Algebra

Pirates

Proofs

7 Marsi

Festën e mësuesit

Ne do ta festojmë

Lulet më të bukura

Ne do ti dhurojmë.

Vijmë tek ti mësues

Buzëqeshjen ta dhurojmë

Kënga krahët reh si flutur

Midis jush jam e lumtur.

109110111

Math and Arithmetic

Proofs

Irrational Numbers

Proof by contradiction: suppose that root 7 (I'll write sqrt(7)) is a rational number, then we can write sqrt(7)=a/b where a and b are integers in their lowest form (ie they are fully cancelled). Then square both sides, you get 7=(a^2)/(b^2) rearranging gives (a^2)=7(b^2). Now consider the prime factors of a and b. Their squares have an even number of prime factors (eg. every prime factor of a is there twice in a squared). So a^2 and b^2 have an even number of prime factors. But 7(b^2) then has an odd number of prime factors. But a^2 can't have an odd and an even number of prime factors by unique factorisation. Contradiction X So root 7 is irrational.

289290291

Proofs

Statistics

Probability

a) 1/16

b) 1/16

c) 1/256 [this answer was given, but it is unclear what part-c is even asking: The pattern occurs before what pattern? There are many variables which are unspecified and would affect the outcome.]

267268269

Math and Arithmetic

Proofs

Lusin's theorem says that every measurable function f is a continuous function on nearly all its domain.

It is given that f measurable. This tells us that it is bounded on the complement of some open set of arbitrarily small measure. Now we redefine ƒ to be 0 on this open set. If needed we can assume that ƒ is bounded and therefore integrable.

Now continuous functions are dense in L1([a, b]) so there exists a sequence of continuous functions an tending to ƒ in the L1 norm. If we need to, we can consider a subsequence.

We also assume that an tends to ƒ almost everywhere. Now Egorov's theorem tells us that that an tends to ƒ uniformly except for some open set of arbitrarily small measure. Since uniform limits of continuous functions are continuous, the theorem is proved.

229230231

Algebra

Proofs

This proof uses modular arithmetic. If you are unfamiliar with this, the basic principle is that if we have integers a, b, and a nonzero integer c, then a = b (mod c) if a/c and b/c have the same remainder. For example, 8 = 2 (mod 3), because 8/3 and 2/3 have remainder 2.

One property of this relation is that for any integer x and for any nonzero integer y, there exists a unique integer z such that x = z (mod y) and z is between 0 and y inclusive. The upshot of this is that, in most cases, if you know how your relation behaves with all integers between 0 and y, you know how it behaves for all integers.

Consider the quadratic residues mod 8; that is, find all possible values of c if 0 <= c < 8 and x2 = c (mod 8) for some integer x. Plugging in all values from 0 to 7 for x, the only possible values of c are 0, 1, and 4.

Now consider 2n + 1 (mod 8). We know that 2n + 1 is a perfect square, so we know that 2n + 1 = 0, 1, or 4 (mod 8). Thus, 2n = -1, 0, or 3 (mod 8). Since 2n is an even number, and 8 is an even number, 2n can only be congruent to an even number mod 8. Therefore, 2n = 0 (mod 8), and therefore n = 0 (mod 4).

Finally, consider 3n + 1 (mod 8). As before, we note that 3n = -1, 0, or 3 (mod 8). We know that n = 0 (mod 4), so we know that n = 4k for some integer k. Therefore, n is even. Since 8 is also even, we know that n = 0 (mod 8). Therefore, n is divisible by 8. QED.

219220221

Math and Arithmetic

Proofs

Topology

The Baire Category Theorem is, in my opinion, one of the most incredible, influential, and important results from any field of mathematics, let alone topology. It is known as an existence theorem because it provides the necessary conditions to prove that certain things must exist, even if there aren't any examples of them that can be shown. The theorem was proved by René-Louis Baire in 1899 and is a necessary result to prove, amongst other things, the uniform boundedness principle and the open mapping theorem (two of the three most fundamental results from functional analysis), the real numbers being uncountable, and the existence of continuous, yet nowhere differentiable, functions from R to R.

The proof is quite long and involves some pretty advanced math, so to help with the reader's comprehension there is a list of symbols and their meanings at the end of this proof. Also, I've added many related links with definitions and explanations of the terms used in this proof.

The Baire Category Theorem:

If B, D is a nonempty, complete metric space, then the following two statements hold:

1) If B is formulated as the union of countably-many subsets, C1, C2, …, Cp, then at least one of the Cp is somewhere dense.

2) If A1, A2, …, Ap are countably-many, dense, open subsets of B, then ∩pAp is dense in B, i.e. Cl(∩pAp) = B

Proof:

1) If the first statement is false, then there is a countable family {Cp}, p Є P, of subsets of B such that B = ∪pCp, but (Cl Cp)o = Ø for each p Є P. Therefore, for each p, Cl Cp≠ B. Select b1 Є B - Cl C1. There is a positive number m1 < 1, since B - Cl C1 is open, such that N(b1, m1) ⊂ B - Cl C1. Now we set G1 = N(b1, m1/2). Then Cl G1 ⊂ N(b1, m1); hence Cl G1 ∩ Cl C1 = Ø.

Since G1 is a nonempty, open subset of B, that means G1 ⊄ Cl C2. So, choose a b2 Є G1- Cl C2. Since G1 - Cl C2 is open, there is an m2 > 0 such that N(b2, m2) ⊂ G1 - Cl C2. This time we'll require m2 < 1/2 and then set G2 = N(b2, m2/2). Then G2 ⊂ G1 and Cl G2 ∩ Cl C2 = Ø.

If we continue on like this, requiring m3 < 1/3, m4 < 1/4, etc., we'll obtain a decreasing sequence of mp-neighborhoods, G1 ⊃ G2 ⊃ G3 ⊃ … ⊃ Gp ⊃ … such that Cl Gp ∩ Cl Cp = Ø and mp < 1/p. Then Cl G1 ⊃ Cl G2 ⊃ Cl G3 ⊃ … ⊃ Cl Gp ⊃ … and d(Gp) --> 0.

I'm going to use a result from another theorem in topology, not proven here, which says that if G1 ⊃ G2 ⊃ G3 ⊃ … ⊃ Gp ⊃ … , d(Gp) --> 0, and ∩pGp ≠ Ø, then the metric space B, D is complete. Therefore, ∩p Cl Gp ≠ Ø. So, if we pick a g Є ∩pGp, then g Є Cp for some p, since ∪pCp = B. However, that would imply that g Є Cl Cp ∩ Cl Gp which is impossible because Cl Cp and Cl Gp are disjoint. So, for 1), Q.E.D.

2) To start, we're going to suppose that {Ap}, p Є P, is a countable family of dense, open subsets of B. To prove that ∩pAp is dense, all that we need to prove is that every neighborhood of any element of B meets ∩pAp. In other words, for any selected g Є Band any m > 0, we'll show that N(g, m) ∩ (∩pAp) ≠ Ø.

If we set T = Cl N(g, m/2), then T ⊂ N(g, m). Now we'll show that T ∩ (∩pAp) ≠ Ø. We know that T is a subspace of the closed metric space, B, D, and that Titself is closed. So, using an earlier theorem that won't be proved here, T is a complete metric space. If we set Gp = T - Ap which is equal to Tp ∩ (B - Ap), we see that the intersection of two closed subsets of B, Gp is closed in both B and T.

Now suppose Gp is somewhere dense. Then there is an element t Є T and a number q > 0 such that N(t, q) ∩ T ⊂ Cl Gp ∩ T = Gp. Therefore, N(t, q) ∩ (T - Gp) = Ø. We can see that t Є T = Cl N(g, m/2). Therefore N(t, q) meets N(g, m/2) at some point z. We then choose q' > 0 such that N(z, q') ⊂ N(t, q) ∩ N(g, m/2). However, since Ap is dense, N(z, q') intersects Ap at a point we'll call z' . Well, then it must be that z' Є N(t, q) ∩ T ⊂ Gp. But, Gp= T - Ap, hence z' Є T - Ap. That implies then that z' ∉ Ap which is a contradiction. Therefore Gp must be nowhere dense in T.

So, by the first statement of the theorem, 1), which we already proved, T ≠ ∪pGp. Thus, there is n element s Є T - ∪pGp. Therefore, since Gp = T - Ap, then s Є T ∩ (∩pAp) and so T∩ (∩pAp) ≠ Ø meaning N(g, m) ∩ (∩pAp) ≠ Ø.

Q.E.D.

List of symbols:

R - The set of real numbers, including rational, irrational, positive, and negative numbers, as well as 0. Not including complex numbers having an imaginary part other than 0i, where i is the imaginary number √(-1).

B, D - The metric space of set B with metric D.

∩pAp - The intersection of all of the subsets A1, A2, …, Ap.

Cl - The closure of whatever set is written after it.

p Є P - p is an element of the set P.

P - The set of all positive integers, not including 0. This set is often referred to as the set of natural numbers and is labeled N, but since at times the natural numbers are said to include 0, I've labeled this set P to avoid ambiguity. Not to mention, I've used N within my label for neighborhood.

∪pCp - The union of all of the of the subsets C1, C2, …, Cp.

( )o - The interior of whatever set is in the parentheses.

Ø - The empty set; i.e. the set with nothing in it.

N(b1, m1) - The neighborhood of point b1 within distance m1.

⊂ - … is a subset of …

⊃ - … is a superset of …

d( ) - The diameter of whatever set is in the parentheses.

--> 0 - The limit of whatever comes before the symbol "-->" goes to 0.

185186187

Calculus

Proofs

I think the following piecewise function satisfies the two criteria: when x is rational: f(x)=x

when x is irrational: f(x)=x*, where x* is the largest rational number smaller than x.

I think not. When x is irrational, there is no largest rational number less than x. No matter what rational number you pick, there is a larger one that is less than x. For example, between 3.1415926 and pi, there is 3.14159265.

The usual answer is the one given by Weierstrass, which is the sum of an infinite series of functions. The first term in the series is a periodic sawtooth (piecewise linear) function, which is = x from x=0 to x=1, and then descends back to 0 between x=1 and x=2 (i.e., it is = -x+2 in that interval). It repeats that pattern between x=2 and x=4, and so on. The second term is just like it, but with 1/10 the frequency and 1/10 the amplitude, and so on. The first function is continuous everywhere and differentiable except at x= an integer. The sum of the first 2 is continuous everywhere and differentiable except for the multiples of 1/10, and so on. It turns out that the series converges to a function that is continuous everywhere and differentiable nowhere.

By the way, if you can take the derivative of a function at a given point, it is said to be differentiable, not derivable at that point.

173174175

Math and Arithmetic

Algebra

Proofs

Linear Algebra

First let's be clear on the definitions.

A matrix M is orthogonal if MT=M-1

Or multiply both sides by M and you have

1) M MT=I

or

2) MTM=I

Where I is the identity matrix.

So our definition tells us a matrix is orthogonal if its transpose equals its inverse or if the product ( left or right) of the the matrix and its transpose is the identity.

Now we want to show why the inverse of an orthogonal matrix is also orthogonal.

Let A be orthogonal. We are assuming it is square since it has an inverse.

Now we want to show that A-1 is orthogonal.

We need to show that the inverse is equal to the transpose.

Since A is orthogonal, A=AT

Let's multiply both sides by A-1

A-1 A= A-1 AT

Or A-1 AT =I

Compare this to the definition above in 1) (M MT=I)

do you see how A-1 now fits the definition of orthogonal?

Or course we could have multiplied on the left and then we would have arrived at 2) above.

157158159

Math and Arithmetic

Geometry

Proofs

AAS (apex)

146147148

Physics

Proofs

Flag of the United States

In Einstein's theory of gravitation, which is also referred to as General Relativity, space-time is warped (i.e., curved) by the presence of mass. It is not meaningful to speak of folding time alone, but rather one speaks of bending space-time. To make a significant bend in space-time, a very massive object (such as a star) or a very dense object (such as a black hole) is required. In 1915, Einstein succeeded in writing down an equation that describes how much space-time is curved in the presence of mass. That equation is a critical part of his General Theory of Relativity. It is not a simple equation to understand, since it requires first some understanding of tensors and Riemannian geometry. You will have to study math and physics for several years in order to understand it.

147148149

Algebra

Proofs

Abstract Algebra

Cayley's theorem:Let (G,$) be a group. For each g Ð„ G, let Jg be a permutation of G such that

Jg(x) = g$x

J, then, is a function from g to Jg, J: g --> Jg and is an isomorphism from (G,$) onto a permutation group on G.

Proof:We already know, from another established theorem that I'm not going to prove here, that an element invertible for an associative composition is cancellable for that composition, therefore Jg is a permutation of G. Given another permutation, Jh = Jg, then h = h$x = Jh(x) = Jg(x) = g$x = g, meaning J is injective.Now for the fun part!

For every x Ð„ G, a composition of two permutations is as follows:

(Jg â—‹ Jh)(x) = Jg(Jh(x)) = Jg(h$x) = g$(h$x) = (g$h)$x = Jg$h(x)

Therefore Jg â—‹ Jh = Jg$h(x) for all g, h Ð„ G

That means that the set Ð‚ = {Jg: g Ð„ G} is a stable subset of the permutation subset of G, written as Ð–G, and J is an isomorphism from G onto Ð‚. Consequently, Ð‚ is a group and therefore is a permutation group on G.

Q.E.D.

139140141

Motorcycles

Computer Terminology

Math and Arithmetic

Proofs

cubic centimeters

129130131

Math and Arithmetic

Geometry

Proofs

A three dimensional oval is simply called an egg, or more mathematically, an ovoid. A three dimensional ellipse (a more symmetric oval) is called a prolate spheroid, or oblate spheroid, depending on how the ellipse is rotated.

515253

Proofs

QED from the Latin "quod erat demonstrandum", meaning "that which was to be demonstrated", normally put at the end of a mathematical proof

122123124

Proofs

Math and Arithmetic

Prime Numbers

2+3+5+7+11+13+17+19+23+29+31+37+41+43+47 = 328

107108109

Algebra

Proofs

Abstract Algebra

The Fundamental Theorem of Algebra:

If P(z) = Σnk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0, then P(z0) = 0 for some z0 Є C. Descriptively, this says that any nonconstant polynomial over the complex number space, C, can be written as a product of linear factors.

Proof:

First off, we need to apply the Heine-Borel theorem to C. The Heine-Borel theorem states that if S is a closed and bounded set in an m-dimensional Euclidean space (written as Rm), then S is compact.

From above, P(z) = Σnk=0 akzk where ak Є C, n ≥ 1, and an ≠ 0. Let m = inf{|P(z)| : z Є C} where inf is the infinum, or the greatest lower bound of the set.

From the triangle inequality, |P(reit)| ≥ rn(|an| - r-1|an-1| - … - r-n|a0|),

so limr --> ∞ |P(reit)| = ∞. Therefore there is a real number R that |P(reit)| > m + 1 whenever r > R.

If S = {reit : r ≤ R}, then S is compact in C, by the Heine-Borel Theorem; and let m = inf{|P(z)| : z Є S}. |P| is a continuous and real-valued function in S, so, using the result from another proof not done here, it has a minimum value on S; i.e., there is a value for z0 Є S that makes |P(z0)| = m. So, if m = 0 then the theorem is proved.

We're going to show that m = 0 by proving that m can't equal anything else, and since we know m exists, it has no choice but to be zero. So, suppose m ≠ 0 and let Q(z) = P(z + z0)/P(z0), z Є C.

Q is therefore a polynomial with degree n and |Q(z)| ≥ 1 for all z Є C.

Q(0) = 1 so Q(z) can be expressed via P's series as:

Q(z) = 1 + bkzk + … + bnzn where k is the smallest positive integer ≤ n such that bk ≠ 0.

Since |-|bk|/bk| = 1, there exists a t0 Є [0, 2π/k] such that eikt0 = -|bk|/bk.

Then Q(reit0) = 1 + bkrkeikt0 + bk+1rk+1ei(k+1)t0 + … + bnrneint0

= 1 - rk|bk| + bk+1rk+1ei(k+1)t0 + … + bnrneint0.

So, if rk|bk| < 1 then |Q(reit0)| ≤ 1 - rk(|bk| - r|bk+1| - … - rn-k|bn|).

That means that if we pick a small enough r, we can make |Q(reit0)| ≤ 1 which contradicts the statement above that |Q(z)| ≥ 1 for all z Є C. Therefore m ≠ 0 doesn't hold and P(z0) = 0

Q.E.D.

Another proofSuppose P has no zeroes. Then we can define the function f(z) = 1 / P(z), and f is analytic. By the proof above, P(z) tends to infinity as z tends to infinity; hence f(z) tends to 0 as z tends to infinity. So there is a disc S such that f, restricted to the outside of S, is bounded. Also by the proof above, f is bounded inside the disc as well; therefore f is bounded. Now we apply a theorem called Liouville's Theorem, which says that any analytic function which is defined on all of C and is bounded must be a constant. So f is a constant; therefore P is constant. But we were assuming that P is not constant, so this is a contradiction.(To prove Liouville's Theorem: Suppose M is a bound for the function f, i.e. |f(z)| < M for all z. Suppose a and b are complex numbers, and we want to show f(a) = f(b). Use the theorem that f(a) = integral of f(z)/(z-a) / (2 * pi * i) around the circle of radius R and centre 0. Then, if R is sufficiently large:

|f(b) - f(a)|

= | integral, around circle, of (f(z) * (1/(z-b) - 1/(z-a))) | / (2*pi)

= | integral around circle of (f(z) * (b-a) / ((z-a)(z-b)) ) | / (2*pi)

<= (M * |b-a| / ((R-|a|)(R-|b|)) ) * (2*pi*R) / (2*pi)

The last line uses the formula |integral| <= |pathlength| * |maximum value|. Then we get |f(b) - f(a)| <= M * |b-a| * R / ((R-|a|)(R-|b|)). Letting R tend to infinity, we can prove that |f(b)-f(a)| is as small as we like; therefore f(a) = f(b).

)

105106107

Math and Arithmetic

Algebra

Proofs

7

3x + 4 = 19

3x = 15

x = 5

101112

Proofs

Prime Numbers

The HCF of two co-prime numbers is 1.

293031

Math and Arithmetic

Geometry

Proofs

Answer: A regular quadrilateral is one with equal sides and equal angles, so it is a square. To negate this definition, we say an irregular quadrilateral is one where the sides are unequal or the angles are unequal OR BOTH. In simpler terms, we could say it is a quadrilateral which is not a square.

575859

Math and Arithmetic

Proofs

Abstract Algebra

Schur's theorem:Let (J,+) be an abelian group with more than one element, and let K be a primitive ring with endomorphisms, E. Then the centralizer, C, of K, in the ring Î¾(E), which is defined as the set of all endomorphisms of (J,+), is a subdivision of Î¾(E). Proof:First off, it needs to be stated that C is a non-zero set because it contains the identity function, I, which obviously fits the definition of a centralizer:

CK(J) = {x Ð„ K: jx = xj for all j Ð„ J}.

Also, using the established theorem that "if J is a subset of ring K, then C(J) is a subring of K, and if an invertible element a of K belongs to C(J), then k-1 Ð„ C(J)," we need only show that if g Ð„ C* (C* being the set of non-zero elements of C), then g is an automorphism of E. Assuming non-triviality, g â‰? 0, and there exists

b Ð„ E such that g(b) â‰? 0. For each h Ð„ E there exists m Ð„ K such that m(g(b)) = y since K is primitive. Thus: g(m(b)) = m(g(b)) = y, showing that g is surjective.

Finally to show g is also injective and thus an automorphism, we take a non-zero element w belonging to the kernel of g. For each z Ð„ E some endomorphism would exist u Ð„ K such that u(w) = z as K is primitive. Therefore:

g(z) = g(u(w)) = u(g(w)) = u(0) = 0, the zero endomorphism which is a contradiction. Hence g is injective and an automorphism of E.

Q.E.D.

858687

Math and Arithmetic

Proofs

Numbers

Irrational Numbers

Yes, here's the proof.

Let's start out with the basic inequality 9 < 14 < 16.

Now, we'll take the square root of this inequality:

3 < âˆš14 < 4.

If you subtract all numbers by 3, you get:

0 < âˆš14 - 3 < 1.

If âˆš14 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent âˆš14. Therefore, âˆš14n must be an integer, and n must be the smallest multiple of âˆš14 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply âˆš14n by (âˆš14 - 3). This gives 14n - 3âˆš14n. Well, 14n is an integer, and, as we explained above, âˆš14n is also an integer, so 3âˆš14n is an integer too; therefore, 14n - 3âˆš14n is an integer as well. We're going to rearrange this expression to (âˆš14n - 3n)âˆš14 and then set the term (âˆš14n - 3n) equal to p, for simplicity. This gives us the expression âˆš14p, which is equal to 14n - 3âˆš14n, and is an integer.

Remember, from above, that 0 < âˆš14 - 3 < 1.

If we multiply this inequality by n, we get 0 < âˆš14n - 3n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus âˆš14p < âˆš14n. We've already determined that both âˆš14p and âˆš14n are integers, but recall that we said n was the smallest multiple of âˆš14 to yield an integer value. Thus, âˆš14p < âˆš14n is a contradiction; therefore âˆš14 can't be rational and so must be irrational.

Q.E.D.

858687

Math and Arithmetic

Algebra

Geometry

Proofs

When proving an identity, you may manipulate only one side of the equation throughout. You may not use normal algebraic techniques to manipulte both sides. Let's begin with the identity you wish to prove. cos2x - sin2x ?=? 2cos2x - 1 We know that sin2x + cos2x = 1 (Pythagorean Identity). Therefore, sin2x = 1 - cos2x. Substituting for sin2x, we may write cos2x - (1 - cos2x) ?=? 2cos2x - 1 cos2x - 1 + cos2x ?=? 2cos2x - 12cos2x -1 = 2cos2x - 1 The identity is proved. (Note that once the identity is proved, you may remove the question marks from around the equal sign.)

848586

Math and Arithmetic

Algebra

Proofs

1, 3, 5, 9, 15, 45

313233

Numerical Analysis and Simulation

Math and Arithmetic

Algebra

Proofs

to prove x|-|y≤|x-y| have to look at 4 cases:

a) x>0, y>0

b) x<0, y<0

c) x>0, y<0

d) x<0, y>0

(to save typing it out over and over again, i have shortened absolutes to abs)

For: a) the values dont matter both sides will always have the same value.

b) let x=-a, y=-b. Because of the abs around x and y, the left will be a-b and the right b-a so the left and right will have the same number but opposite signs, so after taking the abs on both sides, they will always be equal to one another.

c) let x=a, y=-b. The left will be a-b and the right a+b. if b<a then both sides are positive and the statement holds. if a<b then left is negative and right positive, but because you're starting from above 0 when you take away b (on the left side) once you take the abs left will be less than the right. when a=b the left is just 0 and right is positive so statement still holds.

d) let x=-a, y=b. left is a-b again, but the right is-a-b so always negative. if a<b then a-b is negative but not as low as -a-b so left is less than right after taking abs. if b<a then a-b is positive, but moving closer to 0 and -a-b moves further away from 0 so left is less than right again after taking abs. If a=b then then the left is always 0 and after taking abs the right is 2a therefore the statement holds for all 4 cases. Q.E.D.

777879

Math and Arithmetic

Mathematical Finance

Roman Numerals

Proofs

decimeter

212223

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