2510.31 – The Add-One-Subtract-One Operation

The operation is defined by

xΔy=(x+1)(y+1)−1x \Delta y = (x+1)(y+1) - 1

Which of these is false?

  1. xΔy=yΔxx \Delta y = y \Delta x for all real xx and yy.

  2. xΔ(y+z)=(xΔy)+(xΔz)x \Delta (y + z) = (x\Delta y) + (x \Delta z) for all real xx, yy, and zz.

  3. (x−1)Δ(x+1)=(xΔx)−1(x-1)\Delta (x+1) = (x\Delta x) - 1 for all real xx.

  4. xΔ0=xx \Delta 0 = x for all real xx.

  5. xΔ(yΔz)=(xΔy)Δzx \Delta(y\Delta z) = (x\Delta y)\Delta z for all real xx, yy, and zz.

Solution

Here’s a simpler version of the operation:

xΔy=(x+1)(y+1)−1=xy+x+y+1−1=xy+x+y\begin{aligned}x \Delta y &= (x+1)(y+1) - 1 \\&= xy + x + y +1 - 1 \\&= xy + x + y\end{aligned}

  1. Either formula makes it clear that the order of xx and yy in xΔyx \Delta y doesn’t matter.

  2. False:
    xΔ(y+z)=x(y+z)+x+(y+z)=xy+xz+x+y+z\begin{aligned}x \Delta (y + z) &= x (y + z) + x + (y + z) \\&= xy + xz + x + y + z\end{aligned}
    (xΔy)+(xΔz)=xy+x+y+xz+x+z=xy+xz+2x+y+z\begin{aligned}(x \Delta y) + (x \Delta z) &= xy + x + y + xz + x + z \\&= xy + xz + 2 x + y + z\end{aligned}

  3. True:
    (x−1)Δ(x+1)=(x−1)(x+1)+(x−1)+(x+1)=x2−1+2x\begin{aligned}(x-1) \Delta (x+1) &= (x-1)(x+1) + (x - 1) + (x + 1) \\&= x^2 - 1 + 2 x\end{aligned}
    (xΔx)−1=x2+x+x−1(x \Delta x) - 1 = x^2 + x + x -1

  4. True:
    xΔ0=x⋅0+x+0=x.x \Delta 0 = x \cdot 0 + x + 0 = x.

  5. True:
    xΔ(yΔz)=x(yΔz)+x+(yΔz)=x(yz+y+z)+x+(yz+y+z)=xyz+xy+xz+yz+x+y+z\begin{aligned}x \Delta (y \Delta z) &= x (y \Delta z) + x + (y \Delta z) \\&= x(yz + y + z) + x + (yz + y + z) \\&= xyz + xy + xz + yz + x + y + z\end{aligned}
    (xΔy)Δz=(xΔy)z+(xΔy)+z=(xy+x+y)z+(xy+x+y)+z=xyz+xz+yz+xy+x+y+z\begin{aligned}(x \Delta y) \Delta z &= (x \Delta y) z + (x \Delta y) + z \\&= (xy + x + y) z + (xy + x + y) + z \\&= xyz + xz + yz + xy + x + y + z\end{aligned}