Cantor's diagonal slash has been used to argue that the infinity of reals is uncountable, whereas the infinity of integers is countable (hence the two infinities are different). I assume that readers are familiar with Cantor's diagonal slash argument. What happens if we apply Cantor's diagonal slash to the integers? How can this be done? Think of the infinite binary tree, which starts at ground level as the trunk and splits in two (bifurcates) as we ascend; if we take the left branch then we assign 0 to the first binary digit, if the right branch then 1. At the next level we do the same....0 if we go left, 1 if we go right.... and we ascend the binary tree all the way to infinity. In this way every possible infinite sequence of binary digits is represented somehow or other (mapped out) by a path up the tree; and every path up the tree corresponds to a unique infinite sequence of binary digits. For those who are curious, the number of routes up the tree is equal to 2^n (where n is the number of levels), and the number of nodes (branches) in the tree is equal to (2^n)-1. Now we can arrange all of these paths in an infinite x infinite binary matrix, of form similar to the infinite x infinite matrix of real numbers that Cantor used for his diagonal slash. What happens if we perform Cantor's diagonal slash on this infinite binary matrix? According to Cantor, we produce a new infinite binary sequence which is NOT contained within the original matrix (nor is it contained on the infinite binary tree). But how can this be? The matrix contains every possible route up the binary tree, there IS no other route not contained in the matrix. What does this show? That Cantor's diagonal slash argument is meaningless when applied to infinite matrices.