Do you mean replace them both with $2ab - a - b + 1$?

No, we erase both of them and write$ 2ab-a-b+1$

Number $\frac{1}{2}$ is on blackboard If $a=\frac{1}{2}$ then $2ab-a-b+1=\frac{1}{2}$ so number $\frac{1}{2}$ is always on blackboard

@above , i dont think so , could you provide a complete proof. Because we must find all values of t.

@2above I think that is right edit: sniped.

Oh yes i understand ! Is it the only possible value?

yes bcs its always on the board, so it will be at the end.

On the blackboard, there are $2016$ numbers $\frac1{2016},\frac2{2016},\ldots,\frac{2016}{2016}$. One can perform the following operation: Choose any two numbers on the blackboard, say $a,b$, and replace them by $2ab-a-b+1$. After doing $2015$ operations, there will be only one number $\alpha$. Find all possible values of $\alpha$.

It will be (2a-1).....(2z-1)/2

rama1728 wrote: It will be (2a-1).....(2z-1)/2 You have a small typo. It is NOT wise to solely rely on the 1/2 thing, or they will get stuck the next time there is no 1/2 among the numbers. They should think why the operation is commutative. It's easily observed by 2(2ab-a-b+1)-1=(2a-1)(2b-1)

$$2ab - a - b + 1=\frac{(2a-1)(2b-1)+1}{2}$$Invariant: $$\frac{(2\frac{1}{2016}-1)...(2\frac{2016}{2016}-1)+1}{2}$$