Difference between revisions of "1969 Canadian MO Problems/Problem 2"
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | Determine which of the two numbers <math> | + | Determine which of the two numbers <math>\sqrt{c+1}-\sqrt{c}</math>, <math>\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>c\ge 1</math>. |
== Solution == | == Solution == | ||
| − | Multiplying and dividing <math> | + | Multiplying and dividing <math>\sqrt{c+1}-\sqrt c</math> by its conjugate, |
| − | <math> | + | <math>\sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.</math> |
| − | Similarly, <math> | + | Similarly, <math>\sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c-\sqrt{c-1}}</math>. We know that <math>\frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c-\sqrt{c-1}}</math> for all positive <math>c</math>, so <math>\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}</math>. |
---- | ---- | ||
| − | + | {{CanadaMO box|year=1969|num-b=1|num-a=3}} | |
| − | |||
| − | |||
, 
is greater for any 
.
by its conjugate,
. We know that 
for all positive 
, so 
.
