Is sqrt39 a rational number?
3 Answers
Explanation:
Primes do not have factors other than 1 and itself. So, a prime cannot
be a square of an integer. And so,
Also, product of two distinct irrationals is irrational.
Here,
= an irrational number.
Explanation:
Here's a sketch of a proof by contradiction...
Suppose
Without loss of generality, we may suppose that
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
Since
Similarly
So
Then:
39 q^2 = p^2 = (39k)^2 = 39*39k^2
Dividing both ends by
q^2 = 39k^2
So:
q^2/k^2 = 39
and so:
q/k = sqrt(39)
Now:
p > q > k > 0
So
So there is no pair of positive integers
Explanation:
Here's another method to prove
Suppose
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
25x+4x^2 = 156+25x
Subtracting
4x^2 = 156
Dividing both sides by
x^2 = 39
and hence
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.