Which of the rational numbers 4/9, -5/6, -7/-12 and 11/-24 is the smallest?
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Integers are numbers that can be positive, negative or zero in mathematics. Integers can be fractions, but they can’t be fractions. These numbers are used in addition, subtraction, multiplication, and division, among other numerical operations. Integers have numbers such as 1, 24, 5,8, -67, -12, and so on. Integers are represented by the capital alphabet “Z.” A set of whole numbers, a set of natural numbers which are also known as counting numbers, and their inverses are all examples of sets of integers. Integers are real numbers’ subordinates. Integers have numbers like -100, -12,-1, 0, 2, 1000, 594379834, and so on. Show
Integers comprise of the following types of numbers which are stated below:
Vital Properties of IntegersThe primary properties of Integers are given below:
Different Types of Integers:The Number ZeroNeither a positive nor a negative integer, zero exists. It is a number that is neither positive nor negative. This is due to the fact that zero has no sign in Positive Integers. Positive IntegersPositive integers are known as positive integers. A plus sign is used to denote positive integers. In a number line, all positive integers are on the right side of zero. As a result, any positive integer is greater than zero. Negative integersNegative Integers are a type of negative integer in the number system. Negative integers are shown and can be recognized with the help of a minus sign before every number. They are always smaller in value than zero and are placed on the left side of the number line. Also Read Faculty Specialising in Xi'an Jiaotong University China What do you mean by CompositeNumbers?Composite numbers are those numbers that are divisible by more than two variables. Composite numbers are all-natural numbers that are not prime numbers and are separated by more than two numbers. 6 is an example of a composite number since it can be divided by 1, 2, 3, and even 6. Different Types of Composite NumbersThe following are the two main types of composite numbers: Odd Composite Numbers (also known as Composite Odd Numbers) and a type of even composite number. Odd Composite NumbersThe odd composite numbers are all odd integers that aren’t prime numbers. Composite odd numbers include 15, 21, 25, 27, 31, and so on. Take the numbers 1, 2, 3, 4, 9, 10, 11, 12, and 15, 16 for example. Just 9 and 15 are odd composites in this set of numbers. This is because they serve and match all the set properties of odd composite numbers. Even Composite Numbers (also known as Composite Even Numbers)Even composite numbers are all even integers that are not prime numbers. 4, 6, 10, 12, 14, 18, and so on are examples of composite even numbers in the entire number system. Just 4, 10, and 12 are composites in the given numbers set 1, 2, 3, 4, 9, 10, 11,12, and 15 since only those two numbers obey the composite number conditions. Fun facts
Also Read 4 Trends in Education That Will Blow Your Mind ConclusionIntegers and composite numbers are a very vast concept. They hold a very important place in the entire number system. Integers and composite numbers can be mastered easily with the help of Cuemath online resources that are free, accessible, and very effective. Cuemath is an online learning platform that makes learning easy and fun for students. Cuemath helps students understand why behind what of every topic. PWrong wrote:Presumably there is also a "pentaroot" and "hexaroot" and so on. If we redefined elementary functions to include all of these roots, down to infinity, I wonder if we could possibly express pi or e using these functions. Whoa! I never thought of that! An exact and finite formula for pi and e, instant Fields Medal! To bobxp: Yes, you're right, there is no logical need for the names addition, multiplication, etc. However, the first three are already deeply embedded in our language (can you imagine telling a 3-year old: "hyper 2, 1, 2 equals 4" instead of "2 plus 2 equals 4"?). As for tetration, well, look at the name of this forum  I guess its similar with other things in math too. For example, we don't need to say "squaring" and "cubing". We could just say "raising to the second power" and "raising to the third power". As for the iinc (by the way, why did you choose that name) function, I think its a great idea. Now, I'll try to post what I can think of at the moment, and ask some of my own questions. It turns out that 1+2+3+...+n can be expressed simply using multiplication: 1+2+3+...+n = n*(n+1)/2 However, I don't know of a way to express 1*2*3*...*n simply in terms of exponentiation. Still, I do know of this: http://mathworld.wolfram.com/StirlingsA ... ation.html Admittedly, its nowhere near as crisp and elegant as n*(n+1)/2, and its not even an exact answer. Its possible that an exact answer is not possible in terms of exponentiation, but perhaps tetration can do the job! Let's take a look at the third series: iinc(1,3) = 1 iinc(2,3) = 2 iinc(3,3) = 8 iinc(4,3) ~= 2.41785*10^24 I'm quite sure that iinc(n,3) also grows faster than (n tetra c). The tower of exponents gets taller and taller with iinc, while it stays at the same height for (n tetra c). Numerical examples seem to show that the height of the power tower is much more important than the numbers in the tower. I'm also quite sure iinc(n,3) grows faster than (c tetra n) because they both have n levels of exponents, but the exponents grow in the iinc function while the exponents remain constant in the (c tetra n) function. This is analogous to the factorial case. The factorial function grows faster than any polynomial (analogue to n tetra c) or exponential (analogue to c tetra n) function. However, iinc(n,3) grows slower than (n tetra n). This is obvious when I put them into exponential form, bearing in mind that the heights of the power towers are equal (I assume that you know what order to perform the exponentiations, so I leave out the parentheses): iinc(n,3) = 1^2^3^4^5^...^n n tetra n = n^n^n^n^n...^n We can express all this using what I think are elegant inequalities. Here, I use the "<" symbol to mean "grow more slowly than", rather than its usual meaning: (n^c) < (c^n) < iinc(n,2) < (n tetra 2) (n tetra c) < (c tetra n) < iinc(n,3) < (n penta 2) I could probably do a similar analysis for the 1#2#3#4#..., but I'm feeling tired right now. I think the results should be similar. Maybe you could try that. There shouldn't be any problem with setting a lower bound other than 1 for any of these functions, as far as I can tell.jinyduTetronian Posts: 721Joined: Thu Jun 10, 2004 5:31 am Top Sorry for the double post, that last one was really getting too long. I have some other ideas about tetration too. Top
I meant a standard (and hopefully accurate) "tetration notation" that can represent and operate on numbers far larger than computers can currently handle. Suppose your computer has 1 TB (one terabyte) of storage space, which is far beyond what most PCs have at the moment. 1 byte = 8 bits, therefore, you have about 8 * 10^12 bits of memory. Now, suppose you could use all of that memory to represent a single large number, using the current system. Each bit could represent one binary digit of the number. Thus, the largest number you could represent would be about 2^(8*10^12), or if you prefer base ten, 10^(2.41*10^12). While that number may seem large, it is in fact less than (11 tetra 3), hardly a giant in the universe of numbers that can be conveniently represented using tetration. In practice, no current PC can store numbers anywhere close to that size due to things like program limitations (and the fact that most people want their PC to have an operating system and other software). So I was thinking about a new system of number representation using tetration. It should be able to: 1) Handle numbers as large as a giggol, (10 tetra 100) 2) Allow simple comparison of two numbers, so that a user can easily see which one is larger 3) Perform computations on numbers (basic operations, as well as algorithms like the Newton Method)jinyduTetronian Posts: 721Joined: Thu Jun 10, 2004 5:31 am Top
I did understand that, but I can't solve the equation. The best I could do algebraically is this: x^x + x - 3 = 0 x^x = 3-x Taking tetraroots of both sides: x=sqrtrt(3-x) But I have no idea where to go from there.
I tried for a few weeks and couldn't find anything decent either. It would be really nice to have something like (a+b)^2 = a^2 + 2ab + b^2, for tetration. One rule I do have is this. (a+b)^^2 = (a+b)[sup](a+b)[/sup] = (a+b)[sup]a[/sup](a+b)[sup]b[/sup] But you can't do much else other than use the binomial theorum. I really don't want to do that, as you'd end up with a huge pile of ugly factorials, and you probably wouldn't find an answer anyway.
You'd have to go the other way round, otherwise it would always equal 1. Try something like iinc(4) = 4^(3^(2^(1)))=26144 Jinydu, I can't work out how you got your answers.
I don't know if you noticed equation 9 on the page you linked to, about Stirling's approximation: n! = n^n * e^-n * sqrt(2 pi n) Looks like tetration has already done the job! And have a look at this limit for e. It's equation 5 on http://mathworld.wolfram.com/e.html Not only does it use tetration 4 times, but the limit is apparently connected to prime numbers somehow! And speaking of primes, I've considered a version of the prime numbers that involves exponentiation instead of multiplication. A prime number can't be expressed as a product of more than two factors, i.e. it can't be expressed as a*b for integers other than 1 and itself. What about numbers that can't be expressed as a^b, with similar restrictions? Obviously, there are many more "exponential primes" than "multiplicative primes". However I did some graphs a while back, and they appear to be scattered at random. I also tried, "tetrational primes", but nearly every number is a tetrational prime so it's difficult to look for patterns.
I know how to differentiate x^^n in terms of the derivative of x^^(n-1). You can do this over and over until you get a proper expression. By the definition of tetration: x^^n = x[sup](x^^(n-1))[/sup] Now we turn this into an exponential function by taking e to the log: = e ^ln(x[sup](x^^(n-1))[/sup]) = e [sup]x^^(n-1) * ln(x)[/sup] Now we differentiate the top part using the product rule, and multiply by the original function. = [ d/dx(x^^(n-1))*lnx + x^^(n-1)*d/dx(lnx) ]e[sup]x^^(n-1) * ln(x)[/sup] = [ d/dx(x^^(n-1))*lnx + x^^(n-1)/x ]x^^n Hope that helps. You should be able to do that for x^^1000, provided you have a supercomputer handy. On the subject of large numbers, apparently someone recently broke a record for the largest finite number ever used in physics (or rather, the smallest number). Using quantum physics, they calculated the approximate probability that quantum fluctuations will result in the big bang happening, again. The probability is 1:10^10^100, but they didn't say which order to read it in! Apparently it's so small, that it doesn't really matter what units you measure in. So the chance of it happening within a hundred years, somewhere in this galaxy, is almost the same as the chance that there's another universe brewing in your fridge, right now. Creepy stuff. If there's too many replies to this, it should probably go in a different topic. But it might put into perspective how useless tetration is when only applied to the integers. This post is getting really long too, so I'll do a double post to explain my continuous/differentiable tetration. PWrongPentonian Posts: 1599Joined: Fri Jan 30, 2004 8:21 amLocation: Perth, Australia
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I've been exposed! Ok, I cheated a bit: iinc(2) = 2 iinc(3) = 2^3 iinc(4) = 2^(3^4) As you can see, I ignored the initial 1, to avoid always getting the answer 1. Still, its interesting to note that 1 + 2 + 3 + ... + n = n + (n-1) + (n-2) + ... + 1 1 * 2 * 3 * ... * n = n * (n-1) * (n-2) * ... * 1 but 1^(2^(3^(...^n))))) not = n^((n-1)^((n-2)^...^(1))))) Thus, it seems your way of defining iinc(n,3) is consistent, while mine isn't.
Not exactly, its only an approximation. For example, when n = 2, the formula gives approximately 1.919, instead of the correct answer, 2. However, the approximation gets better as n increases. For n = 20, the formula is off by about 0.41577% (you can try it yourself).
You have got a point there. But do you think its possible to express e exactly in closed form (no limits as n approaches infinity) using tetration? That would really be interesting.
Maybe I'm asking for too much, but is it possible to get a simple, closed formula for the derivative. I mean, the derivative of x^n is simply n*x^(n-1). The reason is to allow efficient use of Newton's Method to solve equations with tetration. For instance x^1000 = c is easy to solve numerically but x tetra 1000 = c would be a real pain Also, I've found another difficulty with extending tetration, but maybe your idea will do the trick. This problem follows quickly from the definition fo tetration. Note: log (p,q) = log (base p of q) x tetra n = x^(x tetra (n-1)) taking the log (base x) of both sides: (x tetra (n-1)) = log (x, (x tetra n)) We already know that: x tetra 1 = x Thus: x tetra 0 = log (x,x) = 1 x tetra -1 = log (x,1) = 0 x tetra -2 = log (x,0) which is undefinedjinyduTetronian Posts: 721Joined: Thu Jun 10, 2004 5:31 am Top
That's already been taken care of: See http://mathworld.wolfram.com/GammaFunction.html. 5.5! = Gamma(6.5) = Integral (((ln(1/t))^(5.5))dt,0,1) ~= 288 If you scroll down to the middle of the page, Equation 52 gives an exact formula Gamma(n/2), where n is odd. It turns out that there is an exact answer for 5.5!: 5.5! = 10395pi/64 Take a look at http://mathworld.wolfram.com/Superfactorial.html for two definitions of "Superfactorial". The second definition (by Sloane and Plouffe) evidently grows more slowly, so I'll analyze it first: 1$ = 1! = 1 2$ = 1! * 2! = (1)*(2*1) = (2) * (1)^2 = 2 3$ = 1! * 2! * 3! = (1)*(2*1)*(3*2*1) = 3 *(2)^2 * (1)^3 = 12 n$ = 1! * 2! * ... * n! = (1)*(2*1)*...*(n*(n-1)*(n-2)...) = n * (n-1)^2 * (n-2)^3 * ... (2)^(n-1) * (1)^n n$ < (n^n)*n = (n tetra 2)*n Now, I'll look at the first definition, by Pickover (I'm guessing on what order I should exponentiate. I chose the one that gives the highest value). 1$ = 1! = 1 2$ = 2!^2! = 2^2 = 4 3$ = 3!^3!^3!^3!^3!^3! = 6^6^6^6^6^6 n$ = (n! tetra n!) = (n! penta 2)jinyduTetronian Posts: 721Joined: Thu Jun 10, 2004 5:31 am Top I don't think we'd fit in on that forum. They all sound like real pure mathematicians, so they might not appreciate amatuers like us.
I found this page a while back, and it sounds like noone else has improved on it yet. http://users.forthnet.gr/ath/jgal/math/exponents4.html This website is a bit complicated, so I'll give you the gist of it. Galidakis found a continuous extension for n^^x. Instead of using the hyperroot function, he simply adds an extra n, and puts the fractional part of x at the top of the "power tower". n^^2.0 = n^(n^(n^0)) since n^0 = 1: n^^2.0 = n^n n^^2.5 = n^n^(n^0.5) = n^n^(n^0.5) The function is continuous because as x approaches an integer, it approaches the same value from either direction. 2^^2.9999 = 2^2^(2^.9999) = 15.99573 2^^3.0001 = 2^2^2^(2^.0001) = 16.00295 His function is continuous for all complex n, and real x. Because the base is complex, I tend to use z instead of n, and t instead of x. Anyway, when I found this I realised that his function wasn't differentiable. I've been working on a differentiable version for a while, and I've just finished. All I did was put an arbitrary function of t at the top of the tower, instead of t. i.e. z^^t = z^z^z^...^(z^f(t)) Then I found some conditions for the "top-function" as I call it, so that z^^t is differentiable for all z. I won't got through the whole process I went through, but I'll tell you the main points. The derivative respect to t is: [z^^t * z^^(t-1) * z^^(t-2) * ... * z^z^f(t) * z^f(t)] * f'(t) * (ln(z))^int(t) (how about that for a superfactorial?) The conditions I found basically ensure that the limits are the same on both sides. 1. Tetration is continuous iff f(0) = 0 f(1) = 1 2. The derivative is continuous iff f'(1) = f'(0)*ln(z) note that this condition is in terms of z. This is so that I also found a condition for the second order derivative. 3. The second derivative is continuous iff f''(1) = f''(0)*ln(z) +(f'(0)*ln(z))^2 I've yet to find conditions for the 3rd order derivatives, but it's Then, I found a suitable polynomial top-function to satisfy the conditions. Interestingly, the simplest top-function for continuity is linear, the top-function for the 1st derivative is a quadratic, and the top-function for the 2nd derivative is a cubic. This would all be great, but it turns out that the functions themselves are huge and inelegant. They're definitely correct; I've checked them by graphing. It's not neccessary to post the whole thing, but the coefficients, especially for the 2nd derivative, are ugly combinations of ln(z), full of square roots and a lot of unexpected numbers. It's also very difficult to simplify, even though there are a few repeating bits. It just strikes me as odd that a polynomial with a special property looks so arbitrary. PWrongPentonian Posts: 1599Joined: Fri Jan 30, 2004 8:21 amLocation: Perth, Australia
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That would lead to the strange property: x tetra -n = x^(1/n) Then, x tetra 1 = x tetra -(-1) = x^(1/-1) = x^-1 = 1/x x tetra 2 = x tetra -(-2) = x^(1/-2) = 1/(x^(1/2)) = 1/s(qrt(x)) Thus, it doesn't seem to be consistent with the definition of tetration for positive integers. ----------------------------------------------------------------------------------- Now, today I had an interesting talk with my Math teaching assistant. He told the class about how to differentiate the funcion, f(x) = x^x. First, he quickly brushed through the "conventional" way of doing it: f(x) = x^x = e^(x*ln x) f'(x) = (x*ln x)' * e^(x*ln x) f'(x) = (1+ln x)*(x^x) Then, he talked about his friend's approach to the problem. According to his friend, people commonly make 2 kinds of mistakes when trying to differentiate f(x): 1) They think of the exponent as a constant (that is, the problem can be thought of as x^a). Therefore: f'(x) = x*(x^(x-1)) = x^x Apparently, that answer is incorrect. 2) They think of the base as a constant (that is, the problem can be thought of as a^x). Therefore: f'(x) = (x^x)*(ln x) This answer is also incorrect. However, something very interesting happens when you add both incorrect answers together: x^x + (x^x)*(ln x) = (1+ln x)*(x^x) = f'(x) In this case, two wrongs do make a right! I was very surprised when I heard this, and asked my TA to try differentiating something much more general: h(x) = f(x)^g(x) First, he did it the "conventional way" (I admit its quite hard to read using text box formatting): Let y = h(x) ln y = g(x)*ln(f(x)) (1/y)*h'(x) = g'(x)*ln(f(x)) + g(x)*(f'(x)/f(x)) h'(x) = (f(x)^g(x))*((f'(x)*g(x)/f(x)) + ln (f(x))*g'(x)) Then, he tried making both types of "mistakes": 1) g(x) is a constant. Let g(x) = a (f(x)^a)' = a*(f(x)^(a-1))*f'(x) = (f(x)^g(x))*((f'(x)*g(x))/f(x)) 2) f(x) is a constant. Let f(x) = a (a^g(x))' = g'(x)*(a^g(x))*(ln a) = (f(x)^g(x))*(ln (f(x))*g'(x)) Adding them together (and using the distributive property): (f(x)^g(x))*((f'(x)*g(x))/f(x) + ln (f(x))*g'(x)) Remarkably, this is exactly the correct answer. Furthermore, I have verified everything using the Mathematica. I've also tried applying this "technique" to addition (h(x) = f(x) + g(x)) and multiplication (h(x) = f(x)*g(x)). The technique correctly gave the Sum and Product Rules. The obvious thing to do with the (proven) Exponentiation Rule is to apply it to tetration. Let f(x) = x, g(x) = (x tetra n-1) Then h(x) = (x tetra n) (x tetra n)' = (x tetra n)*((x tetra n-1)/x + (ln x)*(x tetra n-1)') x tetra 1 = x. So we already know that (x tetra 1)' = 1. Thus, we can derive the derivatives for the next few values of n: (x tetra 2)' = (x tetra 2)*(x/x + ln (x)*1) = (x tetra 2)*(1+ln x) (x tetra 3)' = (x tetra 3)*((x tetra 2)/x + (ln x)*(x tetra 2)') = (x tetra 3)*((x tetra 2)/x + (ln x)*(x tetra 2)*(1+ln x)) This is not as good as a fully fledged formula for (x tetra n)' in terms of x, n and tetration + "lower" operations, but it's still progess.jinyduTetronian Posts: 721Joined: Thu Jun 10, 2004 5:31 am Top
I hope so. I'm having trouble finding the n'th derivative of tetration, but I think I've found a common pattern in the conditions for the top-function. I'd really like tetration to be continuous for all derivatives, so for that I'd need an infinite polynomial.
Nope. My eventual plan was to find the n'th derivative of z^^t, then use it to find an infinite polynomial for the top-function, (using a sum to infinity to express it). If I could do that, I would define x^^y using only this function. Unfortunately, it turns out to be impossible. It's not really a matter of a single polynomial. The top-function has to be in terms of the ln of z. The quadratic for instance, is f(x) = ax^2 +bx + c From my conditions, you can show that c=0 a+b= 1 2a + b=blnz From these, b = 2/(lnz+1) a = 1 - 2/(lnz+1) So the function itself is f(x) = [ 1 - 2/(lnz+1) ]x^2 + [ 2/(lnz+1) ]x This is complicated enough, but the cubic is enormous. To make it more simple, I let w = lnz. From the conditions, you can find similar equations to those above, and then end up with a quadratic involving c. But each coefficient is a polynomial of w. So you can find c in terms of w with the quadratic formula, but the expression is huge. Once you do that, you can find expressions for b and a. I just had a go at the quartic, and naturally it's even worse. I ended up with a cubic involving d, again with the coefficients in terms of w. I know vaguely how to solve cubics, so I'll have a go after the final exams, when I'll have plenty of time. Besides, even if I do that, and the next one, I'd then have to solve a quintic, which Godel proved is impossible using elementary functions.
Not at the moment. At least, it doesn't satisfy them using the quadratic or cubic top-function. If it does for any other, I'll be very surprised. Maybe I should use that as a starting point, rather than assuming that tetration is differentiable.
Frankly, I have no idea. Galidakis only defined his function for reals. Since the top-function uses the fraction part, it doesn't really have any meaning for values outside 1 and 0. This is just speculation, but I think tetration might be connected with probability. For one thing, we have numbers that have to be between 1 and 0. Like probabilities, they have no meaning otherwise. Furthermore, one equation that keeps popping up for all the top functions is that the coefficients all sum to 1 (just like probabilities sum to one). That derivative technique is very interesting. Do you think that would work for any operation? PWrongPentonian Posts: 1599Joined: Fri Jan 30, 2004 8:21 amLocation: Perth, Australia
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Well, first of all, it was Abel, not Godel, who proved that the general quintic is unsolvable by radicals. Also, not all quintics are unsolvable by radicals. For example: x^5 -32 = 0 is solvable (obviously, x = 2). What Abel proved is that not all quintics can be solved using the following operations: +, -, x, /, ^2, ^3, ^4, ^5, sqrt, cbrt, 4th rt and 5th rt. Thus, its still possible that your quintic could be a solvable quintic. However, I realize I'm being a nitpicker. Even if you could solve the quintic, you would then have to deal with a 6th degree polynomial, and even then, that would be a long way from a general formula.
I don't know. My teaching assistant just told me about it yesterday. I guess it might depend on what you mean by an "operation". According to Mathworld (http://mathworld.wolfram.com/Operation.html): "Let A be a set. An operation on A is a function from a power of A into A. More precisely, given an ordinal number (alpha), a function from (A^alpha) into A is an (alpha)-ary operation on A. If (alpha=n) is a finite ordinal, then the n-ary operation f is a finitary operation on A." Do you I understand the above? No. The way I have used the word "operation" so far is eerily similar to what a function is. After all, addition, multiplication, exponentiation and tetration can be thought of as (2-variable) functions: f(x,y) = x+y g(x,y) = x*y h(x,y) = x^y k(x,y) = x tetra y Anyway, about the exponentiation rule, I can also use it to try and differentiate (n tetra x). Keep in mind that in order for (n tetra x) to be differentiable at all, it must be a continuous function (at least in a certain interval)! In other words, if this derivative is correct, then it is possible to define (a tetra b) for all real values of b within a certain (possibly infinite!) interval. What is rational number small answer?A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating and non-recurring.
Which number is rational?Rational Numbers: Any number that can be written as a ratio (or fraction) of two integers is a rational number.
Are all integers rational numbers?All natural numbers, whole numbers, and integers are rationals, but not all rational numbers are natural numbers, whole numbers, or integers.
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