San José State University

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Thayer Watkins
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 The Irrationality of the Square Root of Any Integer or Other Rational Number that is not a Ratio of Squared Integers

The fact that the square root of 2 cannot be expressed as a ratio of integers was an astounding mathematical result in ancient Greece. The proof of this is simple; see for instance Irrationality of √2. This also applies for the square roots of 3, 5, 6 and so forth. However it is rather tedious to construct separate proofs for all cases in which the square root of a number is irrational.

The Unique Factorization Theorem for Integers, which is often called, The Fundamental Theorem of Arithmetic, says that any integer can be expressed as a unique product of powers of primes. Unique means that there is only one set of primes and their exponents whose product gives the integer. This theorem is proven, among other places, at Unique Factorization.

Let q be an integer. Unless q is the square of an integer √q is irrational. Suppose √q is equal to n/m, where n and m are integers. Then

#### q = n²/m² and hence n² = qm²

Suppose Πpiki and Πpjkj are prime factorizations of n and m, respectively. Then

Thus

#### Πpi2ki = qΠpj2kj

Every factor that appears on the right-hand side (RHS) of the above equation must appear also on the LHS. All of the factors of m² can be cancelled out. This leaves only q on the RHS and on the LHS there is a product of the form Πps2ks which is the square of Πpsks. Thus

#### q = (Πpsks)² or, equivalently q = v² where v=Πpsks,an integer

Therefore if q is not the square of an integer then √q cannot be expressed as a rational number; i.e., it is irrational.

## A Generalization

Let Q be a rational number and suppose √Q is equal to n/m, where n and m have no factor in common. Let the factorizations of n and m be

#### n = Πpiki Πpjkj

Since √Q=n/m, Q is equal to n2/m2.

Let u=Πpt2kt be the product of the common factors of n² and m². Then n²/u and m²/u are of the forms

#### Πpi2ki = (Πpiki)2 and Πpj2kj = (Πpjkj)2

Thus there exists integers r and s such that

#### g(n²/u) = gr² = hs² = h(m²/u) and hence g/h = r²/s² = (r/s)²

The significant part of the result is that unless the rational number is the ratio of two squared integers its square root is irrational. Thus without any further proof we know √(2/3) is irrational and likewise for √(1/2).

An even simpler proof is as follows. If √Q is rational then there are r and s, integers, such that

#### √Q = r/s and hence Q = r²/s²

Thus if Q=n/m then

#### n/m = r²/s²

Since n and m have no factor in common and r and s, and hence r²/s², have no factor in common it follows that

#### n = r² and m = s²

That is to say, the square root of a rational number is rational only if that rational number is the ratio of two squared integers.