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zmudz

  • one year ago

Let \(f:\mathbb R \to \mathbb R\) be a function such that for any irrational number \(r\), and any real number \(x\) we have \(f(x)=f(x+r)\). Show that \(f\) is a constant function.

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  1. Zarkon
    • one year ago
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    what have you tried

  2. Zarkon
    • one year ago
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    the proof is quite short

  3. zmudz
    • one year ago
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    I am having trouble even getting started. I don't know what I'm supposed to do. Any help is greatly appreciated.

  4. Zarkon
    • one year ago
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    choose x=0 then f(0)=f(0+r)=f(r) for all r irrational so f is constant on the irrational numbers now you need to extend this to the rational numbers consider x=a-r where a is a rational number

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