00:01
So in this problem we want to show that the set of irrational numbers is uncountable.
00:07
So first we start with, let's consider the irrational numbers in the open interval 0 comma 1.
00:17
Now such an irrational number has a binary representation of the form.
00:29
R equals so if r is an irrational number in the open interval 0 comma 1, it can be represented in binary is b1 over 2 plus b2 over 2 squared plus dot a dot plus the general term is bk over 2 to the power k plus dot possibly up to infinity so where the bks are 0 or 1 now if there are only finitely many ones among the bis then the series in star 1 would be finite.
01:15
For example, say if bn is the last one, then r would be equal to b1 over 2 plus b2 over 2 squared plus dot -da -dot up to bn over 2 raise to the power n, right? and this would then be equal to 2 to the power n minus 1 times v1 plus 2 to the 4 -n minus 2 times b2 plus dot dot dot plus bn over 2 raise to the power n.
01:43
Now this, this is a rational number, so we can't have that.
01:49
So for an irrational number, r, the representation in star 1 must have infinitely many ones among the bis.
02:06
Now then, the sequence b1 up to bk up to going on infinity is a binary sequence.
02:17
That is each term is either one or zero, and containing infinitely many ones.
02:25
So if we were to represent an irrational number as a binary, in the binary representation, there would be infinitely many ones.
02:35
And if we just take out the bis and form a sequence out of them, then that sequence obviously would contain infinitely many ones corresponding to an irrational number.
02:48
And then each such a number.
02:50
Sequence corresponds to an irrational number in 0 .1 via the representation in star 1.
03:05
Now let's denote by s the set of all binary sequences with infinitely many ones.
03:15
So the idea is that we want to show that s is uncountable.
03:20
Now the cardinality of s is the same as the cardinality of the set of all irrational numbers in the interval 0 comma 1.
03:36
Now let f be the set of all binary sequences with finitely many ones in them.
03:46
So we are talking about infinitely, well, infinitely long binary sequences, but with only finitely many ones.
03:53
That means it has to have infinitely many series, but only finitely many ones.
03:57
So then we can show that f is countable.
04:04
Note that if for a sequence in f, the last one occurs at the nth term, there are two raised to the four n -1 such sequences.
04:20
We can order them alphabetically.
04:26
So if the last one occurs in the nth term, then we have n -minus one more terms preceding that.
04:34
And for each such, each of the two raised to the power, so each of the n minus one terms preceding the anneth term, we have two choices.
04:44
They can be there, each of them can be either zero or one.
04:47
So altogether we have two raised to the poor n minus one choices.
04:51
So two raise to the poor n minus one binary sequences such that the last one occurs at the nth term.
04:59
And once we have that, two raise to the power n minus one such sequences, we can order them alphabetically...
