Prove that the sum $ \sqrt{1001^2 + 1}+\sqrt{1002^2 + 1} \ + ... + \sqrt{2000^2 + 1}$ is irrational

Question

Prove that the sum $$ \sqrt{1001^2 + 1}+\sqrt{1002^2 + 1} \ + ... + \sqrt{2000^2 + 1}$$ is irrational.

The textbook has the solution too but I'm unable to understand it.

The strategy is divided into two parts:-

Proving that the sum is not an integer
Proving that it is a zero of a monic polynomial

First part is simple to understand and prove. $$n^2 +1 < n^2 +2 < n^2 + 2n\left(\frac{1}{n}\right) + \left(\frac{1}{n}\right)^2 = (n+\frac{1}{n})^2\,.$$ So $$S = 1001 +a_1 + 1002 + a_2 + \ ... \ +2000 + a_{1000}\,, $$ and $$0 <a_1 + a_2 + ...+a_{1000} < \frac{1}{1001} (1000) < 1\,.$$

But how is the second part going to help in proving the result?

Is it $\sqrt{1001^2 + 1}+\sqrt{1002^2 + 1} \ + ... + \sqrt{2000^2 + 1}$? — Robert Z, Jul 23 '20 at 16:00
The second part probably uses the result: "If $P(x)$ is a polynomial with integer coefficients, and $\frac{a}{b}$ is a rational root of $P(x)$ then $b$ divides the leading coefficient of $P(x)$ and $a$ divides the constant term of $P(x)$". Can you prove this result? — Fawkes4494d3, Jul 23 '20 at 16:01
@Fawkes4494d3: the result is true only if the fraction is irreducible. — Bernard, Jul 23 '20 at 16:04
@Bernard of course, the fraction has to be irreducible. Thank you for adding that. :-) — Fawkes4494d3, Jul 23 '20 at 16:12
See also https://artofproblemsolving.com/community/q2h164239p915807. An explicit polynomial with integer coefficients with $S$ as a root is given there. — Batominovski, Jul 23 '20 at 16:39

Robert Z · Accepted Answer · 2020-07-23T16:17:08.700

2

Since $\sqrt{n^2+1}$ is an algebraic integer for $n=1001,\dots,2000$ and algebraic integers are closed under addition, it follows that $$S=\sqrt{1001^2 + 1}+\sqrt{1002^2 + 1} \ + ... + \sqrt{2000^2 + 1}$$ is algebraic integer too. Now use the fact that any rational algebraic integer IS an integer. This contradicts the fact that $S$ is NOT an integer (as you have already shown). Hence $S$ is not rational.

edited Jul 23 '20 at 16:17

answered Jul 23 '20 at 16:12

Robert Z

145,942

score 2 · Answer 2 · answered Jul 23 '20 at 16:12

A root of a monic polynomial with integer coefficients is called an algebraic integer. For example, $\sqrt{n}$ is an algebraic integer, because it is a root of $X^2 - n$. An important result is that the algebraic integers form a ring, i.e. the sum and product of algebraic integers is an algebraic integer. See here for my favourite proof of this fact.

Prove that the sum $ \sqrt{1001^2 + 1}+\sqrt{1002^2 + 1} \ + ... + \sqrt{2000^2 + 1}$ is irrational

2 Answers2