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

Wade Bullock

Wade Bullock

Answered

2022-07-03

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

Answer & Explanation

Aryanna Caldwell

Aryanna Caldwell

Expert

2022-07-04Added 11 answers

[ 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

rjawbreakerca

Expert

2022-07-05Added 5 answers

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

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