Robots on a Plane

A proof that, should you ever be attacked by a giant robotic bishop while on a chessboard, you can survive by standing in the right spot.

P(n): at any time n\in\mathbb{N},x+y is even.

Basis

P(1): (1,1),(1,-1),(-1,1),(-1,-1)

x+y = 2,0,0,-2.

Thus, P(1) is true.

Inductive Step

Assume 1\leq k and P(k) is true, we show that P(k+1) is true.

The coordinates (a,b) resemble the location of the robot at time k.

Because P(k) is true, a+b is even.

Thus, \exists l\in\mathbb{Z}\ni a+b=2l because P(k) is true.

At time k+1 there are three cases:

Case 1: The robot moves up and to the right from (a,b)

Therefore, P(k+1) coordinates are (a+1,b+1)

thus, x+y=(a+1)+(b+1)

\Rightarrow x+y=a+b+2

\Rightarrow x+y=2l+2 by substitution

\Rightarrow x+y=2(l+1) where l+1\in\mathbb{Z}

Therefore, x+y is even.

Case 2: The robot moves either left and up or right and down from (a,b)

Therefore, P(k+1) coordinates are either (a+1,b-1) or (a-1,b+1).

Thus, x+y=(a+1)+(b-1) or x+y=(a-1)+(b+1)

\Rightarrow x+y=2l

Therefore, x+y is even.

Case 3: The robot moves left and down from (a,b)

Therefore, P(k+1) coordinates are (a-1,b-1)

Thus, x+y=(a-1)+(b-1)

x+y=a+b-2

x+y=2l-2

x+y=2(l-1) where (l-1)\in\mathbb{Z}

Therefore, x+y is even.

 Conclusion

Therefore, P(k)\Rightarrow P(k+1).

Thus, no matter which way the robot moves, x+y will always be even.

And thus, no matter how much time the robot has to move around, it will never be able to reach the point (1,0), since 1+0=1 is odd.