irrationality of k ^ (1/2) (1 Viewer)

[cheesegrater]

HELP ?!?
prove that the square root of k is irrational for any integer k which is not the square of an integer?

im well stumped,i proved that it is rational for the square of an integer but that seemed pretty useless any ideas?

 

[KeypadSDM]

Reductio ad absurdum.

 

[cheesegrater]

elaborate

 

[KeypadSDM]

Let Sqrt[k] = p/q
where (p, q) = 1
(p and q are coprime and integral)
Thus k = p^2/q^2
p^2 = k * q^2

Now p^2 is a multiple of k, and from this, we deduce that p must also be a multiple of k.

Thus we can rewrite p as
a * k
:. (ak)^2 = k * q^2
a^2 * k = q^2
From this, we deduce that q also is a multiple of k

But this is a contradiction from our initial assumption that Sqrt[k] can be represented in the form p/q, where p and q are irreducible integers.

Thus Sqrt[k] cannot be a rational number.

EDIT: This isn't ENTIRELY true. Take a look at the case where k is a square number. Although k is a factor of p^2, it may not be a factor of p, as Sqrt[k] can be a factor.

 

[cheesegrater]

cool thanks for that