Where can i find Proof square root of 5 is irrational?

Answer 1

The proof that the square root of 5 is irrational is exactly the same as the well-known proof that the square root of 2 is irrational - except using 5 in place of 2. We can prove a more general result: the square root of any prime is irrational.

First of all, we require the lemma:

for any prime p, and integer x,

p|x2 ⇒ p|x

That is, if x2 is divisible by p, then so is x.

Proof:

The prime factorization of x2 necessarily contains p at least once, since it is divisible by p. But it also has to contain an even power of every prime, since it is the prime factorization of a square. Therefore, it contains p at least twice, and its square root, x, contains p at least once: that is, x is divisible by p.

Now, given a prime p, assume that its square root is rational. Then, it may be written in the form a/b, where a and b have no common factors (that is, the fraction a/b is in lowest terms). This is always possible for any nonzero rational number. Since this quantity is the square root of p, its square equals p, that is

(a/b)2 = p

a2/b2 = p

a2 = pb2

Now, pb2 is a multiple of p, so a2 must be too. And, using the result above, this means that a must be a multiple of p also. Thus, there exists an integer c such that

a = PC

Then,

(PC)2 = pb2

p2c2 = pb2.

Since p is not zero, we may divide both sides by p to obtain

PC2 = b2

That is, b2 is divisible by p also, and thus b is divisible by p.

Since a and b were both divisible by p, the fraction a/b could not have been in lowest terms, which contradicts our initial assumption. Therefore, the square root of p cannot possibly be a rational number. Since 5 is prime, the proof is complete.