Suppose the positive integer is odd. First Al writes the numbers on the blackboard. Then he picks any two numbers erases them and writes instead . Prove that an odd number will remain at the end.
In the beginning there are even integers there are odd integers.
When we replace two the parity of the odd integers stays odd at each step. So we have 2 integers left 1 is odd and 1 is even. And when we subtract them we have an odd integer.
I said the sum initially is which is odd. The sum is reduced by each time which is even. So the parity stays constant, and the sum is . The parity is odd at the beginning. So it is odd at the end.
I said the sum initially is which is odd. The sum is reduced by each time which is even. So the parity stays constant, and the sum is . The parity is odd at the beginning. So it is odd at the end.
It is similar to mine. But I find your approach more elegant.