Maths Olympiad Prep

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Number theory Difficulty 6.2 National olympiad Prove it

Example 10 Let bb be a real number 1\geqslant 1, we have
b2+1=[b,2˙b]\sqrt{b^{2}+1}=[b, \dot{2} b]

When bb is a real number 2\geqslant 2, then we have
b21=[b1,1,2(b1)].\sqrt{b^{2}-1}=[b-1,1,2(b-1)] .

Solution

By (2) and Definition 4, (39) holds. By (13) and Definition 4, we know (40) holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.