1982 AHSME Problems/Problem 7

If the operation $x \star y$ is defined by $x \star y = (x+1)(y+1) - 1$ , then which one of the following is FALSE?

$\text{(A)} \ x \star y = y\star x \text{ for all real } x,y. \\ \text{(B)} \ x \star (y + z) = ( x \star y ) + (x \star z) \text{ for all real } x,y, \text{ and } z.\\ \text{(C)} \ (x-1) \star (x+1) = (x \star x) - 1 \text{ for all real } x. \\ \text{(D)} \ x \star 0 = x \text{ for all real } x. \\ \text{(E)} \ x \star (y \star z) = (x \star y) \star z \text{ for all real } x,y, \text{ and } z.$

Solution

(A) is true because multiplication is commutative.

(B) is false because we have $x\star (y + z) = (x+1)(y+z+1) - 1$ and $x\star y + x\star z = (x+1)(y+1) - 1 + (x+1)(z+1) - 1$ $(x+1)(y+z+2) - 2$ $(x+1)(y+z+1) - (x+1) - 2$ $[(x+1)(y+z+1) - 1] + x,$ which does not match with the previous expression.

(C) is true because $(x-1)\star (x+1) = ((x-1)+1)((x+1)+1) - 1$ $x(x+2) - 1$ $x^2 + 2x - 1$ $[(x+1)^2 - 1] - 1$ $(x\star x) - 1$ for all $x$ .

(D) is true because $x\star 0 = (x+1)(0+1) - 1 = (x+1) - 1 = x$ for all $x$ .

(E) is true because multiplication is associative (and the plus-ones inside the parentheses and the minus-ones outside cancel out).

Therefore, our answer is $\boxed{\textbf{(B)}}$ , and we are done.

1982 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1982 AHSME Problems/Problem 7

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