Prove that the sum is irrational.
I've planned my approach in 2 ways:
1. Must show that the sum is not an integer.
2. Must show it is a zero of a monic polynomial.
But how to go about it?
It appears to me that since the first term doesn't cancel the remaining terms, then all you have to do is prove that the first term is irrational.
Any whole number that's squared which you add one to has to be irrational when you take its square root. I think you can work through the rest.
This is equivalent to $\displaystyle \sum_{x= 1001}^{2000} \sqrt{1+x^2} $. And $\displaystyle \sqrt{x^2+1} $ is not a square number which implies it is irrational.
Since it is monotonic increasing, the sum of the irrationals can't be rational (e.g. we can't have expressions of the form $\displaystyle (2-\sqrt{2}) +(4+\sqrt{2}) $).
Okay so had another look at this.
So to prove the sum is not an integer.
So,
Let
An overestimate for the sum would be
An underestimate for the sum would be
But since
Thus the sum is not an integer.
Use induction to prove it is a zero of a monic polynomial.
Base case:
Let the sum be written as
Let .
Thus
Thus is a zero.
Now for the inductive hypothesis, assume is true.
Thus is true.
Now we wish to prove that is also true.
We will denote this as .
Let this be a zero of a monic polynomial
So using the inductive hypothesis, is a zero , namely
so
Now what...?
if you are familliar with basic concepts of algebraic number theory, you should know this much that the set of integral elements (an elements is called integral if it's a root of a monic polynomial
with integer coefficient) is a ring and hence it's closed under addition. now each $\displaystyle \sqrt{a_i}$ is integral and thus $\displaystyle A=\sum_{i=1}^n \sqrt{a_i}$ is integral too. it's also a basic fact that $\displaystyle \mathbb{Z}$ is integrally closed, that is an
integral element which is also a rational number has to be an integer. but, as you almost showed, $\displaystyle A$ is not an integer and so it cannot be rational. Q.E.D.