Thread: Proof of the irrationality of non-integer square roots on Decorum

Proof of the irrationality of non-integer square roots by Arancaytar. Forum: Debate.

Arancaytar administrator 879 posts 2007-12-02 01:22:40	There does not seem to be a better forum to put this. The proof of the irrationality of sqrt(2) is a well-known example of a proof by contradiction. Below, I've expanded this proof to work for all prime numbers, not just 2. Let n be prime. Let r be the square root of n. Suppose that r is rational. This means that r = p/q where p and q are natural and coprime. It follows that r^2 = p^2/q^2 and thus n = p^2/q^2 n * q^2 = p^2 Therefore, p^2 is divisible by n. p^2 mod n = 0. The prime factors of a natural number are identical to the prime factors of its square, because the square n^2 of a prime has no other factors than 1, n and n^2. Therefore, if a number is prime, and it is a factor of a square, then it must also be a factor of the root of the square. Therefore since p^2 mod n = 0 and n is prime, p mod n must also be 0. If a natural number has a prime factor, then the square of this prime will be a factor of its square. It follows from p^2 mod n = 0 and n being prime that p^2 mod n^2 = 0. That means a natural number g exists such that p^2 = g * n^2. Recall now that n * q^2 = p^2. From substituting for p^2, it follows that n * q^2 = n^2 * g q^2 = n * g Thus, n is a factor if q^2. Repeating the earlier reasoning, it follows that n is a factor of q. It was demonstrated that the natural number n is a factor of both p and q. However, p and q are defined to have no common factors. It follows that no numbers p and q can fulfill the conditions without contradiction. Therefore, the square root of any prime is irrational. Badabing. This part was pretty simple. But now I tried to expand this for any number with a non-natural square root. I got pretty far, but then hit a roadblock which I'm not sure I navigated correctly. The steps I'm unsure about are in bold. Perhaps someone less rusty in Math than I am could check that I didn't make any logical errors there or anywhere else? I'm sure that even if it's correct, I have a few redundancies in there that could be made far more elegant... Let n be a natural number that is not the square of a natural number. This means that at least one prime factor of n has an odd power. Or, more technically, that there is a prime p and a natural number q such that n is divisible by p^(2q-1) but not by p^(2q). In other words, n = g * p^(2q-2) p, Where g is a natural number not divisible by p. Let r be the square root of n. It follows that r = sqrt(g) * p^(q-1) * sqrt(p) It is known that sqrt(p) is irrational as p is prime. The product of an irrational number and any rational (and thus any natural) number is always irrational. Therefore, p^(q-1) * sqrt(p) is irrational. The product of an irrational number c and another irrational number d is only rational if c is the product of d and a rational number. g is natural, but not the product of any natural number and p. Therefore sqrt(g) is not the product of any rational number and sqrt(p). Therefore, sqrt(g) * p^(q-1) * sqrt(p) is irrational. It follows that the square root of any natural number with a non-natural square root is irrational. This means there is no natural number whose square root is rational but not natural. QED.
jib user 641 posts 2007-12-02 09:55:58	Natural number n = (p/q)^2, where p and q are coprime n = p^2 / q^2 n * q^2 = p^2 Therefore p^2 is a multiple of q^2 Therefore p^2 has some prime factors which q^2 also has Since the prime factors of a number are the same as the prime factors of its square, this implies that p has some prime factors which q also has. This is a contradiction, because p and q were defined to be coprime Therefore there is no rational number p/q such that p and q are coprime and (p/q)^2 is a natural number. Either I have no idea what I'm talking about (as usual), or I have just proved the same thing as you without needing all that stuff about irrational numbers. Badabing. (I am aware that what you actually asked for was for people to check your proof, not to write their own proofs. I read your proof, and it does seem to be correct. I think the bold part could be expanded into a few more steps to make it clearer, but I'm too tired to write it out at the moment, and also I possibly have no idea what I'm talking about.)
Liempt user 158 posts 2007-12-03 05:49:40	The product of an irrational number c and another irrational number d is only rational if c is the product of d and a rational number. A hint: If c*d is an rational number but c and d are irrational then either c or d is the product of the other variable and a rational number. This is only a conditional, and not a biconditional.
Arancaytar administrator 879 posts 2008-03-03 12:08:01	I just realized something. The first proof ("primes have no rational roots") uses the following known statement: "The prime factors of a natural number are identical to the prime factors of its square." Of course, both are true. But I'm wondering if they're not somehow equivalent, meaning that the reasoning is circular... Oh wait. The second statement would only be invalidated by a prime number with a natural square root. And if n^2 is prime, then it violates the basic requirements of a prime, namely that it has no other factors than n^2 and 1. There, not circular after all.
Arancaytar administrator 879 posts 2008-03-03 12:21:32	either c or d is the product of the other variable and a rational number Wait. If c is the product of d and a rational number r, then d is also the product of c and the rational number 1/r. They seem to follow from each other. Or did you mean there is a counter-example where the above applies, but c*d is still not rational? Otherwise I don't see why this is not biconditional...
Tiberius Gracius user 12 posts 2008-04-08 01:46:08	A classier proof. Let p(x)=x^2-q, q a prime. Then by the rational root theorem the only possible rational roots are q, 1, -1, and q. None of these work. So the roots are irrational.