Let f : [ 0 , 1 ] → R be continuous with f ( 1 ) = f ( 0 ). Prove that if h ∈ ( 0 , 1 2 ) is not of the form 1 n , then there does not necessarily exist | x − y | = h satisfying f ( x ) = f ( y ). Provide an example that illustrates this using h = 2 5 .So I was given the hint that I can use a modified s i n function, however I'm not really sure how I would go about that. Preferably, an example not using that would be great.My thoughts so far are that I use a proof by contradiction saying that ∃ x , y ∈ [ 0 , 1 ] such that f ( x ) = f ( 0 ) and | x − y | = h. And I need to get to the point where h = 1 n , which would create the contradiction (I assume that's the endpoint?). However, how would I use h = 2 5 to prove that? Is it trivial in that h = 1 n is false? Also, it isn't given that n ∈ N , so maybe I shouldn't assume that?

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Let   f  :  [  0  ,  1  ]  →      R   be continuous with   f  (  1  )  =  f  (  0  ). Prove that if   h  ∈      (    0    ,          1      2        )   is not of the form       1    n  , then there does not necessarily exist       |    x  −  y      |    =  h satisfying   f  (  x  )  =  f  (  y  ). Provide an example that illustrates this using   h  =      2    5  .So I was given the hint that I can use a modified   s  i  n function, however I&#039;m not really sure how I would go about that. Preferably, an example not using that would be great.My thoughts so far are that I use a proof by contradiction saying that   ∃   x  ,  y  ∈  [  0  ,  1  ] such that   f  (  x  )  =  f  (  0  ) and       |    x  −  y      |    =  h. And I need to get to the point where   h  =      1    n  , which would create the contradiction (I assume that&#039;s the endpoint?). However, how would I use   h  =      2    5   to prove that? Is it trivial in that   h  =      1    n   is false? Also, it isn&#039;t given that   n  ∈      N  , so maybe I shouldn&#039;t assume that?

Mira Spears · Accepted Answer

What would it take to modify   sin  ⁡  x to make it satisfy   f  (  0  )  =  f  (  1  )? Well,   sin  ⁡  0  =  sin  ⁡  π, so how about  f  (  x  )  =  sin  ⁡      (    π    x    )      ?Now you just need to show that  sin  ⁡      (    π    x    )    ≠  sin  ⁡      (    π    x    +                  2        π            5        )  for any   x  ∈  [  0  ,      3    5    ], or equivalently,  sin  ⁡  x  ≠  sin  ⁡      (    x    +                  2        π            5        )  for any   x  ∈  [  0  ,            3      π        5    ].A little trig should give you a formula in the tangent satisfied by any   x for which the above is true. Show that the only solutions fall outside of   [  0  ,            3      π        5    ] and you are done.

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