Famous theorems of mathematics From Wikibooks, the open-content textbooks collection (Redirected from The Book of Mathematical Proofs ) There are no reviewed revisions of this page, so it may not have been checked for quality. Jump to: navigation , search This book or module is under construction. An editor will return to it shortly. You can help assist in its construction . Please see the discussion page for any active discussion on this. Contents [hide ] 1 Preface 2 Table of Contents o 2.1 High School o 2.2 Undergraduate o 2.3 Postgraduate 3 Old table of Contents 4 Further Reading 5 Resources o 5.1 Manual of style o 5.2 External links etc [edit ] Preface
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Famous theorems of mathematicsFrom Wikibooks, the open-content textbooks collection
(Redirected from The Book of Mathematical Proofs)
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This book or module is under construction. An editor will return to it shortly.You can help assist in its construction. Please see the discussion page for any active discussion on this.
Contents
[hide]
1 Preface 2 Table of Contents
o 2.1 High School
o 2.2 Undergraduate
o 2.3 Postgraduate
3 Old table of Contents
4 Further Reading
5 Resources
o 5.1 Manual of style
o 5.2 External links etc
[edit] Preface
Mathematics deals with proofs. Whatever statement, remark, result etc one uses in mathematics it is considered meaningless until is accompanied by a rigorous mathematical proof. This book is intended to contain the proofs (or sketches of proofs) of many famous theorems in mathematics in no particular order. It should be used both as a learning resource, a good practice for acquiring the skill for writing your own proofs is to study the existing ones, and for general references.
It is not however intended as a companion to any other wikibook or wikipedia articles but can complement them by providing them with links to the proofs of the theorems they contain.
One note here. There are usually many ways to solve a problem. Many times the proof used comes down to the primary definitions of terms involved. We will follow the definition given by the first major contributor.
[edit] Table of Contents
[edit] High School
Pythagoras Theorem Euclid's proof of the infinitude of primes
√2 is irrational
sin 2 Θ+cos 2 Θ=1
[edit] Undergraduate
e is irrational π is irrational
Fermat's little theorem
Fermat's theorem on sums of two squares
Sum of the reciprocals of the primes diverges
Bertrand's postulate
Law of large numbers
Spectral Theorem
[edit] Postgraduate
Fermat's last theorem Brouwer fixed-point theorem
Jordan Curve Theorem
Prime number theorem
[edit] Old table of Contents
This section contains the table of content of the book as according to its original intentions. The material here should be either incorporated in the existing book or discarded.
ProofWiki.org - The online compendium of mathematical proofs
GNU Free Documentation License
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Famous theorems of mathematics/Pythagoras TheoremFrom Wikibooks, the open-content textbooks collection
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The Pythagoras Theorem or the Pythagorean theorem, named after the Greek mathematician Pythagoras states that:
In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).
This is usually summarized as follows:
The square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.
If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the equation:
or, solved for c:
If c is already given, and the length of one of the legs must be found, the following equations can be used (The following equations are simply the converse of the original equation):
or
This equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle it reduces to the Pythagorean theorem.
o 4.2 List of primitive Pythagorean triples up to 100
o 4.3 The existence of irrational numbers
o 4.4 Distance in Cartesian coordinates
[edit] History
The history of the theorem can be divided into four parts: knowledge of Pythagorean triples, knowledge of the relationship between the sides of a right triangle, knowledge of the relationship between adjacent angles, and proofs of the theorem.
Megalithic monuments from circa 2500 BC in Egypt, and in Northern Europe, incorporate right triangles with integer sides. Bartel Leendert van der Waerden conjectures that these Pythagorean triples were discovered algebraically.
Written between 2000 and 1786 BC, the Middle Kingdom Egyptian papyrus Berlin 6619 includes a problem whose solution is a Pythagorean triple.
During the reign of Hammurabi the Great, the Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC, contains many entries closely related to Pythagorean triples.
The Baudhayana Sulba Sutra, the dates of which are given variously as between the 8th century BC and the 2nd century BC, in India, contains a list of Pythagorean triples discovered algebraically, a statement of the Pythagorean theorem, and a geometrical proof of the Pythagorean theorem for an isosceles right triangle.
The Apastamba Sulba Sutra (circa 600 BC) contains a numerical proof of the general Pythagorean theorem, using an area computation. Van der Waerden believes that "it was certainly based on earlier traditions". According to Albert Bŭrk, this is the original proof of the theorem; he further theorizes that Pythagoras visited Arakonam, India, and copied it.
Pythagoras, whose dates are commonly given as 569–475 BC, used algebraic methods to construct Pythagorean triples, according to Proklos's commentary on Euclid. Proklos, however, wrote between 410 and 485 AD. According to Sir Thomas L. Heath, there is no attribution of the theorem to Pythagoras for five centuries after Pythagoras lived. However, when authors such as Plutarch and Cicero attributed the theorem to Pythagoras, they did so in a way which suggests that the attribution was widely known and undoubted.
Around 400 BC, according to Proklos, Plato gave a method for finding Pythagorean triples that combined algebra and geometry. Circa 300 BC, in Euclid's Elements, the oldest extant axiomatic proof of the theorem is presented.
Written sometime between 500 BC and 200 AD, the Chinese text Chou Pei Suan Ching (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) gives a visual proof of the Pythagorean theorem — in China it is called the "Gougu Theorem" (勾股定理) — for the (3, 4, 5) triangle. During the Han Dynasty, from 202 BC to 220 AD, Pythagorean triples appear in The Nine Chapters on the Mathematical Art, together with a mention of right triangles.
The first recorded use is in China, known as the "Gougu theorem" (勾股定理) and in India known as the Bhaskara Theorem.
There is much debate on whether the Pythagorean theorem was discovered once or many times. Boyer (1991) thinks the elements found in the Shulba Sutras may be of Mesopotamian derivation.
[edit] Proofs
This is a theorem that may have more known proofs than any other; the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs.
[edit] Proof using similar triangles
Proof using similar triangles.
Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles.
Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A,
meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..: As
so
These can be written as
Summing these two equalities, we obtain
In other words, the Pythagorean theorem:
[edit] Euclid's proof
Proof in Euclid's Elements
In Euclid's Elements, Proposition 47 of Book 1, the Pythagorean theorem is proved by an argument along the following lines. Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata:
1. If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent. (Side - Angle - Side Theorem)
2. The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
3. The area of any square is equal to the product of two of its sides.
4. The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3).
The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.
The proof is as follows:
1. Let ACB be a right-angled triangle with right angle CAB.2. On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order.
3. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
4. Join CF and AD, to form the triangles BCF and BDA.
Illustration including the new lines
1. Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
2. Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
3. Since AB and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC.
4. Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD.
5. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
6. Therefore rectangle BDLK must have the same area as square BAGF = AB2.
7. Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
8. Adding these two results, AB2 + AC2 = BD × BK + KL × KC
9. Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
10. Therefore AB2 + AC2 = BC2, since CBDE is a square.
This proof appears in Euclid's Elements as that of Proposition 1.47.
[edit] Garfield's proof
James A. Garfield (later President of the United States) is credited with a novel algebraic proof[1] using a trapezoid containing two examples of the triangle, the figure comprising one-half of the figure using four triangles enclosing a square shown below.
Proof using area subtraction.
[edit] Similarity proof
From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones. By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the larger square is the sum of the areas of the two smaller squares.
[edit] Proof by rearrangement
A proof by rearrangement is given by the illustration and the animation. In the illustration, the area of each large square is (a + b)2. In both, the area of four identical triangles is removed. The remaining areas, a2 + b2 and c2, are equal. Q.E.D.
A square created by aligning four right angle triangles and a large square.
This proof is indeed very simple, but it is not elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself and Banach-Tarski paradox. Actually, this difficulty affects all simple Euclidean proofs involving area; for instance, deriving the area of a right triangle involves the assumption that it is half the area of a rectangle with the same height and base. For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above).
A third graphic illustration of the Pythagorean theorem (in yellow and blue to the right) fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller.[1]
[edit] Algebraic proof
An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by an angle corresponding with the side of length C.
The A-side angle and B-side angle of each of these triangles are complementary angles, so each of the angles of the blue area in the middle is a right angle, making this area a square with side length C. The area of this square is C2. Thus the area of everything together is given by:
However, as the large square has sides of length A + B, we can also calculate its area as (A + B)2, which expands to A2 + 2AB + B2.
(Distribution of the 4)
(Subtraction of 2AB)
[edit] Proof by differential equations
One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse in the following diagram and employing a little calculus.
Proof using differential equations.
As a result of a change in side a,
by similar triangles and for differential changes. So
upon separation of variables.
which results from adding a second term for changes in side b.
Integrating gives
When a = 0 then c = b, so the "constant" is b2. So
As can be seen, the squares are due to the particular proportion between the changes and the sides while the sum is a result of the independent contributions of the changes in the sides which is not evident from the geometric proofs. From the proportion given it can be shown that the changes in the sides are inversely proportional to the sides. The differential equation suggests that the theorem is due to relative changes and its derivation is nearly equivalent to computing a line integral.
These quantities da and dc are respectively infinitely small changes in a and c. But we use instead real numbers Δa and Δc, then the limit of their ratio as their sizes approach zero is da/dc, the derivative, and also approaches c/a, the ratio of lengths of sides of triangles, and the differential equation results.
[edit] Converse
The converse of the theorem is also true:
For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b.
This converse also appears in Euclid's Elements. It can be proven using the law of cosines, or by the following proof:
Let ABC be a triangle with side lengths a, b, and c, with a2 + b2 = c2. We need to prove that the angle between the a and b sides is a right angle. We construct another triangle with a right angle between sides of lengths a and b. By the Pythagorean theorem, it follows that the hypotenuse of this triangle also has length c. Since both triangles have the same side lengths a, b and c, they are congruent, and so they must have the same angles. Therefore, the angle between the side of lengths a and b in our original triangle is a right angle.
A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Where c is chosen to be the longest of the three sides:
If a2 + b2 = c2, then the triangle is right. If a2 + b2 > c2, then the triangle is acute.
If a2 + b2 < c2, then the triangle is obtuse.
[edit] Consequences and uses of the theorem
[edit] Pythagorean triples
A Pythagorean triple has 3 positive numbers a, b, and c, such that a2 + b2 = c2. In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths. Evidence from megalithic monuments on the Northern Europe shows that such triples were known before the discovery of writing. Such a triple is commonly written (a, b, c). Some well-known examples are (3, 4, 5) and (5, 12, 13).
[edit] List of primitive Pythagorean triples up to 100
One of the consequences of the Pythagorean theorem is that irrational numbers, such as the square root of 2, can be constructed. A right triangle with legs both equal to one unit has hypotenuse length square root of 2. The Pythagoreans proved that the square root of 2 is irrational, and this proof has come down to us even though it flew in the face of their cherished belief that everything was rational. According to the legend, Hippasus, who first proved the irrationality of the square root of two, was drowned at sea as a consequence.
[edit] Distance in Cartesian coordinates
The distance formula in Cartesian coordinates is derived from the Pythagorean theorem. If (x0, y0) and (x1, y1) are points in the plane, then the distance between them, also called the Euclidean distance, is given by
More generally, in Euclidean n-space, the Euclidean distance between two points, and , is defined, using the Pythagorean theorem, as:
Wikipedia has related information atPythagoras Theorem
Cite error: <ref> tags exist, but no <references/> tag was found
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Famous theorems of mathematics/√2 is irrationalFrom Wikibooks, the open-content textbooks collection
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This is a proof by contradiction, so we assumes that and hence for some a, b that are coprime.
This implies that . Rewriting this gives .
Since the left-hand side of the equation is divisible by 2, then so must the right-hand side, i.e., 2 | a2. Since 2 is prime, we must have that 2 | a.
So we may substitute a with 2a', and we have that .
Dividing both sides with 2 yields , and using similar arguments as above, we conclude that 2 | b.
Here we have a contradiction; we assumed that a and b were coprime, but we have that 2 | a and 2 | b.
Hence, the assumption were false, and cannot be written as a rational number. Hence, it is irrational.
[edit] Another Proof
The following reductio ad absurdum argument is less well-known. It uses the additional information √2 > 1.
1. Assume that √2 is a rational number. This would mean that there exist integers m and n with n ≠ 0 such that m/n = √2.
2. Then √2 can also be written as an irreducible fraction m/n with positive integers, because √2 > 0.
3. Then , because .
4. Since √2 > 1, it follows that m > n, which in turn implies that m > 2n – m.
5. So the fraction m/n for √2, which according to (2) is already in lowest terms, is represented by (3) in strictly lower terms. This is a contradiction, so the assumption that √2 is rational must be false.
Similarly, assume an isosceles right triangle whose leg and hypotenuse have respective integer lengths n and m. By the Pythagorean theorem, the ratio m/n equals √2. It is possible to construct by a classic compass and straightedge construction a smaller isosceles right triangle whose leg and hypotenuse have respective lengths m − n and 2n − m. That construction proves the irrationality of √2 by the kind of method that was employed by ancient Greek geometers.
[edit] Notes
As a generalization one can show that the square root of every prime number is irrational. Another way to prove the same result is to show that x2 − 2 is an irreducible polynomial in the field of
rationals using Eisenstein's criterion.
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Famous theorems of mathematics/Euclid's proof of the infinitude of primesFrom Wikibooks, the open-content textbooks collection
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The Greek mathematician Euclid gave the following elegant proof that there are an infinite number of primes. It relies on the fact that all non-prime numbers --- composites --- have a unique factorization into primes.
Euclid's proof works by contradiction: we will assume that there are a finite number of primes, and show that we can derive a logically contradictory fact.
So here we go. First, we assume that that there are a finite number of primes:
p1, p2, ... , pn
Now consider the number M defined as follows:
M = 1 + p1 * p2 * ... * pn
There are two important --- and ultimately contradictory --- facts about the number M:
1. It cannot be prime because pn is the biggest prime (by our initial assumption), and M is clearly bigger than pn. Thus, there must be some prime p that divides M.
2. It is not divisible by any of the numbers p1, p2, ..., pn. Consider what would happen if you tried to divide M by any of the primes in the list p1, p2, ... , pn. From the definition of M, you can tell that you would end up with a remainder of 1. That means that p --- the prime that divides M --- must be bigger than any of p1, ..., pn.
Thus, we have shown that M is divisible by a prime p that is not on the finite list of all prime. And so there must be an infinite number of primes.
These two facts imply that M must be divisible by a prime number bigger than pn. Thus, there cannot be a biggest prime.
Note that this proof does not provide us with a direct way to generate arbitrarily large primes, although it always generates a number which is divisible by a new prime. Suppose we know only one prime: 2. So, our list of primes is simply p1=2. Then, in the notation of the proof, M=1+2=3. We note that M is prime, so we add 3 to the list. Now, M = 1 +2 *3 = 7. Again, 7 is prime. So we add it to the list. Now, M = 1+2*3*7 = 43: again prime. Continuing in this way one more time, we calculate M = 1+2*3*7*43 = 1807 =13*139. So we see that M is not prime.
Viewed another way: note that while 1+2, 1+2*3, 1+2*3*5, 1+2*3*5*7, and 1+2*3*5*7*11 are prime, 1+2*3*5*7*11*13=30031=59*509 is not.
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Famous theorems of mathematics/Pythagorean trigonometric identityFrom Wikibooks, the open-content textbooks collection
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The Pythagorean trigonometric identity is a trigonometric identity expressing the Pythagorean theorem in terms of trigonometric functions. Along with the sum-of-angles formulae it is the basic relation among the sin and cos functions from which all others may be derived.
Contents
[hide]
1 Statement of the identity 2 Proofs and their relationships to the Pythagorean theorem
o 2.1 Using right-angled triangles
o 2.2 Using the unit circle
o 2.3 Using power series
o 2.4 Using the differential equation
[edit] Statement of the identity
Mathematically, the Pythagorean identity states:
(Note that sin2 x means (sin x)2.)
Two more Pythagorean trigonometric identities can be identified. They are derived as follows.
Like (1), they also have simple geometric interpretations as instances of the Pythagorean theorem.
[edit] Proofs and their relationships to the Pythagorean theorem
[edit] Using right-angled triangles
Using the elementary "definition" of the trigonometric functions in terms of the sides of a right triangle,
the theorem follows by squaring both and adding; the left-hand side of the identity then becomes
which by the Pythagorean theorem is equal to 1. Note, however, that this definition is only valid for angles between 0 and ½π radians (not inclusive) and therefore this argument does not prove the identity for all angles. Values of 0 and ½π are trivially proven by direct evaluation of sin and cos at those angles.
To complete the proof, the identities of Trigonometric symmetry, shifts, and periodicity must be employed. By the periodicity identities we can say if the formula is true for -π < x ≤ π then it is true for all real x. Next we prove the range ½π < x ≤ π, to do this we let t = x - ½π, t will now be in the range 0 < x ≤ ½π. We can then make use of squared versions of some basic shift identites (squaring conveniently removes the minus signs).
All that remains is to prove it for −π < x < 0; this can be done by squaring the symmetry identities to get
[edit] Using the unit circle
If the trigonometric functions are defined in terms of the unit circle, the proof is immediate: given an angle θ, there is a unique point P on the unit circle centered at the origin in the Euclidean plane at an angle θ from the x-axis, and cos θ, sin θ are respectively the x- and y-coordinates of P. By definition of the unit circle, the sum of the squares of these coordinates is 1, whence the identity.
The relationship to the Pythagorean theorem is through the fact that the unit circle is actually defined by the equation
Since the x- and y-axes are perpendicular, this fact is actually equivalent to the Pythagorean theorem for triangles with hypotenuse of length 1 (which is in turn equivalent to the full Pythagorean theorem by applying a similar-triangles argument).
The trigonometric functions may also be defined using power series, namely (for x an angle measured in radians):
Using the formal multiplication law for power series we obtain
Note that in the expression for sin2, n must be at least 1, while in the expression for cos2, the constant term is equal to 1. The remaining terms of their sum are (with common factors removed)
by the binomial theorem. The Pythagorean theorem is not closely related to the Pythagorean identity when the trigonometric functions are defined in this way; instead, in combination with the theorem, the identity now shows that these power series parameterize the unit circle, which we used in the previous section. Note that this definition actually constructs the sin and cos functions in a rigorous fashion and proves that they are differentiable, so that in fact it subsumes the previous two.
[edit] Using the differential equation
It is possible to define the sin and cos functions as the two unique solutions to the differential equation
satisfying respectively y(0) = 0,y'(0) = 1 and y(0) = 1,y'(0) = 0. It follows from the theory of ordinary differential equations that the former solution, sin, has the latter, cos, as its derivative, and it follows from this that the derivative of cos is −sin. To prove the Pythagorean identity it suffices to show that the function
z = sin2x + cos2x
is constant and equal to 1. However, differentiating it and applying the two facts just mentioned we see that z' = 0 so z is constant, and z(0) = 1.
This form of the identity likewise has no direct connection with the Pythagorean theorem.
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Famous theorems of mathematics/e is irrationalFrom Wikibooks, the open-content textbooks collection
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The series representation of Euler's number e
can be used to prove that e is irrational. Of the many representations of e, this is the Taylor series for the exponential function ey evaluated at y = 1.
[edit] Summary of the proof
This is a proof by contradiction. Initially e is assumed to be a rational number of the form a/b. We then analyze a blown-up difference x of the series representing e and its strictly smaller bth partial sum, which approximates the limiting value e. By choosing the magnifying factor to be b!, the fraction a/b and the bth partial sum are turned into integers, hence x must be a positive integer. However, the fast convergence of the series representation implies that the magnified approximation error x is still strictly smaller than 1. From this contradiction we deduce that e is irrational.
[edit] Proof
Suppose that e is a rational number. Then there exist positive integers a and b such that e = a/b.
Define the number
To see that x is an integer, substitute e = a/b into this definition to obtain
The first term is an integer, and every fraction in the sum is an integer since n≤b for each term. Therefore x is an integer.
We now prove that 0 < x < 1. First, insert the above series representation of e into the definition of x to obtain
For all terms with n ≥ b + 1 we have the upper estimate
which is even strict for every n ≥ b + 2. Changing the index of summation to k = n – b and using the formula for the infinite geometric series, we obtain
Since there is no integer strictly between 0 and 1, we have reached a contradiction, and so e must be irrational.
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Famous theorems of mathematics/Fermat's little theoremFrom Wikibooks, the open-content textbooks collection
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Fermat's little theorem (not to be confused with Fermat's last theorem) states that if p is a prime number, then for any integer a, ap − a will be evenly divisible by p. This can be expressed in the notation of modular arithmetic as follows:
A variant of this theorem is stated in the following form: if p is a prime and a is an integer coprime to p, then ap
− 1 − 1 will be evenly divisible by p. In the notation of modular arithmetic:
Famous theorems of mathematics/Law of large numbersFrom Wikibooks, the open-content textbooks collection
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Given X1, X2, ... an infinite sequence of i.i.d. random variables with finite expected value E(X1) = E(X2) = ... = µ < ∞, we are interested in the convergence of the sample average
[edit] The weak law
Theorem:
Proof:
This proof uses the assumption of finite variance (for all i). The independence of the random variables implies no correlation between them, and we have that
The common mean μ of the sequence is the mean of the sample average:
Using Chebyshev's inequality on results in
This may be used to obtain the following:
As n approaches infinity, the expression approaches 1. And by definition of convergence in probability (see Convergence of random variables), we have obtained
