91Ó°ÊÓ

Diagonalization is a technique to demonstrate that there are some languages that cannot be decided by a Turing machine. This is used as a way of showing that the (infinite) set of real numbers is larger than the (infinite) set of integers.

We show this by contradiction. Suppose that the given language is and the language T is accepted by Turing machine are were countable, and let f be the correspondence with N . We will show that there exists some b∈R that is not in the image of f, violating the assumption that f is onto. To do this, we start by ordering all the elements of according to the values of their pre-images:

Because f is a correspondence, all elements of must appear somewhere on this list. We now define a real number b∈ that does not appear on this list, demonstrating a contradiction. Recall that real numbers can be infinitely long. Choose the language’s digit of b such that it is different from some other language. Suppose that b appears in the language’s position of this list (that is that f(j) = b). But this cannot be true, because we know that b differs from language in the other position! This is called diagonalization, because we define the contradictory value by$T=\left\{\left(i,j,k\right)\right|i,j,k∈N\}.$appropriately setting its value along the diagonal.

$$\begin{gathered} \text{1mapsto(1,1,1)} \\ \text{2mapsto(2,2,2)} \\ \text{´Ú:µ÷°Õ→N}\left(i,j,k\right),T→2^i{3}^j{5}^k\left\} \\ \text{itiscertainlyinjective.Therefore}\right|T|â\copyright ½|N|â\copyright ½\left|T\right|â\copyright ½\left|N\right| \\ T=\{\left(i,j,k\right)|i,j,k∈N\}. \end{gathered}$$

Each element in is an infinite sequence so, here$c=\left\{1,1,0,0,......\right\}$define thus for each, $n=1,2,3....$is differs from the sequence and bit so, here it is not equal f(n) for any n . Which is a contradiction. Hence,$T=\left\{\left(i,j,k\right)\right|i,j,k∈N\}.$is countable.