# 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 == | ||

− | 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>. |

− | - | + | {{Old CanadaMO box|num-b=1|num-a=3|year=1969}} |

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