A question about how to express a fraction as 1 q 1 + 1 q 2 + ⋯ + 1 q N Let x be a positive rational number, strictly between 0 and 1. Prove that there is a finite strictly increasing list of positive integers 2 ≤ q 1 &lt; q 2 &lt; ⋯ &lt; q N such that x = 1 q 1 + 1 q 2 + ⋯ + 1 q N .I have tried many methods, such as mathematical induction. I know it is obvious that 1 m is right for assumption, but when I begin to prove 2 m I feel it is hard to find some patterns for 2 m . Thus, maybe my thoughts are wrong. Also, I still tried to change the fraction so that the numerator will be smaller and smaller. But I still cannot find a way to lower the numerator. Can someone help me solve the question? Or, can someone give me some hints. I will appreciate you very much!

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A question about how to express a fraction as             1              q        1              +            1              q        2              +  ⋯  +            1              q        N            Let x be a positive rational number, strictly between 0 and 1. Prove that there is a finite strictly increasing list of positive integers   2  ≤      q    1    &amp;lt;      q    2    &amp;lt;  ⋯  &amp;lt;      q    N   such that  x  =            1              q        1              +            1              q        2              +  ⋯  +            1              q        N              .I have tried many methods, such as mathematical induction. I know it is obvious that       1    m   is right for assumption, but when I begin to prove       2    m    I feel it is hard to find some patterns for        2    m  .  Thus, maybe my thoughts are wrong. Also, I still tried to change the fraction so that the numerator will be smaller and smaller. But I still cannot find a way to lower the numerator. Can someone help me solve the question? Or, can someone give me some hints. I will appreciate you very much!

mseralge · Accepted Answer

Let   q  =  m      /    n rational, with   q  ∈  (  0  ,  1  ), and hence      1    k    ≤      m    n    &amp;lt;      1          k      −      1      for some integer   k  &amp;gt;  1 . If       1    k    =      m    n  , we have nothing to prove. Otherwise, we have      m    n    −      1    k    =            k      m      −      n              k      n        =            m      ′              n      ′        &amp;gt;  0. But             1              k        −        1              &amp;gt;            m      n       implies that   n  &amp;gt;  m  (  k  −  1  ) , or    m  &amp;gt;  k  m  −  n  =      m    ′    &amp;gt;  0 Hence, after we subtracted the the nearest and smallest fraction of the form   1      /    k , the remainder       m    ′        /        n    ′   has smaller numerator.Hence, in at most m steps this procedure ends, and thus we obtain the sought for representation.

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