This discussion focuses on the proofs of irrationality, specifically for numbers like √2 and π. The proof for √2 utilizes a method of infinite descent and contradiction, demonstrating that assuming √2 is rational leads to a contradiction, thus confirming its irrationality. Additionally, the discussion highlights a theorem that states if a continuous function f has integer-valued iterated anti-derivatives at specific points, then the positive real number c is irrational, with π being a prime example. The distinction between irrational and transcendental numbers is also clarified, emphasizing that while all transcendental numbers are irrational, the reverse is not necessarily true.
PREREQUISITESMathematicians, students of mathematics, and anyone interested in number theory, particularly those studying irrational and transcendental numbers.
Transcendental" always implies "irrational
That is a flipping awesome proof!Irrational said:http://en.wikipedia.org/wiki/Square_root_of_2"
One proof of the number's irrationality is the following proof by infinite descent. It is also a proof by contradiction, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.
1. Assume that \sqrt{2} is a rational number, meaning that there exists an integer a and an integer b such that a / b = \sqrt{2}.
2. Then \sqrt{2} can be written as an irreducible fraction a / b such that a and b are coprime integers and (a / b)^2 = 2.
3. It follows that a^2 / b^2 = 2 and a^2 = 2 b^2. ((a / b)^n = a^n / b^n)
4. Therefore a^2 is even because it is equal to 2 b^2. (2 b^2 is necessarily even because it is 2 times another whole number; that is what "even" means.)
5. It follows that a must be even as (squares of odd integers are also odd, referring to b) or (only even numbers have even squares, referring to a).
6. Because a is even, there exists an integer k that fulfills: a = 2k.
7. Substituting 2k from (6) for a in the second equation of (3): 2b^2 = (2k)^2 is equivalent to 2b^2 = 4k^2 is equivalent to b2 = 2k^2.
8. Because 2k^2 is divisible by two and therefore even, and because 2k^2 = b^2, it follows that b^2 is also even which means that b is even.
9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).
Since there is a contradiction, the assumption (1) that \sqrt{2} is a rational number must be false. The opposite is proven: \sqrt{2} is irrational.
Proofs of transcendental-ness are not as easy.
How does it follow from 1. that it must be irreducible?Irrational said:2. Then \sqrt{2} can be written as an irreducible fraction a / b
Right. That's obvious now. Thanks.Irrational said:think what it's supposed to mean is that if it's rational, it can be written as a fraction.
any fraction can be 'reduced' to it's simplest form (that you can't divide top and bottom by 2/3/.../whatever...)
HallsofIvy said:There is a very interesting theorem that says
"Let c be a positive real number. If there exist a function f, continuous on [0, c] and positive on (0, c), such that f and all of its iterated anti-derivatives can be taken to be integer valued at 0 and c, then c is irrational."
If f(x)= sin(x), then all anti-derivatives can be taken (by taking the constant of integration to be 0) as sin(x), -sin(x), cos(x), or -cos(x). The values of those at 0 and \pi are 0, 1, or -1, all integers. Therefore, \pi is irrational.
Office_Shredder said:I had no idea you could have symbols like pi in the URL for a website