Then, is between them and rational.
I am not familar with analysis on a serious level and this is what the proof requires.also prove that between every two rationals there exists an irrational(this proof requires some thinking!!)
However, I have my own little proof. (In fact 3)
You said between any two rationals,
So I can find a rational,
But we can transform this into an irrational number!
Since the that fraction is rational its continued fractional expansion terminates,
All you do know is attach ones to it,
This will be an infinite continued fraction, which is irrational and be around the value of
We can express any rational in terms of a decimal expansion (since the expansion is not necessarily unique we will prohibit the use the strings of nine).
So we have,
if it exists (that is )[/tex])
Next go to ...
And so on...
Note the statment can not be always true for that would imply that which is false.
Thus there is a place where the fraction for is in between.
After that place write for the decimals,
A number which is bound to be irrational
(Or use Louville's construction).
This will create an irrational number.
If that was not the case then the irrationals would be countable, an impossibility.
Hence there exists one such number, in fact infinitely many, in fact uncountable many.