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14.) In Cantor's diagonalization, we construct a number x between 0 and 1 that's not on...

14.) In Cantor's diagonalization, we construct a number x between 0 and 1 that's not on the supposed list of real numbers between 0 and 1. Recall, to construct x we make x's ith digit (after the decimal point) equal to 1 if the corresponding digit of the ith number on the list is even and we make x's ith digit 0 otherwise.

a) Suppose the list happens to start with the numbers 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 20.2/3, pi/4, sqrt(3) /2, 1/e and 1/13. What are the first eleven digits of x, after then decimal?

b) Explain why x cannot be on the list, even though it's a number between 0 and 1. Why does this lead to a contradiction?

c) Cantor's proof shows that the real numbers between 0 and 1 aren't countable. Suppose we try to use the same argument to show that the rational numbers between 0 and 1 aren't countable? (Of course, we know that they are....) Where does the proof break down?

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Answer #1

Here first we have written down the decimal expressions of the given numbers and then constructed 'from.

x is not in the list because x is different from a_i at i'th place.

We cannot use this argument for rationals because the constructed x may not be rational.

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