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

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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>.
 
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 ==
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== Solution 1 ==
 
Multiplying and dividing <math>\sqrt{c+1}-\sqrt c</math> by its conjugate,
 
Multiplying and dividing <math>\sqrt{c+1}-\sqrt c</math> by its conjugate,
  
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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>.
 
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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== Solution 2 ==
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Considering the derivative of <math>f(x)=\sqrt{x+1}-\sqrt{x}</math>.
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We have <math>f'(x)=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}</math>. Putting under a common denominator, we can see that the top will be negative.
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Thus <math>\boxed{\sqrt{c}-\sqrt{c-1}}</math> is greater.
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~hastapasta
  
 
{{Old CanadaMO box|num-b=1|num-a=3|year=1969}}
 
{{Old CanadaMO box|num-b=1|num-a=3|year=1969}}

Revision as of 21:45, 19 December 2022

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 1

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 $\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}$.

Solution 2

Considering the derivative of $f(x)=\sqrt{x+1}-\sqrt{x}$.

We have $f'(x)=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}$. Putting under a common denominator, we can see that the top will be negative.

Thus $\boxed{\sqrt{c}-\sqrt{c-1}}$ is greater.

~hastapasta

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