Question Number 4775 by Yozzii last updated on 08/Mar/16
$${Let}\:\ast\:{be}\:{a}\:{binary}\:{operation}\:{on}\:\mathbb{Z} \\ $$$${defined}\:{by}\: \\ $$$${x}\ast{y}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{x}+{y}} \right)\right). \\ $$$${Is}\:\ast\:{associative}? \\ $$
Commented by prakash jain last updated on 09/Mar/16
$${x}+{y}=\mathrm{2}{i} \\ $$$${y}+{z}=\mathrm{2}{n}+\mathrm{1} \\ $$$${x}=\mathrm{2} \\ $$$${y}=\mathrm{6} \\ $$$${z}=\mathrm{3} \\ $$$${y}\ast{z}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{10}\right)=\mathrm{5} \\ $$$${x}\ast{y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}+\mathrm{6}+\mathrm{1}+\mathrm{1}\right)=\mathrm{5} \\ $$$${x}\ast\left({y}\ast{z}\right)=\mathrm{2}\ast\mathrm{5}=\mathrm{4} \\ $$$$\left({x}\ast{y}\right)\ast{z}=\mathrm{5}\ast\mathrm{3}=\mathrm{5} \\ $$
Commented by Yozzii last updated on 09/Mar/16
$${Thanks}.\: \\ $$