Hey, I was wondering about proving that square roots of numbers are rational or irrational.
I know about proving this for the numbers 1(duh), 2 and all multiples of 2 and 4, but what about 3?
How do i prove that 3(or any other number) is irrational or rational?
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Anonymous2007-11-08 12:30
Use the uniqueness of prime decomposition
if there exists natural numbers p/q s.t p^2/q^2 = n, some n in the naturals
=> p^2 = n.q^2
If we decompose n,q,p into primes, we see that unless n = a^2 for some a in the naturals, then one of the prime exponents on the RHS of the equations is odd, and by the uniqueness of prime factorisation, this cannot be true as the exponents on the LHS are all even.
You can prove the uniqueness of prime decomposition quite easily, wiki it.
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Anonymous2007-11-09 2:33
assume x = sqrt(3) is rational
x^2 = 3
x^2 - 3 = 0
use the rational roots theorem to find the only possible rational solutions for this equation are 1,-1,3,-3, none of which solve the equation. so, no rationals solve this equation.
thus sqrt(3) is not rational since it solves the equation but there is no rational solution.
That is really awesome, except for one thing, it relies on a theorem. Therefore cannot be considered a proper proof, however it is incredibly easy, now to find a relationship between all rational and irrational proofs to find out more about the world. Oh god....
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Anonymous2007-11-09 11:56
>>6
One of us doesn't know what a theorem is, and it's not me.
Yeah, a proof that proves only one case of a general theorem using much higher powered maths than is needed, is better than an elementary proof using basic number theory that covers every case.
You're an idiot.
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Anonymous2007-11-09 21:47
This is an answer set forward by Gabriel Haeseron orkut
let k and n be positive integers, if k^(1/n) is not an integer than it is irrational.
proof: supose k^(1/n)=a/b, than k.b^n=a^n. we know k is not the n-th power of an integer, so there must be a prime factor p of k with multiplicity alpha, such that alpha is not a multiple of n. So, p must appear with multiplicity beta in kb^n, such that beta is not a multiple of n, but that is a contradiction because the multiplicity of p in a^n is clearly a multiple of n. That concludes the proof.
Wrong. You're confusing "almost always" with "always". You mean that a random real number will be irrational with probability (exactly) 1. Furthermore, it is almost certainly transcendental, incomputable, and even undefinable.
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Anonymous2007-11-10 21:03
So why not just call it always, the futher into infinty you go, the smaller the odds get, so the odds are infinatly small. you might as well call the odds 0
Wrong. You're confusing "almost never" with "never". You mean that a random real number will be rational with probability (exactly) 0. Furthermore, it is almost never not transcendental, incomputable, or (and) undefinable.
I don't know why, but probability still doesn't strike me right. I have a bit of OCD and it seems like messy guesswork, although I know it has some good results.
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Anonymous2007-11-11 19:07
>>24
I agree, that there's something in probability that disagrees with me on an abstract.
>>17 >>19
Exactly the kind of confusions I came across when first encountering continuous probability distributions. For a distribution positive on [0,1], the probability of being picked, for each number in the set, is zero, but the probability of an individual number being picked is 1. LOL.
I might get onto it next term, or "semester" or whatever, if I choose to study the utterly and entirely dull-looking course titled "Statistics". I've had a small brush with it, and it was just as bad as probability was last year.