Is there a way to show the sum of any different square root of prime numbers is irrational? For example, $$\sqrt2+\sqrt3+\sqrt5 +\sqrt7+\sqrt{11}+\sqrt{13}+\sqrt{17}+\sqrt{19}$$ should be a irrational number.
One approach I used is to let the sum be a solution of an even polynomial $f(x)$with integer coefficients and prove by induction that by adding another $\sqrt{p_{k+1}}$. The new polynomial can be written as $$f(x+\sqrt{p_{k+1}})f(x-\sqrt{p_{k+1}})$$
where $$f(x+-\sqrt{p_{k+1}})=P(x)+- Q(x)\sqrt{p_{k+1}},$$
where $P(x)$ is an even plynomial and $Q(x)$
is an odd polynomial.
The new polynomial can be written as $$P^{2}(x)- Q^{2}(x)p_{k+1}.$$
Assume it has a rational solution $a$, we must have$$P(a)=Q(a)=0.$$
My calculation stopped here since I can't find any contradiction result from this. Can anyone continue this proof, or has other better way to solve this? Thanks!
No comments:
Post a Comment