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coleppard
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What is a proof that a number is irrational.
For instance, how do we know the PI goes on forever without a pattern?
For instance, how do we know the PI goes on forever without a pattern?
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 [tex]\sqrt{2}[/tex] is a rational number, meaning that there exists an integer a and an integer b such that [tex]a / b = \sqrt{2}[/tex].
2. Then [tex]\sqrt{2}[/tex] can be written as an irreducible fraction [tex]a / b[/tex] such that [tex]a[/tex] and [tex]b[/tex] are coprime integers and [tex](a / b)^2 = 2[/tex].
3. It follows that [tex]a^2 / b^2 = 2[/tex] and [tex]a^2 = 2 b^2[/tex]. ([tex](a / b)^n = a^n / b^n[/tex])
4. Therefore [tex]a^2[/tex] is even because it is equal to [tex]2 b^2[/tex]. ([tex]2 b^2[/tex] is necessarily even because it is 2 times another whole number; that is what "even" means.)
5. It follows that [tex]a[/tex] 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 [tex]k[/tex] that fulfills: [tex]a = 2k[/tex].
7. Substituting [tex]2k[/tex] from (6) for a in the second equation of (3): [tex]2b^2 = (2k)^2[/tex] is equivalent to [tex]2b^2 = 4k^2[/tex] is equivalent to [tex]b2 = 2k^2[/tex].
8. Because [tex]2k^2[/tex] is divisible by two and therefore even, and because [tex]2k^2 = b^2[/tex], it follows that [tex]b^2[/tex] is also even which means that [tex]b[/tex] is even.
9. By (5) and (8) [tex]a[/tex] and [tex]b[/tex] are both even, which contradicts that [tex]a / b[/tex] is irreducible as stated in (2).
Since there is a contradiction, the assumption (1) that [tex]\sqrt{2}[/tex] is a rational number must be false. The opposite is proven: [tex]\sqrt{2}[/tex] 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 [tex]\sqrt{2}[/tex] can be written as an irreducible fraction [tex]a / b[/tex]
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 [itex]\pi[/itex] are 0, 1, or -1, all integers. Therefore, [itex]\pi[/itex] is irrational.
Office_Shredder said:I had no idea you could have symbols like pi in the URL for a website
An irrational number is a number that cannot be expressed as a ratio of two integers. It is a non-terminating, non-repeating decimal.
To prove that a number is irrational, we must show that it cannot be expressed as a ratio of two integers. This can be done through various methods, such as the proof by contradiction or the proof by uniqueness.
Some examples of irrational numbers include pi (3.141592...), the square root of 2 (1.414213...), and Euler's number (2.718281...).
Proving that a number is irrational helps us understand the properties and characteristics of numbers. It also helps us solve mathematical problems and equations that involve irrational numbers.
No, a number cannot be both rational and irrational. A number is either rational or irrational, but not both. If a number can be expressed as a ratio of two integers, it is rational. Otherwise, it is irrational.