Does f ( [ 0 , 1 ] ) must always contain an irrational number?

Question

Does   f  (  [  0  ,  1  ]  ) must always contain an irrational number?

Aryanna Caldwell · Accepted Answer

[  0  ,  1  ] is uncountable, and so because   f is injective, so will   f  (  [  0  ,  1  ]  ). If   f  (  [  0  ,  1  ]  ) did not contain an irrational number, then it would lie in       Q  , and so be countable. But that would give us a contradiction.Because   [  0  ,  1  ]  ∖      Q   has the same cardinality as   [  0  ,  1  ], there is in fact a bijection from   [  0  ,  1  ] onto   [  0  ,  1  ]  ∖      Q  .

rjawbreakerca · Accepted Answer

f  (  x  )  =      {                            {          π          +          x          }                          x           is of the form           n          π          +          q           for           n           non-negative integer and           q           rational                                      x                           otherwise                         where   {  x  } is the fractional part of   x, though strictly speaking this is a bijection   [  0  ,  1  )  →  [  0  ,  1  )  ∖      Q

Does f ( [ 0

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