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

Latest revision as of 20:05, 13 August 2023

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

Solution 3

Plugging in $1$ for both of the expressions, we get that $\sqrt{c+1} + \sqrt{c} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1$ and $\sqrt{c} - \sqrt{c-1} = \sqrt{1} - \sqrt{0} = 1$ . Since $\sqrt{2} - 1 < 1$ , $\boxed{\sqrt{c} - \sqrt{c-1}}$ is greater -andliu766

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Determine which of the two numbers <math>\displaystyle \sqrt{c+1}-\sqrt{c}</math>, <math>\displaystyle\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>\displaystyle 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 ==
+== Solution 1 ==
-Multiplying and dividing <math>\displaystyle \sqrt{c+1}-\sqrt c</math> by its conjugate,
+Multiplying and dividing <math>\sqrt{c+1}-\sqrt c</math> by its conjugate,
-<math>\displaystyle \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>\displaystyle \sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c-\sqrt{c-1}}</math>. We know that  <math>\displaystyle \frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c-\sqrt{c-1}}</math> for all positive <math>\displaystyle c</math>, so <math>\displaystyle \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>.
-----
+== Solution 2 ==
-* [[1969 Canadian MO Problems/Problem 1|Previous Problem]]
+Considering the derivative of <math>f(x)=\sqrt{x+1}-\sqrt{x}</math>.
-* [[1969 Canadian MO Problems/Problem 3|Next Problem]]
-* [[1969 Canadian MO Problems|Back to Exam]]
+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}}

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