1989 AHSME Problems/Problem 18

Problem

The set of all real numbers for which $x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}$ is a rational number is the set of all

(A) integers $x$ (B) rational $x$ (C) real $x$

(D) $x$ for which $\sqrt{x^2+1}$ is rational

(E) $x$ for which $x+\sqrt{x^2+1}$ is rational

Solution

Rationalizing the denominator of $\frac{1}{x+\sqrt{x^2+1}}$ , it simplifies: $\frac{1}{x+\sqrt{x^2+1}}$ = $\frac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}$ = $\frac{x-\sqrt{x^2+1}}{-1}$ = $-(x-\sqrt{x^2+1})$ . Substituting this into the original equation, we get $x + \sqrt{x^2+1} - (-(x-\sqrt{x^2+1})) = x+\sqrt{x^2+1} + x - \sqrt{x^2+1} = 2x$ . 2x is only rational if x is rational $\mathrm{(B)}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1989 AHSME Problems/Problem 18

Problem

Solution