Difference between revisions of "1969 Canadian MO Problems/Problem 2"

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<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>
 
<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>\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>.
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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>.
  
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{{Old CanadaMO box|num-b=1|num-a=3|year=1969}}
{{CanadaMO box|year=1969|num-b=1|num-a=3}}
 

Latest revision as of 01:18, 15 February 2010

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 Followed by
Problem 3


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