Question

Is `sqrt39` a rational number?

Answer 1

`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.

Answer 2

`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)`

Answer 3

`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.