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I'm stuck on this question, please help.

The binary operation $*$ is defined on $z$ by $x*y=xy-x-y+c$ for all $x, y, c$ belonging to $\Bbb Z$, $c$ is a constant. Given that $*$ is associative, what is the value of $c$?

I know associative operations are like $m*(n*p)=(m*n)*p$, so I was assuming $x*(y*z)=(x*y)*z$.

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  • $\begingroup$ Proceed. What does $x*(y*z)$ equal? And $(x*y)*z$? $\endgroup$ – Git Gud Mar 8 '15 at 22:38
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Corrected version:

HINT: There’s only one reasonably straightforward way to begin: expand $x*(y*z)$ and $(x*y)*z$ according to the definition, set them equal to each other, and see what you can discover about $c$. After you eliminate identical terms on both sides of the equation, you should have an equation that does not contain $y$. Now remember that this equation has to hold for all integers $x$ and $y$; what does that force $c$ to be?

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Yes, you are correct about associative operations. (But remember $\mathbb Z$ is the set of all integers; you are free to use $z$ as a variable.)

Hint: it might be easier instead to consider $$x * y = (x-1)(y-1)+c'$$ where $$c'=c-1$$

Then set $$(x*y)-1=(x-1)(y-1)+c''$$ where $$c''=c'-1$$

Then $*$ can be mapped to the ordinary multiplication of the integers one less than its operands, which is associative. What must $c''$ be?

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From the symmetry of:

$$(1)\qquad x∗y=xy−x−y+c$$

and the double usage of $y$ in:

$$(2)\qquad x∗(y∗z)=(x∗y)∗z$$

we know that $y$ will drop out when we substitute $(1)$ into $(2)$.

As:

$$((xy−x−y+c)*z)=(xy−x−y+c)z - (xy−x−y+c) - z + c$$

$$= −xz + cz + x - c - z + c +y(\cdots)$$

$$= −xz + cz + x - z +y (\cdots)$$

, by symmetry the left-hand side is:

$$= −zx + cx + z - x + y(\cdots)$$

with the $y(\cdots)$ being the same. So we have:

$$−xz + cz + x - z = −zx + cx + z - x$$

and $xz=zx \;\text {because}\; x,y \in \mathbb{Z}$, so:

$$cz + x - z = cx + z - x$$

$$c(z-x) = 2(z - x)$$

But because $z$ can equal $x$, we can have division by zero, so $c$ is undefined.

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