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→Method of proof: add proof outline for the pigeonhole principle |
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One can divide the interval <math>[0, 1)</math> into <math>n</math> smaller intervals of measure <math>\frac{1}{n}</math>. Now, we have <math>n+1</math> numbers <math>x_0,x_1,...,x_n</math> and <math>n</math> intervals. Therefore, by the pigeonhole principle, at least two of them are in the same interval. We can call those <math>x_i,x_j</math> such that <math>i < j</math>. Now:
: <math>|(j-i)\alpha-(m_j-m_i)|=|j\alpha-m_j-(i\alpha-m_i)|=|x_j-x_i|
Dividing both sides by <math>j-i</math> will result in:
: <math>\left|\alpha-\frac{m_j-m_i}{j-i}\right|
And we proved the theorem.
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