I think it's just a site for basic learners besides I didn't know there was an article on there. What are you talking about?
An interesting analogy on one pi "Imagine a race of talking ants. The ants can compress the infinite digits of pi in an interesting way. For example, let us imagine that the ants can speak by manipulating their crude jaws. The first ant in the long parade of ants screams out the first digit, "3". The next yells the number on its back, a "1". The next yells a "4", and so on. Further imagine that each ant speaks its digit in only half the time of the preceding ant. Each ant has a turn to speak. Only the most recent digit is spoken at any instant. If the first digit of pi requires 30 seconds to speak (due to the ant's cumbersome jaws and little brain), the entire ant colony will speak all the digits of pi in a minute! (Again, this is because the infinite sum 1/2 minute + 1/4 minute + 1/8 minute + ... is equal to 1 minute.) Astoundingly, at the end of the minute, there will be a quick-talking ant that will actually say the "last" digit of pi! The geometer God, upon hearing this last digit, may cry, "That's impossible, because pi has no last digit!" ................." Any like that mind twist? Now to really get the tinker tinking, play with phi and pi, together.
Its nothing more than the well known convergent sum \(\sum_{n=1}^{\infty} \frac{1}{2^{n}} = 1\) Anyone whose gone convergent sums will have come across that. It only sounds like a 'mind twist' because you're associating the digits of a well known irrational number, pi, with the terms in the sum and done reworking of the well known 'Xeno paradox', which isn't a paradox at all when you know the difference between the place of a digit in summation and the value of the contribution that terms makes to the sum. Infact the decimal expansion of any number is written in the form \(\sum_{n=1}^{\infty}a_{n}10^{-n}\) where \(a_{n}\) is the n'th digit in the expansion. For pi you have \(a_{1} = 1\), \(a_{2} = 4\) etc. Its elementary to prove that for \(0 \leq a_{n} \leq 9\) the sum \(\sum_{n=1}^{\infty}a_{n}10^{-n}\) is finite. You do realise all you've done is rephrase Xeno's paradox, right? There's nothing any more special about pi and phi than any other pairing of irrational numbers. Those two have nice properties or expressions when written in certain ways but no more than many other numbers. But feel free to dazzle us all with your amazing grasp of numerology.
Bishadi---you're on thin ice. I'd ban you right now, but I'll just give you a warning. I don't want this to be misinterpreted as stemming from another conversation in another forum. That being said, if your next post in this forum has any insults in it, expect a week long ban.
"Pi is an irrational and transcendental number meaning it will continue infinitely without repeating." Every transcendental number is irrational, but not vice versa. The second half of the sentence characterizes all irrational numbers, even if not transcendental.
That's the best defintion I've ever seeen on this forum of the word "irrational" but what exactly does transecendental mean?
An irrational number can't be written as \(\frac{a}{b}\) for a and b integers. An algebraic number is a number which is the root of a polynomial with integer coefficients. For instance, to use Bishadi's example the golden ratio phi satisfies \(x^{2} - x - 1 = 0\). That's a polynomial and so phi is algebraic. Its not rational though. Similarly \(\sqrt{2}\), which obviously satisfies \(x^{2} = 2\). The rationals are those which satisfy linear polynomials, \(bx-a = 0\). All rationals are algebraic. A transcendental number is a non-algebraic number so you can't write down a polynomial with integer coefficients which they are a root of. Clearly pi is a root of \(x-\pi\) but that polynomial has a non-integer coefficient, pi. No matter how high a term you include, like \(a_{0} + a_{1}x + a_{2}x^{2} + \ldots + a_{10000000}x^{10000000}\) if \(a_{n}\) are whole numbers pi will not be a root of the polynomial. It came up in another thread but I think it bears repeating that irrational or transcendental doesn't mean the digits are random. Things like pi, e and sqrt(2) seem so but the first number to be proven transcendental was \(\sum_{n=0}^{\infty}10^{-n!}\) which has only 0s and 1s in its decimal expansion (base 10) so obviously not 'random'. The issue of whether pi's digits are 'random' (ie when viewed as a set of integers 0 to 9 they have the properties found in random sets, no preference for some particular digit etc) is an open problem. The first trillion digits are known and they seem to be.
The other even MORE bonkers assertions on boards such as this is that, since (as far as is known) \(\pi\) cannot be written as a finite string, then it must be infinite. This clearly insane, since \(3 < \pi < 4\) In an attempt to raise the level a little: your poly \(x-\pi\) is reducible, since \(x-\pi = (\sqrt{x}- \sqrt{\pi})(\sqrt{x}+ \sqrt{\pi})\) with zeros at \(\pm\sqrt{\pi}\). So, granted that the zero of \(x-\pi\) is \(\pi\) (self-evidently it is), and noting that \(\pm\sqrt{\pi}^2 = \pi\), is it the case that, for any reducible poly, the product of roots is the root of products? In fact, does this generalize?
This is akin to a trick I remember figuring out when I first started learning about polynomials. If you have, say \(x^4-16=0\), you can (always, I guess write it as \(y^2-16=0\), then factor. It seems that you've just redefined \(y \equiv \sqrt{x}\) which is a well defined coordinate transformation for all x, then factored the polynomial. The roots of the old polynomial are at \(\pm x_0\) so the roots of the new polynomial should be at \(\pm \sqrt{x_0}\). It looks like the only assumption is that the coordinate transformation is well-defined, so it seems that it should be general.
Ah yes, the classic "There's an infinite number which is less than 4 and greater than 3" one. But then we all know 2+2=5 for exceptionally large values of 2 Please Register or Log in to view the hidden image! In what sense is it reducible? Its not reducible over the ring \(\mathbb{R}[x]\). Splitting polynomials into non-polynomials is a dangerous game to play. I'm not entirely sure what you're getting at here. It is clear to see that given \(f_{n}(x) = x+a_{n}\) then \(F(x) \equiv \Pi_{n}f_{n}(x)\) is such that \(F(0) = \Pi_{n}a_{n}\). Is this what you mean? A linear polynomial in \(\mathbb{R}[x]\) is irreducible, up to an overall factor in \(\mathbb{R}\) (so only consider cases where the coefficient of x is 1). Over \(\mathbb{R}[x^{\frac{1}{n}}]\) then you allow the possibility of factorising a linear polynomial into n factors and thus if you set \(x^{\frac{1}{n}} = y\) then you get n times as many solutions for y than for x but all of them will be such that \(y^{n}\) is always the same. By going from \(\mathbb{R}[x]\) to \(\mathbb{R}[x^{\frac{1}{n}}]\) you are simply including the n'th roots of the original polynomial roots. I feel I'm missing something because I'm not sure what the point you're trying to make is :shrug:
Algebraic numbers (which are countable) are solutions of polynomial equations with integer (or rational) coefficients. Transcendental numbers are non-algebraic.
Wrong. Some irrational numbers are algebraic and some are not: http://en.wikipedia.org/wiki/Algebraic_number Irrational number are not countable. Hence there are irrational number that are algebraic but the irrational numbers are not countable. In mathematics, an irrational number is any real number which cannot be expressed as a fraction m/n, where m and n are integers, with n non-zero and is therefore not a rational number. Informally, this means that an irrational number cannot be represented as a simple fraction. It can be proven that irrational numbers are precisely those real numbers that cannot be represented as terminating or repeating decimals, although mathematicians do not take that to be the definition. As a consequence of Cantor's proof that the real numbers are uncountable (and the rationals countable) it follows that almost all real numbers are irrational.[1] Perhaps the best-known irrational numbers are π, e and √2.[2][3][4] http://en.wikipedia.org/wiki/Irrational_number
Grassman numbers are something different altogether. They anticommute, that is \(\left\{ a,b \right\} = ab + ba = 0\)
