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.x \Delta y = (x+1)(y+1) - 1 = xy + x + y +1 - 1 = xy + x + y.

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

B. False:

xΔ(y+z)=x(y+z)+x+(y+z)=xy+xz+x+y+z.x \Delta (y + z) = x (y + z) + x + (y + z) = xy + xz + x + y + z.

(xΔy)+(xΔz)=xy+x+y+xz+x+z=xy+xz+2x+y+z.(x \Delta y) + (x \Delta z) = xy + x + y + xz + x + z = xy + xz + 2 x + y + z.

C. True:

(x−1)Δ(x+1)=(x−1)(x+1)+(x−1)+(x+1)=x2−1+2x.(x-1) \Delta (x+1) = (x-1)(x+1) + (x - 1) + (x + 1) = x^2 - 1 + 2 x.

(xΔx)−1=x2+x+x−1.(x \Delta x) - 1 = x^2 + x + x -1.

D. True:

xΔ0=x⋅0+x+0=x.x \Delta 0 = x \cdot 0 + x + 0 = x.

E. 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}