Difference between revisions of "1969 Canadian MO Problems/Problem 2"
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== 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 1 == |
− | 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>. |
− | ---- | + | == Solution 2 == |
− | + | Considering the derivative of <math>f(x)=\sqrt{x+1}-\sqrt{x}</math>. | |
− | + | ||
− | + | 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. | |
+ | |||
+ | Thus <math>\boxed{\sqrt{c}-\sqrt{c-1}}</math> is greater. | ||
+ | |||
+ | ~hastapasta | ||
+ | |||
+ | == Solution 3 == | ||
+ | Plugging in <math>1</math> for both of the expressions, we get that <math>\sqrt{c+1} + \sqrt{c} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1</math> and <math>\sqrt{c} - \sqrt{c-1} = \sqrt{1} - \sqrt{0} = 1</math>. Since <math>\sqrt{2} - 1 < 1</math>, <math>\boxed{\sqrt{c} - \sqrt{c-1}}</math> is greater | ||
+ | -andliu766 | ||
+ | |||
+ | {{Old CanadaMO box|num-b=1|num-a=3|year=1969}} |
Latest revision as of 21:05, 13 August 2023
Contents
Problem
Determine which of the two numbers ,
is greater for any
.
Solution 1
Multiplying and dividing by its conjugate,
Similarly, . We know that
for all positive
, so
.
Solution 2
Considering the derivative of .
We have . Putting under a common denominator, we can see that the top will be negative.
Thus is greater.
~hastapasta
Solution 3
Plugging in for both of the expressions, we get that
and
. Since
,
is greater
-andliu766
1969 Canadian MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 3 |