Suppose 2^(1/2) = a/b in lowest terms, where a,b are integers. 2 = a^2/b^2. Implies 2b^2 = a^2. Thus a^2 is even, and so must be a. So express a as 2m, where m is some integer. a^2 = 4m^2. This implies 2b^2 = 4m^2 implies b^2 = 2m^2. So b^2 and b are also even. But a/b was in lowest terms! Contradiction.

Thus 2^(1/2) is irrational. QED.

Technically, this is a proof that the positive rationals are not closed under exponentiation. If we do not care about closure under exponentiation, we do not have to assume the existence of the irrationals in the algebraic structure we are working with.

It is said that one of Pythagoras' students discovered this proof. However, Pythagoras' entire world view was based on natural numbers and their ratios, the rational numbers. He believed that the universe was founded upon these, and so the concept of irrational numbers was heresy to him. He had the student murdered (drowned, which was the usual form of execution for Pythagoreans who had broken the group's laws) to prevent the idea from spreading. Thus, the person who laid the foundation for western science also (in spirit) laid the foundation for the dark ages, religious intolerance and the Inquisition.

Keep in mind that there are no direct written records of Pythagoras' life and work, and all information about him is indirect or conjecture, so this story may be untrue.

SPUI is, essentially, correct. What this proof establishes is that there is no rational number a/b such that (a/b)^2 = 2. So, if the square root of 2 exists, it must be irrational. In a theoretical sense, there is no reason to assume that the square root of 2 must exist, and there are certainly plenty of examples in abstract algebra of algebraic structures in which square roots and other radicals do not exist. However, the Pythagoreans had a very practical reason for wanting to discover the square root of 2. Their math was mostly geometry, and as anyone who knows the Pythagorean Theorem is aware, sqare roots are an important part of geometry. So, with tongue resting lightly in cheek, the result could rephrased thus:

Any system of numbers complete enough to measure the hypotenuse of a triangle is going to have irrational numbers in it.

Since mathematics would be very limited without the ability to measure the lengths of things, what we really have is the proof of the necessity of irrational numbers.

Little known fact: The square root of two is also known as Pythagoras' constant.

Now that we know there are irrationals in the set of real numbers, one might ponder how many other irrational numbers exist. One existance of an irrational constant implies that there are at least infinitely many other irrationals. This follows from the fact that given an irrational k and a rational r, kr is irrational. (Proof: Suppose there exists integers a != 0, b, c, d such that k * c / d = a / b. Then k = ad/bc. Contradiction.)

Also, the entire set of irrational numbers is uncountably many.

Proof: Suppose there exists an indexing on the set of all irrationals K, such that K = {kn}. Since the set of rational numbers is countably infinite, there exists an indexing Q = {qn}. The set of real numbers R then has an indexing: rn = kn/2 if n even, rn = q(n-1)/2 if n odd. This contradicts the fact that the real numbers are uncountable. QED.
A direct proof of the existence of irrational real numbers:

Let S = {x ∈ R : x2 < 2}. S is bounded above (be eg. 3/2) so by the completeness of the real numbers S has a supremum, call it x. It is trivial to check from the definition of the supremum that neither x2 < 2 nor x2 > 2 are possible. Therefore x2 = 2, and we call x the positive square root of 2.
Now suppose that x = p/q for some integers p,q. Without loss of generality p, q are coprime. Then p2 = 2q2, so p is even. Hence q2 = p2/2, so q is even. This contradicts the choice of p, q coprime. Hence the real number x is irrational.

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