Is sqrt39 a rational number?

3 Answers
Sep 27, 2016

sqrt 39 is irrational.

Explanation:

Primes do not have factors other than 1 and itself. So, a prime cannot

be a square of an integer. And so, sqrt(prime) is irrational.

Also, product of two distinct irrationals is irrational.

Here,

sqrt 39=sqrt 3 sqrt 13, where 3 and 13 are primes.

= an irrational number.

Sep 27, 2016

sqrt(39) is irrational

Explanation:

Here's a sketch of a proof by contradiction...

Suppose sqrt(39) = p/q for some positive integers p, q with p > q > 0.

Without loss of generality, we may suppose that p, q is the smallest such pair of integers.

Then we have:

p^2/q^2 = 39

So:

p^2 = 39 q^2 = 3 * 13 q^2

Since the right hand side is divisible by 3, p^2 is divisible by 3.

Since 3 is prime, p must also be divisible by 3.

Similarly p must be divisible by 13.

So p is divisible by 3*13 = 39 and there is some positive integer k such that p = 39k

Then:

39 q^2 = p^2 = (39k)^2 = 39*39k^2

Dividing both ends by 39 we find:

q^2 = 39k^2

So:

q^2/k^2 = 39

and so:

q/k = sqrt(39)

Now:

p > q > k > 0

So q, k is a smaller pair of positive integers satisfying q/k = sqrt(39), contradicting our hypothesis.

So there is no pair of positive integers p, q with p/q = sqrt(39)

Jul 23, 2018

sqrt(39) is irrational

Explanation:

Here's another method to prove sqrt(39) is irrational:

Suppose x > 0 satisfies:

x = 6+1/(4+1/(6+x))

Then:

x = 6+1/(4+1/(6+x))

color(white)(x) = 6+(6+x)/(25+4x)

color(white)(x) = (156+25x)/(25+4x)

Multiplying both ends by (25+4x) we find:

25x+4x^2 = 156+25x

Subtracting 25x from both sides, this becomes:

4x^2 = 156

Dividing both sides by 4 we find:

x^2 = 39

and hence x = sqrt(39)

So:

sqrt(39) = 6+1/(4+1/(6+sqrt(39)))

color(white)(sqrt(39)) = 6+1/(4+1/(12+1/(4+1/(12+1/(4+1/(12+...))))))

Since this is a non-terminating continued fraction, it is not expressible as a terminating - i.e. rational - one.