All Natural Numbers Are Whole Numbers, All Whole Numbers Are Integers All Whole Numb: Fill & Download for Free

When I first looked at the problem statement what I saw wassomething something[math]\dfrac{1}{x^n=-1=0} \tag*{}[/math]Yes, a mathematical abracadabra but - nonetheless.Though I have no way of knowing the level of the reader’s involvement with complex analysis, please allow me to pass a subjective judgement anyway - here is a simple path to victory via complex numbers.Simple because there exists an absolutely beautiful and elegant connection between contour integration in a complex plane and improper real integrals.If you’ve never heard about complex integrals prior consider this to be your baptism by fire as I will try to entice you the best I can muster - as long as your mind won’t fluster (and may be after working through this answer you will become a complex integration advancer).The overall computational (pseudo) algorithm numb to some gory details runs as follows:switch from a real to a complex integration variableidentify the troublesome points where things may go haywire (like division by zero, for example)fish for an advantageous collection of contours that make up a (finite by square area) closed loop that doesn’t self-intersect - such loops are called, I kid you not, simplemake sure that some contour in your collection enters into an intimate relationship with the (real) [math]x[/math]-axis - eventually we’ll walk to infinity (and beyond) along itcall in the Cauchy Integral Theorem (which has the underpinnings in the Green’s Theorem) or the Cauchy Integral Formula or bothwalk the walk - unfold the pieces that make up the complex variable [math]z[/math] over the contours that you’ve picked earliermanipulate the integrals that pop outmake one of the contour’s parameters infinitely large (tend it to infinity or infinities) while making the other (contour’s) parameter very small (tend it to zero)equate the imaginary and/or the real parts as neededeventually - and that is the astonishing bit - solve purely arithmetically for the entire improper real integral is if it were a good old single (atomic and indivisible) unknownLet’s see how this may work out in practice.Bluntly replace [math]x[/math] with [math]z[/math] and, temporarily, run away from [math]\mathbb{R}[/math] to [math]\mathbb{C}[/math]:[math]\displaystyle \oint_K \dfrac{dz}{1+z^n} \tag{1}[/math]The circle that hugs the integral sign above says that we are integrating over some contour (or path) named [math]K[/math].Here I will commit a small crime by inhaling my understanding into an otherwise unspecified situation: we will take it that [math]n[/math] is a strictly positive integer:[math]n > 0, \; n = 1, 2, 3, 4, \ldots \tag*{}[/math]We now have an [math]n[/math]-th degree polynomial in the denominator. Surely, the divide-by-zero troublesome guest is knocking on our door. But here, pal. We see you coming from a mile away:[math]z^n + 1 = 0 \tag*{}[/math][math]z^n = -1 \tag{2}[/math]If we switch to a form of complex numbers known as polar, then:[math]z = r\cdot e^{i\theta} \tag*{}[/math]and:[math]-1 = e^{i(2k + 1)\pi}, \; k \in \mathbb{Z} \tag*{}[/math]after which (2) becomes:[math]r^n \cdot e^{in\theta} = 1 \cdot e^{i(2k + 1)\pi} \tag{3}[/math]Here [math]r[/math] is a strictly not negative real number, therefore:[math]r^n = 1 \tag*{}[/math][math]r = 1 \tag*{}[/math]Loosely think of [math]r[/math] as a radius. Then, since [math]r = 1[/math], our numbers [math]z[/math] that satisfy (2) dwell on a circumference of a unit circle (a circle with the radius equal to [math]1[/math]). Errr …. Can you be more specific? Sure. Let’s equate the exponents from (3) since we are done with the radii:[math]e^{in\theta} = e^{i(2k + 1)\pi} \tag*{}[/math]Take the natural logarithm of both sides of the above equation:[math]in\theta = i(2k + 1)\pi \tag*{}[/math]Cancel out the [math]i[/math]s:[math]n \theta = (2k + 1)\pi \tag*{}[/math][math]\theta_k(n) = \dfrac{(2k + 1)\pi}{n} \tag*{}[/math]You see, a number is a number is a number - and a complex number is no exception. Technically - a number is just some abstract entity. An entity that has interpretations. One of the interpretations is geometric. And one of the geometric interpretations of a complex number is that of a point on a complex plane.Even more technically - to each unique and distinct complex number there corresponds a unique and distinct point in a complex plane and conversely (this must be proved). To see why and how the above complex exponent is connected to a circle, scroll through these Quora answers.At this point our inner mathematicians are knocking on the door anxiously:- What the hell is going on in there? Is everything up to snuff and above water? Let us in!OK, I admit - some hand-waving is going on because we (in reality) must prove that out of [math]\mathbb{Z}[/math] for [math]k[/math] we are only interested in the first [math]n[/math] numbers of the form:[math]k = 0, 1, 2, \ldots , n - 1 \tag*{}[/math]because after that we will just be going in circles. But I’m on the recruiting path, so for now we will take it on faith that our dangerous guys, the distinct roots [math]\zeta[/math][math]_k(n)[/math], are of the following shape:[math]\zeta_k(n) = \text{exp}\Big(i\dfrac{2k+1}{n}\pi\Big) \tag{4}[/math]where the [math]k[/math]s run through the set mentioned above.Since we agreed that [math]n[/math] is an integer (whole number), it follows that our [math]\zeta_k(n)[/math]s are sprinkled uniformly along the circumference of a unit circle (below I’ve shown a case for [math]n = 6[/math]):We see that for [math]n \geqslant 3[/math] the [math]\zeta[/math][math]_k(n)[/math]s form a regular [math]n[/math]-gon.Next, what should our contour of integration, [math]K[/math], be?Relax. Try the different ones and don’t be afraid to fail. This is the fun part.Mess things up - with style.Some usual suspects to try are - a right triangle, a rectangle, a circle.The precious mathematical nuggets come in small packages. The little gemstones that have a great value. One such diamond in the rough is the quantifier any: the Cauchy Integral Theorem states that it holds true for any proper contour.And if such a contour is any then we might just as well pick the most convenient for us.Let’s think about this one together by analyzing the structure of the guys from the wrong side of town.We see that they have a generally varying magnitude [math]n[/math] at the bottom. But for a current problem we may very well fix it. Then the piece [math]\pi \cdot n^{-1}[/math] becomes atomic and constant and the only variance comes from the upstairs magnitude that looks distinctly like an odd number.We recall a theorem proved early on in a Number Theory course - the Division Algorithm. From that theorem it follows that if any natural number is divided by [math]2[/math] then only two remainders may result - either [math]1[/math] or [math]0[/math]. If the remainder is [math]1[/math] then we call the original numbers odd, and if the remainder is [math]0[/math] then we call the originals [math]even[/math].In our case the bad guys cycle through the odd numbers. Between any two consecutive odd numbers there lives one even number.Now, we have to pick a contour that runs along the [math]x[/math]-axis and ideally we want to flow around the bad guys which officially are known as singularity points.If such a singularity point were on the [math]x[/math]-axis then we could just semi-circle around it and call it good. We would then use the Cauchy Integral Theorem, make the radius of the semi-circle tiny and be home free.But our bad guys are not on the [math]x[/math]-axis. So we want to make sure that we do grab a bad guy but - one and only one bad guy. Then we can use the Cauchy Integral Formula (and the Cauchy Integral Theorem).One way to ensure that we grab only one singularity point is to use our earlier odd-even observation. If we go for some pizza, or a slice of it to be exact, then we manufacture such a slice in a very peculiar manner:its side cuts, [math]OA[/math] and [math]OC[/math], will be the radiiits outer crust (filled with some melted cheese goodness) will be a mathematically perfect circular arc [math]ABC[/math]its central angle - an angle under which [math]ABC[/math] is seen from [math]O[/math] - will be [math]2 \cdot \dfrac{\pi}{n}[/math]We shall walk around our pizza slice counterclockwise:Now we “unfold the pieces that make up the complex variable [math]z[/math] over the contours”, noting that overall our contour integral can be represented as an arithmetic sum of three integrals - one per path (this must be proved):[math]\displaystyle \oint_K \dfrac{dz}{1+z^n} = I_1(OA) + I_2(ABC) + I_3(CO) \tag*{}[/math]A traditional, high school, rendering of a complex number has five pieces to it: [math]x[/math], a plus, [math]i[/math], a multiplication sign and [math]y[/math]. We operate on just two of those: the [math]x[/math] and the [math]y[/math].Over the [math]OA[/math] path: here the imaginary part vanishes: [math]z = x[/math]. Therefore, [math]dz = dx[/math] and the [math]x[/math] varies from [math]0[/math] to [math]A[/math] (a real number which we will soon send to positive infinity):[math]\displaystyle I_1 = \int_0^A \dfrac{dx}{1+x^n} \tag{5}[/math]Over the [math]ABC[/math] path: switch to the polar form: [math]z = Ae^{i\theta}[/math]. Therefore:[math]dz = iAe^{i\theta}d\theta \tag*{}[/math]and the [math]\theta[/math] varies from [math]0[/math] to [math]2\cdot \dfrac{\pi}{n}[/math]:[math]\displaystyle I_2 = \int_0^{2\cdot \frac{\pi}{n}} \dfrac{iAe^{i\theta}d\theta}{1+A^ne^{in\theta}} \tag{6}[/math]Over the [math]CO[/math] path: it is just a straight line and it is convenient to use the polar form again (loosely think - the angle is fixed and only the distance from origin varies):[math]z = r \cdot \text{exp}\Big(i\dfrac{2\pi}{n}\Big) \tag*{}[/math][math]dz = \text{exp}\Big(i\dfrac{2\pi}{n}\Big) dr \tag*{}[/math][math]0 \leqslant r \leqslant A \tag*{}[/math]and we integrate/walk from [math]A[/math] to [math]0[/math] (the order of the integration boundaries is important):[math]\displaystyle I_3 = \int_A^0 \dfrac{e^{i\frac{2\pi}{n}}dr}{1+r^ne^{i2\pi}} \tag{7}[/math]Notice that the fraction of [math]n[/math] in the bottom exponent disappeared.Consider [math]I_2[/math]. In it, the real number [math]A[/math] is just a constant. Therefore, if we send [math]A[/math] to positive infinity then [math]I_2[/math] will simply vanish because [math]A^{n-1}[/math] races toward infinity very aggressively. Formally:[math]\displaystyle \lim_{A \to + \infty} I_2 = 0 \tag*{}[/math]Consider [math]I_3[/math]. In it, change the current direction of integration to its opposite: instead of walking from [math]A[/math] to [math]0[/math] go from [math]0[/math] to [math]A[/math] and flip the integral’s sign:[math]\displaystyle I_3 = - \int_0^A \dfrac{e^{i\frac{2\pi}{n}}dr}{1+r^ne^{i2\pi}} \tag*{}[/math]Factor out the constant (and don’t forget that [math]e^{i2\pi} = 1[/math]):[math]\displaystyle I_3 = - e^{i\frac{2\pi}{n}} \int_0^A \dfrac{dr}{1+r^n} \tag*{}[/math]Rename the dummy integration variable in [math]I_3[/math] if you want (to [math]x[/math]), it doesn’t matter, but if we send [math]A[/math] to a positive infinity in it and in [math]I_1[/math] then the general shape of [math]I_3[/math] will match that of [math]I_1[/math] exactly:[math]\displaystyle I_1 = \int_0^{\infty} \dfrac{dx}{1+x^n} \tag*{}[/math][math]\displaystyle I_3 = - e^{i\frac{2\pi}{n}} \int_0^{\infty} \dfrac{dr}{1+r^n} \tag*{}[/math]Lose the subscripts of [math]I[/math]s, designate the real improper integral sought-after as just [math]I[/math] and treat it purely arithmetically:[math]\displaystyle \oint_K \dfrac{dz}{1+z^n} = I\Big(1 - e^{i\frac{2\pi}{n}}\Big) \tag{8}[/math]Isn’t it marvelous?And, technically, we haven’t computed a single integral yet.Some more arithmetic manipulations: in the difference (in parentheses) in (8) factor out just one copy of the exponent:[math]\Big(1 - e^{i\frac{2\pi}{n}}\Big) = e^{i\frac{\pi}{n}}\Big(e^{-i\frac{\pi}{n}} - e^{i\frac{\pi}{n}}\Big) \tag*{}[/math]and see the new difference for what it really is:[math]\Big( e^{-iz} - e^{iz} \Big) = - \Big(e^{iz} - e^{-iz} \Big) = -2i\sin z \tag*{}[/math]or:[math]\Big(1 - e^{i\frac{2\pi}{n}}\Big) = -2i\sin\Big(\dfrac{\pi}{n}\Big) e^{i\frac{\pi}{n}} \tag{9}[/math]Put (9) back into (8):[math]\displaystyle \oint_K \dfrac{dz}{1+z^n} = -2i\sin\Big(\dfrac{\pi}{n}\Big) e^{i\frac{\pi}{n}} I \tag{10}[/math]Now recall the bad guys that we have found earlier - the roots of negative unity. It follows that we can represent the polynomial in question as a series of products:[math]z^n+1 = (z-\zeta_0)(z-\zeta_1)\ldots(z-\zeta_{n-1}) \tag{10}[/math]and the reciprocal of (10) can be represented as:[math]\dfrac{1}{z^n+1} = \dfrac{C_0}{z-\zeta_0}+\dfrac{C_1}{z-\zeta_1}+ \dots + \dfrac{C_{n-1}}{z-\zeta_{n-1}} \tag{11}[/math]where each [math]C_j[/math] is a constant.Though normally we can recover all the [math]C_j[/math]s via a routine and boring process, this time we don’t have to. Why?An integral of a sum of a finite number of terms is equal to the sum of the integrals of the individual terms (basic stuff).However, from the Cauchy Integral Theorem it follows that all the individual integrals for [math]C_1[/math] through to [math]C_{n-1}[/math] over [math]K[/math] will be zero precisely because we, in a cunning maneuver, excluded all the corresponding singularity points [math]\zeta_1[/math] through [math]\zeta_{n-1}[/math] from [math]K[/math] - that is why we krafted that kontour so karefully. Formally:[math]\displaystyle \sum_{k=1}^{n-1}\oint_K \dfrac{C_k dz}{z-\zeta_k} = 0 \tag*{}[/math]and only the [math]C_0[/math] integral survives:[math]\displaystyle \oint_K \dfrac{dz}{1+z^n} = \oint_K \dfrac{C_0dz}{z-\zeta_0} \tag{12}[/math]But by the Cauchy Integral Formula for the integral on the right hand side of (12) modulo the [math]C_0[/math] we have:[math]\displaystyle \dfrac{1}{2\pi i}\oint_K \dfrac{dz}{z-\zeta_0} = 1 \tag*{}[/math]where [math]f(z) = 1[/math] is certainly analytic in the Cauchy-Riemann sense. Therefore, purely arithmetically:[math]\displaystyle \oint_K \dfrac{dz}{z-\zeta_0} = 2\pi i \tag*{}[/math]and:[math]\displaystyle \oint_K \dfrac{dz}{1+z^n} = 2\pi i \cdot C_0 \tag*{}[/math]Therefore, (10) becomes:[math]\displaystyle 2\pi i \cdot C_0 = -2i\sin\Big(\dfrac{\pi}{n}\Big) e^{i\frac{\pi}{n}} I \tag*{}[/math]or:[math]\displaystyle \pi \cdot C_0 = -\sin\Big(\dfrac{\pi}{n}\Big) e^{i\frac{\pi}{n}} I \tag{13}[/math]I can hear the audience:- Enough already, Andronov! Can we integrate something for god’s sake?Sorry to disappoint you, guys, but - no integration today.We recover the only missing piece, [math]C_0[/math], from our (11) expansion by multiplying its both sides by [math](z-\zeta_0)[/math]:[math]\dfrac{z-\zeta_0}{z^n+1} = C_0 + C_1\dfrac{z-\zeta_0}{z-\zeta_1}+ \dots + C_{n-1}\dfrac{z-\zeta_0}{z-\zeta_{n-1}} \tag{14}[/math]and taking a limit of (14) as [math]z[/math] tends to [math]\zeta_0[/math]:[math]\displaystyle C_0 = \lim_{z \to \zeta_0}\dfrac{z - \zeta_0}{z^n+1} \tag*{}[/math]Poor l’Hopital - no one seems to like using his rule so let us do just that:[math]\displaystyle C_0 = \lim_{z \to \zeta_0}\dfrac{1}{nz^{n-1}} = \dfrac{\zeta_0^{1-n}}{n} \tag*{}[/math][math]C_0 = \dfrac{\zeta_0^{1-n}}{n} \tag*{}[/math]But we’ve computed the [math]\zeta_0[/math] already (in (4)):[math]\zeta_0(n) = \text{exp}\Big(i\dfrac{\pi}{n}\Big) \tag*{}[/math]Therefore:[math]C_0 = \dfrac{1}{n}\text{exp}\Big(i\pi \dfrac{1-n}{n}\Big) \tag*{}[/math][math]C_0 = \dfrac{e^{\dfrac{i\pi}{n}}}{ne^{-i\pi}} = - \dfrac{1}{n} e^{\dfrac{i\pi}{n}} \tag*{}[/math]Lastly, put [math]C_0[/math] into (13):[math]- \dfrac{\pi}{n} e^{\dfrac{i\pi}{n}} = -\sin\Big(\dfrac{\pi}{n}\Big) e^{i\frac{\pi}{n}} I \tag*{}[/math]Kill the [math]-1[/math]s and the exponents and - as promised! - solve arithmetically for [math]I[/math]:[math]I = \dfrac{\pi}{n} \cdot \dfrac{1}{\sin\Big(\dfrac{\pi}{n}\Big)} \tag{15}[/math]How absolutely amazing is that - we’ve managed to compute the improper real integral without actually doing any proper (or is it improper?) real integration!Extra for experts. For the first cut you do not have to read what follows but if you want to go for a solid A+ then the way I’d do these things is I’d go back and attempt to 1) simplify; 2) generalize; 3) invert (as I explain in this Quora post).I see a generalization opportunity in the fact that our limit in [math]I_2[/math] converges to zero at an astonishing rate:[math]\displaystyle \lim_{A \to +\infty}\dfrac{1}{A^{n-1}} = 0 \tag{16}[/math]What if we slow it down a bit in a carefully orchestrated fashion by researching the following limit:[math]\displaystyle \lim_{A \to +\infty}\dfrac{f(A)}{A^{n-1}} = 0 \tag{17}[/math]where [math]f(A)[/math] is some function which is both - fruitful and allows (17) to hold?Maybe there are other solutions out there but an easy one that comes to mind is [math]f(A) = A^p[/math] where an obvious constraint on [math]p[/math] (a non-negative integer) is:[math]\text{max}(p) = n - 2 \tag*{}[/math]because in this worst case scenario:[math]\displaystyle \lim_{A \to +\infty}\dfrac{A^{n-2}}{A^{n-1}} = \lim_{A \to +\infty}\dfrac{1}{A}= 0 \tag*{}[/math]still works - not as fast as in (16) but it does, so who cares.Now we “solve our problem in reverse” and splice the [math]A^p[/math] back into [math]I_2[/math]: we will have to put [math]p[/math] into the exponent in the numerator:[math]A^pe^{ip\theta} = z^p \tag*{}[/math]If we propagate the new factor of [math]z^p[/math] into [math]I_3[/math]‘s numerator then the new exponent there will be:[math]\text{exp}\Big(i\dfrac{2p\pi}{n}\Big) \cdot \text{exp}\Big(i\dfrac{2\pi}{n}\Big) = \text{exp}\Big(i\dfrac{2(p+1)\pi}{n}\Big) \tag*{}[/math]Adjusting the numerator of [math]I_1[/math] we get:[math]\displaystyle I_1 = \int_0^A \dfrac{x^p dx}{1+x^n} \tag*{}[/math]with the complex integral being:[math]\displaystyle \oint_K \dfrac{z^p dz}{1+z^n} \tag*{}[/math]Since the new exponent in (8) will be:[math]\text{exp}\Big(i\dfrac{2(p+1)\pi}{n}\Big) \tag*{}[/math]the sine in (9) will be:[math]\sin\Big(\dfrac{(p+1)\pi}{n}\Big) \tag*{}[/math]The limit for the new [math]C_0[/math] works out to be:[math]\displaystyle C_0 = \lim_{z \to \zeta_0}\dfrac{(p+1)z^p-p\zeta_0z^{p-1}}{nz^{n-1}} = \dfrac{\zeta_0^{p+1-n}}{n} \tag*{}[/math]or:[math]C_0 = - \dfrac{1}{n} e^{\dfrac{i(p+1)\pi}{n}} \tag*{}[/math]The rest of the arithmetic remains the same and we obtain a more general result:[math]\displaystyle \int_0^{\infty}\dfrac{x^pdx}{1+x^n} = \dfrac{\pi}{n} \cdot \dfrac{1}{\sin\Big(\dfrac{(p+1)\pi}{n}\Big)} \tag{18}[/math]given that:[math]\text{max}(p) = n - 2 \tag*{}[/math]which degenerates into the original if [math]p \equiv 0[/math].

Did you want a practical answer or an absolute answer.Answer 1:The practical answer would be as many words as are ever spoken or written in the life of the Universe.Large unmeasuarble but finite.Answer 2:The first level absolute answer is the same infinity as there are countable numbers that is the smallest infinity. Having said that. They could not be written or said in the life time of the Universe.Answer 3:However, the second level of absolute answer is if the Universe really is a multi branching multiverse as some physical theorists believe then we can count all the words spoken and or written in parallel and by the way they are not neccessarily the same in all the uncountable branches of the multiverse.Now as the multiverse is, assuming it exists, uncountably infinite this means the English dictionary then also becomes uncountably infinite then the combinations become also uncountably infinite. So the number of combinations goes up to the second level of infinity, Uncountably infinite.Answer 4:The third absolute answer that I want to put forward today:Because we have the words to describe the uncountable multiverse, this implies in an abstract sense the multiverse exist and all those words exist whether what we consider the physical Universe is actually a multiverse or not.This means regardless of actul physics the list is uncountably infinite.For those of you who do not understand the difference between countable and uncountable infinite here is a brief explanation.The real line is used in mathematical graphs as a line of numbers. Usually on the X and Y sides of a graph. At any point on that line represents a number but not every point on that line can be written in digits even an infinite number of digits. In fact most points cannot.The real line can be broken up into the natural numbers by starting at 0 and adding 1 every time; 0, 1, 2, 3,.... They can become integers by adding in the set of negative numbers starting at -1 and subtracting 1 each time, giving us; ... -3, -2, -1, 0, 1, 2, 3, ... .There are an infinite number of them because you can always add 1 or substract 1. Curiously the infinity of natural numbers that is poitive round numbers is excactly the same as the one for round whole numbers including the negative numbers that is infinite. Because you can define the process of adding in the next element in both cases as adding 1 number at a time to the set of numbers and processes continue 1 step at a time for ever.This is true also for fractions.Now let us look at the numbers in between those points. Let us narrow our view down to one example the line segment 0 to 1. it has two integers at either end and and a load of numbers in between for which we do not have a value.Let us increase our accuracy from 1 to 0.1 and add in those segments.0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0We know have 11 fractions in this range and 10 line segments we still do not know the numbers in those line segments and there are still an infinite number of numbers in those line segments.Select any one of those line segments and increase the accuracy on it as many times as you like, you will always get a point at either end a line segment containing an infinite numbe of numbers.The numbers in those line segments are uncountable. It has also been proved that there are as many numbers in any line segment as any other,e.g. the line segment 0-1 can be move to 2-2000003 by multipplying any number in 0-1 by 2,000,000 and adding 2. each point in the first range maps to a single point in the second. because 0-1 has as many numbers in it as 2-2,000,003. uncountably infinite. they are the same size infinite ranges of real numbers.In practice we usually are only interested in finding numbers to a certain precission so this does not matter. Ignore infinity at your peril. Chaos theory e.g. weather projection or earthquake predictions works only if you use exact numbers. Digital information cannot give you that.