AoPSWiki
Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

1969 Canadian MO Problems/Problem 2

From AoPSWiki

Problem

Determine which of the two numbers \sqrt{c+1}-\sqrt{c}, \sqrt{c}-\sqrt{c-1} is greater for any c\ge 1.

Solution

Multiplying and dividing \sqrt{c+1}-\sqrt c by its conjugate,

\sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.

Similarly, \sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c-\sqrt{c-1}}. We know that \frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c-\sqrt{c-1}} for all positive c, so \sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}.

1969 Canadian MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 Followed by
Problem 3


Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1969_Canadian_MO_Problems/Problem_2"
Visit the AoPS Book Store.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us