbers p,..., p k and natural numbers m,..., m k so that: n = p m p m k k We just have to show that the above expression is unique. Suppose that n can also be written as: n = q t qs ts There are two ways in which the above two prime factorizations are different. The first is if there is a prime number in one that does not appear in the other. However this is impossible, because every prime number that appears in the first, divides the product of primes in the second. So by 0005, it has to be equal to one of them. The same thing can be said about primes appearing in the second factorization. This means that k = s and p = q,..., p k = q k. The second way in which the factorizations may be different is if there are two different exponents. Without loss of generality, suppose that m t
and in particular m > t. Then we may divide both factorizations by p t and get two factorizations for n/p t, one that contains p and one that does not. This is impossible by our previous step. Therefore the exponents must coincide, which means that the prime factorization of any natural number greater than is unique, as desired. Exercise: Write down the prime factorizations of all natural numbers from 2 to 40. Solution: 2 = 2, 3 = 3, 4 = 2 2, 5 = 5, 6 = 2 3, 7 = 7, 8 = 2 3, 9 = 3 2, 0 = 2 5, =, 2 = 2 2 3, 3 = 3, 4 = 2 7, 5 = 3 5, 6 = 2 4, 7 = 7, 8 = 2 3 2, 9 = 9, 20 = 2 2 5 2 = 3 7, 22 = 2, 23 = 23, 24 = 2 3 3, 25 = 5 5, 26 = 2 3, 27 = 27, 28 = 2 2 7, 29 = 29, 30 = 2 3 5, 3 = 3, 32 = 2 5, 33 = 3, 34 = 2 7, 35 = 5 7, 36 = 2 2 3 2, 37 = 37, 38 = 2 9, 39 = 3 3, 40 = 2 3 5
00 First of all, we will say that a natural number greater than is square-free if it is not divisible by the square of any prime number. For example 6,0,2 are square free integers but 8,9,6,8 are not. Note that a square-free natural number greater than is a product of distinct prime numbers. The Möbius function is a function µ : N R which is defined as follows: µ() = If n is not square-free, then µ(n) = 0. If n > is a square free number and r is the number of its prime factors, i.e. n = p p r, then we define µ(n) = ( ) r. Exercise: Compute µ(n) for n 00. Solution: µ() =, µ(2) =, µ(3) =, µ(4) = 0, µ(5) = µ(6) =, µ(7) =, µ(8) = 0, µ(9) = 0, µ(0) = µ() =, µ(2) = 0, µ(3) =, µ(4) =, µ(5) = µ(6) = 0, µ(7) =, µ(8) = 0, µ(9) =, µ(20) = 0 µ(2) =, µ(22) =, µ(23) =, µ(24) = 0, µ(25) = 0 µ(26) =, µ(27) = 0, µ(28) = 0, µ(29) =, µ(30) = µ(3) =, µ(32) = 0, µ(33) =, µ(34) =, µ(35) = µ(36) = 0, µ(37) =, µ(38) =, µ(39) =, µ(40) = 0 µ(4) =, µ(42) =, µ(43) =, µ(44) = 0, µ(45) = 0 µ(46) =, µ(47) =, µ(48) = 0, µ(49) = 0, µ(50) = 0
µ(5) =, µ(52) = 0, µ(53) =, µ(54) = 0, µ(55) = µ(56) = 0, µ(57) =, µ(58) =, µ(59) =, µ(60) = 0 µ(6) =, µ(62) =, µ(63) = 0, µ(64) = 0, µ(65) = µ(66) =, µ(67) =, µ(68) = 0, µ(69) =, µ(70) = µ(7) =, µ(72) = 0, µ(73) =, µ(74) =, µ(75) = 0 µ(76) = 0, µ(77) =, µ(78) =, µ(79) =, µ(80) = µ(8) = 0, µ(82) =, µ(83) =, µ(84) = 0, µ(85) = µ(86) =, µ(87) =, µ(88) = 0, µ(89) =, µ(90) = 0 µ(9) =, µ(92) = 0, µ(93) =, µ(94) =, µ(95) = µ(96) = 0, µ(97) =, µ(98) = 0, µ(99) = 0, µ(00) = 0
002 Euler s number-theoretic (or totient) function φ is defined as follows. For a natural number n, φ(n) is the number of natural numbers from to n that are coprime to n. For example, we have φ() =, φ(2) =, φ(3) = 2 etc. Exercise. Compute φ(n) for n 20. Solution: φ() =, φ(2) =, φ(3) = 2, φ(4) = 2, φ(5) = 4 φ(6) = 2, φ(7) = 6, φ(8) = 4, φ(9) = 6, φ(0) = 4 φ() =, φ(2) = 4, φ(3) = 2, φ(4) = 6, φ(5) = 8 φ(6) = 8, φ(7) = 6, φ(8) = 6, φ(9) = 8, φ(20) = 8 Exercise What is φ(p) if p is a prime number? Justify your answer. Solution: If p is a prime number, then φ(p) = p since all numbers less than p are comprime to p. For example, 5 is prime and all the numbers less than 5(,2,3,4) are coprime to 5, so φ(5) = 5 = 4. Exercise What is φ(p n ) if p is prime and n is a natural number? Justify your answer. Solution: I looked this one up but I only understood some of the reasoning. φ(p n ) = φ(p n ) φ(p n ). If m = p n Yes, but I am afraid you are still not answering the question. Look at your answer to what φ(p) was. You gave φ(p) = p. So far you are not giving an answer here. Let s analyze the problem. We wish to find how many numbers between and p n are coprime to p n. But the only divisor of p n is p. In other words, we need to figure out what numbers between and p n are not divisible by p. Why don t we ask the opposite question? How many numbers between and p n are divisible by p? Then φ(p n ) will be p n minus that number.
OK, so let s figure this out. If a number between and p n is divisible by p then it can be written in the form pk. But since pk p n, we must have k p n. Conversely, if we begin with any number k between and p n, then pk will be a number divisble by p between and p n. In other words there are as many numbers that are divisible by p between and p n as the amount of numbers between and p n. This means that there are exactly p n multiples of p between and p n. This means that φ(p n ) = p n p n.
003 Definition: A function f : N R is called multiplicative if f() and f(nm) = f(n)f(m) for any two relatively prime numbers n and m. A multiplicative function will be called fully multiplicative if f(mn) = f(m)f(n) for any two natural numbers m and n. Multiplicative function are very common in number theory. Proving that a given function f : N R is multiplicative is not always so easy. Let s look at some simple examples of multiplicative functions. We begin with the celebrated tau function τ. For a natural number n we define τ(n) to be the number of divisors of n. For example, τ() =, τ(2) = 2, τ(3) = 2, τ(4) = 3 etc. Actually, there is a simple explicit formula for the function τ. Suppose that n is a natural number greater than with prime factorization: n = p m p m k k where p i are distinct prime numbers and m i are natural numbers. Then a divisor of n has the form: p l p l k k where l i m i for i =,..., k. So the number of divisors of n equals the number of choices of l,..., l k with 0 l i m i for i =,..., k. But there are exactly m i + integers between 0 and m i. So when we form a collection of numbers l,..., l k, then we have m + choices for l, m 2 + choices for l 2, etc. These means that we have (m + ) (m k + ) choices for l,..., l k. This means that we have (m + ) (m k + ) different divisors for n. This implies that: τ(p m p m k k ) = (m + ) (m k + ) Now, if n = p m p m k k and u = q s qst t are relatively prime, then p i
q j for any i and j. This implies that p m p m k k qs qst t factorization of nu. This means that is the prime τ(nu) = (m + ) (m k + )(t + ) (t s + ) = τ(n)τ(u) i.e. τ is multiplicative. Let s look at another function now. For evey natural number n, we define by σ(n) to be the sum of all divisors of n. In particular, σ() =, σ(2) = 3, σ(3) = 4, σ(4) = 7 etc. For a prime number p, we have σ(p) = p + since the only divisors of p are and p. Moreover, for a prime power p n, we have σ(p n ) = + p +... + p n = p n+ p. Now if n is a natural number with prime factorization: n = p m p m k k then it is easy to see that all divisors of n are summands that occur in the expansion of the product: ( + p +... + p m ) ( + p k +... + p m k k ) Conversely all summands in the expansion are divisors of n. This means that: or σ(p m p m k k = ( + p +... + p m ) ( + p k +... + p m k k ) σ(p m p m k p k = pm + pm k+ k p k
As before, if n and u are relatively prime with factorizations p m p m k k q s qst t respectively, then nu has prime factorization: and which means that σ(mn) is given by: nu = p m p m k k qs qst t p m + p pm k+ k p k q s + q qst+ t q t and this equals σ(n)σ(u). Thus σ is multiplicative too. More generally we can define: σ k (n) = d n d k i.e. we take the product of k-th powers of the divisors of n instead of first powers. Exactly as before, it can be proven that σ k is multiplicative. Exercise: Provide all details that prove that σ k is multiplicative. Solution:
004 Definition: Suppose that f is a function defined on (a, + ) for some real number a. Then we say that f(x) = +, if for every M > 0, there lim x + exists x o > 0 such that x > x o implies f(x) > M which is what we wanted. Exercise: Show that Solution: lim x = +. x + We fix M > 0. We need to find an x o > 0 so that if x > x o then x > M. If we solve the inequality for x, x > M 2. Let x o = M 2 so if x > M 2 then x > M.
005 Definition: Let f be a function whose domain of definition contains an interval of the form (a, + ) for some real number a. We will say that f is bounded at infinity if there exists a positive number C and an x o > 0 such that: f(x) C for x > x o Intuitively speaking, the graph of a function which is bounded at infinity is contained in a strip of finite width. Our first order of business is to show that when we multiply a bounded function with a function that tends to zero, then the product tends to 0. Proposition: Suppose that lim f(x) = 0 and that g is bounded at infin- x + ity. Then lim x + f(x)g(x) = 0. Proof: Suppose that x and C are positive real numbers so that g(x) < C for x > x. We now fix ɛ > 0. We need to find x o > 0 so that f(x)g(x) < ɛ for x > x o. Since lim f(x) = 0, there exists a x 2 such that for x > x 2, we have x + f(x) < ɛ C. We now set x o = max(x, x 2 )., Then for x > x o both the conditions g(x) < C and f(x) < ɛ C hold. Therefore: f(x)g(x) < ɛ C C = ɛ for x > x o. This completes our proof. To illustrate, this proposition consider lim x + sin x. We claim that this limit x
is 0. Indeed, lim = 0 as we have already shown. On the other hand x + x sin x for every x R, which means that sin x is bounded at infinitity. Therefore the product sin x must tend to zero by the proposition above. x We will now see that we can find plenty of bounded at infinity functions among functions whose limit at infinity exists. In particular: Proposition: Suppose that f is a function for which lim Then f is bounded at infinity. x + f(x) = l R. Proof:. By the definition of limit, there exists an x o > 0, so that for x > x o we have f(x) l < - we just applied the definition of limit for ɛ =. Then if x > x o we have f(x) l <. But this means that: f(x) = f(x) l + l f(x) l + l < + l This means that f(x) < + l for x > x o, which means that f is bounded at infinity, as desired. It is only natural to have expected a proposition like the above. After all, being bounded at infinity, means that the graph of the function is ultimately contained in a strip of finite width (between C and C where C is the constant in the definition of boundedness at infinity). If a function has a limit at infinity, then its graph is contained in strips of ever decreasing width. Notation: If a function f is bounded at infinity, we will write that f(x) = O() as x tends to infinity.
006 In this unit we will show that the sum of two function that have limit infinity at infinity, also has limit infinity. Proposition: Suppose that Then lim f(x) = + and that lim g(x) = +. x + x + lim f(x) + g(x) = + x + Proof: We fix M > 0. We need to find x o > 0 such that if x > x o, then f(x) + g(x) > M. Since lim f(x) = +, there exists x > 0 such that if x > x, then x + f(x) > M 2. Also, since lim g(x) = +, there exists x 2 > 0 such that if x + x > x 2, then g(x) > M 2. If we take x o = max(x, x 2 ), then for x > x o both conditions f(x) > M 2 g(x) > M 2 hold true, which implies that: and f(x) + g(x) > M 2 + M 2 = M as desired.
007 Proposition: Suppose that Then lim f(x) = + and that lim g(x) = +. x + x + lim f(x)g(x) = + x + Proof: Try to provide the proof of this proposition yourself. Fix M > 0. We need to find an x o > 0 so that if x > x o then f(x)g(x) > M. Since lim f(x) = + We need to find x o > 0 so that if x > x, then x + f(x) > M. Also, since lim gx = +, there exists x 2 > 0 such that if x + x > x 2, then g(x) > M. If we set x o = max(x, x 2 ), then for x > x o both condition f(x) < M and g(x) < M hold true, which implies that: which is what we wanted. f(x)g(x) = M M < M
008 Definition: Let n be a natural number. We will say that the integer numbers a and b are congruent modulo n if n divides the difference a b. In this case we will be writing a b mod n. Let s try to understand what this definition means. According to the theory of Euclidean division, every integer number m can be written as kn+r where 0 r < n. For example, if n = 5, then 7 = 5+2 whereas 2 = ( 3) 5+3. r is referred to as the remainder of the division of m by n. Thus integer numbers can be partitioned into n classes according to the remainder that they leave when divided by n. Let us call these classes congruence classes mod n. Then two numbers a and b are congruent mod n if and only if they have the same remainder when divided by n. Equivalently, if they belong to the same congruence class mod n. Let us give some examples: 22 8 mod 5 Indeed both numbers leave remainder 2 when divided by 5. Also, their difference 22 ( 8) is divisible by 5. All integer numbers are congruent mod There are two congruence classes mod 2. Even numbers are congruent to 0 mod 2, whereas odd numbers are congruent to mod 2. There are three congruence classes mod 3. Every integer number is congruent to either, 0 or mod 3. Note that we could have said 0, or 2. However, it is up to us to choose the representatives from each congruence class and single them out.
009 We prove the following simple properties for congruences. In all of them, we work with a fixed natural modulus n. a a mod n for any integer number a. Indeed n always divides a a = 0. a b mod n = b a mod n. Indeed if n divides a b, then it also divides b a. If a b mod n and b c mod n, then a c mod n. Indeed, the first fact implies that n (a b) and the second that n (b c). But then n must divide their sum (a b) + (b c) = a c. This means precisely that a c mod n. Incidentally, the above three properties tell us that congruence mod n is an equivalence relation, i.e. it is reflexive, symmetric and transitive. If a b mod n and a 2 b 2 mod n, then a + a 2 b + b 2 mod n. Indeed, the two given facts imply that n a b and that n a 2 b 2. Then n divides the sum (a b )+(a 2 b 2 ). But this can be rewritten as (a + a 2 ) (b + b 2 ) which means that a + a 2 b + b 2 mod n. If a b mod n and a 2 b 2 mod n, then a a 2 b b 2 mod n. Indeed, the two given facts imply that n a b and that n a 2 b 2. We now need to show that n divides a a 2 b b 2, But this latter can be written as: a a 2 b b 2 = a a 2 a b 2 + a b 2 b b 2 = a (a 2 b 2 ) + b 2 (a b ) This last expression can be easily seen to be divisible by n, which implies that a a 2 b b 2 mod n. Why is this part of the expression a (a 2 b 2 ) + b 2 (a b ) divisible by n? We proved this in Math 25 but it was a different approach.
Because n (a 2 b 2 ) and n (a b ). Proof: We are given that a is congruent to b mod n and c is congruent to d mod n. So, n (a b) and n (c d). Applying a previous proposition to (a b) and c and (c d) and b, we can say n [(a b)(c)] and n [(c d)(b)]. This can be written as n (ac bc) and n (bc bd). Then n [(ac bc) + (bc bd)]. We simplify to get n (ac bd) and therefore ac bd mod n. If a b mod n and k is a natural number then a k b k mod n. Induction is one way to do this. However using the fact that n a b in order to show that n a k b k can also be done by means of the identity: a k b k = (a b)(a k + a k 2 b +... + b k ) If we wanted to expand the right side more, what would we get (a b)(a k + a k 2 b + a k 3 b?... + b k ) If you expand the left side you will get the left. Try to do it for a concrete example, say k = 5. Write out the identity and then fold the right hand side. After you cancel everything that can be cancelled you will get the left side.
0020 In this unit, we prove: Proposition: Suppose that m and n are two relatively prime natural numbers. If we are integer numbers a and b, then there is a number c that satisfies: c a mod m c b mod n Moreover any two numbers c and c that satisfy these two congruences are congruent mod mn. Proof: Since m and n are coprime, there exist integers x and y, such that mx + ny =. We claim that the number c := any + bmx has the desired properties. Let s first check that c a mod m. If we take the difference c a, then we can write it as: c a = any + bmx a = a(ny ) + bmx = a( mx) + bmx = m(bx ax) i.e. c a is divisble by m which implies that c a mod m. In an identical fashion we can see that c b mod n. Indeed: c b = any + bmx b = any + b(mx ) = any + b( ny) = n(ay by) Thus n c b which shows that c b mod n. Finally we wish to show that if c and c both satisfy the congruences then c c mod mn. In other words we need to show that mn divides c c.
Since m and n are relatively prime, it suffices to show that m c c and n c c. For the first: c c a a mod m i.e. c c mod m For the second, c c b b mod n i.e. c c mod n and this completes the proof. In the proposition c a mod n and c b mod n, does the number c have to be the same number that satisfy both of these two congruences for the relatively prime numbers m, n? Is c and c them same number that satisfy the first part of the proposition? Yes it has to be the same number that satisfies these two simultaneous congruences. In cases like this, it would be a good idea to do some examples. To do so, take two relatively prime moduli e.g. 3 and 5 and also take two random remainders. e.g. 2 and 3. Try to find a number c which is 2 mod 3 and 3 mod 5. Then try to see that all numbers that differ from it by multiples of 5 satisfy the same congruences. In fact these are exactly the numbers that satisfy these congruences. c is supposed to be a number that satisfies the congruences from the first part. The conclusion for this number is that it has to be congruent to c mod (mn).
002 Definition: Suppose that f is a function defined on (a, + ) for some real number a. Then we say that f(x) =, if for every M > 0, there lim x + exists x o > 0 such that x > x o implies f(x) < M which is what we wanted. Exercise: Show that lim x + x3 + x =. We fix M > 0. We need to find an x o > 0 so that if x > x o then x 3 + x < M. Let s suppose x > 2. Then x 3 x > 2 2 x x = 3x. We need to show 3x > M.Fix M > 0. We need to find an x > 0 so that if x > x then 3x > M. If we solve the inequality for x, x > M/3. Let x = M/3 then we have x > M/3 which is equivalent to 3x > M as desired. If we take x o = max(m/3, 2) then for x > x o both conditions 3x > M and x > 2 hold which implies 2 2 x x > M which implies x 3 x > M. But if we multiply by (-), we get x 3 + x < M. The solution as you present it is correct. But there are some points that can be smoothened out: You say Fix M > 0 twice in the course of the proof. We said that M presents our proximity challenge and we have it fixed throughout the proof. You write We need to show 3x > M. You don t try to show this. You try to establish it. In other words, you try to find x o past which, this is true. So let me rewrite the proof with this things smoothened out. Notice how I will avoid mentioning x : We fix M > 0. We need to find an x o > 0 so that if x > x o then x 3 + x < M. We initially set x > 2. Then x 3 x > 2 2 x x = 3x.
Now, it suffices to establish 3x > M. This is true if x > M/3. So we may set x o = M/3. But we would also like to consider the fact that all of this was possible for x > 2. Therefore, we set x o = max(2, M/3). Indeed if x > x o then x 3 x > 4x x = 3x > M and if we multiply by (-), we get x 3 + x < M, as desired.
0022 In this unit we prove a proposition that we have been using quite often. Proposition: Suppose that a and b are two relatively prime numbers that divide the integer number n. Then ab n. Proof: Since a and b are relatively prime, there exist integer number x and y such that: ax + by = If we multiply both numbers of the above equation by n, then we get: n = nax + nby On the other hand, since a n we can write n = au. Also, since b n, we can write m = bv. Now the relation n = nax + nby may be written: n = (bv)ax + (au)by This can be rewritten as: n = ab(vx + uy) which shows that ab n.
0023 In this unit we present an algebraic structure that generalizes the usual operations that we use in mathematics. Take, for example, the case of real numbers. The set R of real numbers is equipped with the operation of addition. This operation assigns to every pair of real numbers, another real number: their sum. This operation has the property that it is associative, it has a neutral element (zero) and every number has an additive inverse (positive). We would like to single out the structures that satisfy similar properties. This prompts us to introduce the notion of group: Definition: A set G equipped with a map : G G G, where we use the notation a b := (a, b) for every a, b G, is called a group if: i) (a b) c = a (b c) for every a, b, c G. ii) There exists a unique element e G such that a e = e a = a for every a G. iii) For every a G, there exists a G such that a a = a a = e. The map is referred to as the group operation. Strictly speaking a group is the pair (G, ) but when the operation is clear from the context, its explicit reference is omitted and we say let G be a group... If in addition, the following property is satisfied: iv) a b = b a for every a, b G then G is called a commutative or abelian group. For a noncommutative group, we frequently use the multiplicative notation. Here the group operation is denoted by or is omitted, e from ii) is denoted by and a is denoted by a. For commutative groups we usually have the additive notation. In this case the group operation is denoted by +, e is denoted by 0 and a by a. Example: Let us denote by R the set of all nonzero real numbers. We consider the operation of multiplication. Then R with the operation of multiplication is a group. Indeed, multiplication is associative, it has a
neutral element (), and each nonzero number has a multiplicative inverse (its reciprocal). Finally the operation is commutative, which means that R equipped with multiplication is an abelian group.
0024 In this unit we will show an equivalence concerning the definition of limit. Proposition: Suppose that f is a function defined on an interval of the form (a, + ), for some a R and l is a real number. Then the following statements are equivalent: lim x + f(x) = l lim (f(x) l) = 0 x + lim f(x) l = 0 x + Proof We examine what quantity has to be made small in each case according to the definition of limit. We have: f(x) l (f(x) l) 0 f(x) l 0 All of these quantities are identical. If one can be made small, the same is true for the other two. The proof is complete.
0025 In this unit, we prove that the Möbius function is multiplicative. Proposition The Möbius function is multiplicative. Proof: We need to show that if m and n are relatively prime natural numbers, then µ(mn) = µ(m)µ(n). If either m = or n =, then there is nothing to prove. Also, if either m or n is not square free (i.e. it is divisible by the square of a prime number), then both µ(mn) and µ(m)µ(n) are zero. Therefore we may suppose that m and n are square free. Then each one of them is a product of distinct prime numbers. Since m and n are relatively prime, mn is also a product of distinct prime numbers. If m is a product of r prime numbers and n is a product of s prime numbers, then mn is a product of r + s prime numbers. This means that: µ(m) = ( ) r µ(n) = ( ) s µ(mn) = ( ) r+s This means that µ(mn) = µ(m)µ(n), as desired.
0026 In this unit, we prove another very important property of the Möbius function. We show that: Proposition: Suppose that n is a natural number. Then: µ(d) = d n { 0 n > n = Proof: If n =, then there is nothing to prove. So let s assume that n >. Suppose that n = p m p m k k is the prime factorization of n. Then only square-free divisors of n contribute non-zero values to the sum µ(d). d n But such divisors are also divisors of p p k. Therefore, without loss of generality, let us suppose that n = p p k, where all primes in this sum are distinct. The sum of all the divisors of n is given by the product: ( + p ) ( + p k ) The sum of all the µ values of the divisors is given by replacing each prime in the above product by. This means that the sum of the µ values of the divisors is: ( + ( )) ( + ( )) k factors which equals 0. This completes the proof Remark: We will revisit this proof after the Newton binomial formula becomes available.
0027 Proposition: If lim f(x) = and lim g(x) =, then x + x + lim (f(x) + g(x)) = x + Proof: Please provide the proof of this fact. We fix M > 0. We need to find an x o > 0 such that if x > x o then (f(x) + g(x)) < M Since lim f(x) = there exists x > 0 such that if x > x then x + f(x) < M/2. Also, since lim g(x) = there exists x 2 > 0 such that x + if x > x 2 then g(x) < M/2. If we take x o = max(x, x 2 ) then for x > x o both conditions f(x) < M/2 and g(x) < M/2 hold true which implies as desired. f(x) + g(x) < ( M/2) + ( M/2) = M
0028 We fix a natural modulus n and we consider the set of congruence classes mod n. There are exactly n such classes. They correspond to the possible remainders of the division by n. More precisely, we have the class consisting of those numbers that are divisible by n, the class of those numbers that leave remainder, when divided by n etc. Let us introduce a notation for these classes. We will denote by [a] the congruence class mod n that contains the number a. Let s look at a concrete example. Suppose that n = 4. Then we have four congruence classes: [0], [], [2] and [3]. The number a in the case of a congruence class is called a representative of the class and it is by no means unique. If we stay in our example, then [] = [5] = [ 3], etc. In other words, we could have been using other representatives in order to represent a given class. The set of congruence classes mod n will be denoted by Z/nZ. To make sure Z/nZ = n, i.e. there are n congruence classes mod n. We willl now introduce an operation in Z/nZ. We will show how to add congruence classes. We define: [a] + [b] := [a + b] Let us try to understand how this works with the example that we introduced above. If we take the sum [2] + [2] then that is defined to be [4]. But [4] is the same congruence class as [0]. Therefore we have: [2] + [2] = [0]. Let s make a table that fully describes the operation: + [0] [] [2] [3] [0] [0] [] [2] [3] [] [] [2] [3] [0] [2] [2] [3] [0] [] [3] [3] [0] [] [2] So this is the addition table for Z/4Z.
Exercise: Give the addition table for Z/6Z. + [0] [] [2] [3] [4] [5] [0] [0] [] [2] [3] [4] [5] [] [] [2] [3] [4] [5] [0] [2] [2] [3] [4] [5] [0] [] [3] [3] [4] [5] [0] [] [2] [4] [4] [5] [0] [] [2] [3] [5] [5] [0] [] [2] [3] [4]
0029 Proposition: If lim f(x) = and lim g(x) =, then x + x + lim (f(x)g(x)) = x + Proof: Please provide the proof of this fact. Fix M > 0. We need to find an x o > 0 such that if x > x o then (f(x)g(x)) < M. Since lim = there exists an x such that if x > x then f(x) < M x + which is equivalent to f(x) > M. Also, since lim g(x) = there x + exists an x 2 such that x > x 2 then g(x) > M. If we set x o = max(x, x 2 ) then for x > x o both condititons, f(x) > M and g(x) > M, hold true which implies f(x)g(x) > M M = M. If we multiply by (-) on both sides of the inequality, we get as desired. (f(x)g(x)) < M
0030 We will give the definition of upper bound and lower bound for a set. Definition: Suppose that A is a subset of R. We will say that the real number z is an upper bound for A if a z for every a A. Similarly we have: Definition: Suppose that A is a subset of R. We will say that the real number z is a lower bound for A if a z for every a A. For example 0 is a lower bound for N. Also and 2 are lower bounds for N. If a set has a lower bound z, then any number less than z is also a lower bound for the same set. Obviously, there are sets that have no lower bound. neither an upper bound nor a lower bound. For example Z has Definition: If a set has a lower bound, then it is called bounded below. If it has an upper bound, then it is called bounded above. If a set is both bounded below and bounded above, then it is called bounded. Exercise: Determine whether the following sets are bounded below or bounded above. If they are provide a lower bound or an upper bound, respectively. [0, ]: Bounded. Lower bound: 0 and upper bound: since the set is a closed interval. (0, + ): Bounded below. Lower bound: 0 and any number less than 0. the set of prime numbers: Bounded below. Lower bound: 2 and any number less than 2. { } n : n N Bounded. Lower bound: 0 and any number less than 0. Upper bound: or any number greater than. {x R : x 2 > 2} Neither bounded above nor below but x, 0, because the inequality would not hold true.