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Exercise 12.1 · Q10

Q.(i) Let AA be Q∖{1}\mathbb{Q}\setminus\{1\}. Define ∗* on AA by x∗y=x+y−xyx*y = x+y-xy. Is ∗* binary on AA? If so, examine the commutative and associative properties satisfied by ∗* on AA.

(ii) Let AA be Q∖{1}\mathbb{Q}\setminus\{1\}. Define ∗* on AA by x∗y=x+y−xyx*y=x+y-xy. Is ∗* binary on AA? If so, examine the existence of identity, existence of inverse properties for the operation ∗* on AA.
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The trick is to notice x∗y−1x*y-1 factorises, which shows the output can never accidentally equal the excluded value 11; after that, commutativity, associativity, identity and inverse are direct algebraic checks.

Step 1. Check ∗* is binary on A=Q∖{1}A=\mathbb Q\setminus\{1\}. For x,y∈Ax,y\in A (so x,y≠1x,y\ne1), x∗y=x+y−xyx*y=x+y-xy is clearly rational. To confirm it avoids 11: x∗y−1=x+y−xy−1=−(xy−x−y+1)=−(x−1)(y−1)x*y-1=x+y-xy-1=-(xy-x-y+1)=-(x-1)(y-1). Since x≠1x\ne1 and y≠1y\ne1, (x−1)(y−1)≠0(x-1)(y-1)\ne0, so x∗y−1≠0x*y-1\ne0, i.e. x∗y≠1x*y\ne1. Hence x∗y∈Ax*y\in A always -- ∗* is binary on AA.

Step 2. Commutative. x∗y=x+y−xy=y+x−yx=y∗xx*y=x+y-xy=y+x-yx=y*x for all x,yx,y (ordinary +,×+,\times are commutative). Commutative.

Step 3. Associative. Expand both groupings.

(x∗y)∗z=(x+y−xy)∗z=(x+y−xy)+z−(x+y−xy)z=x+y−xy+z−xz−yz+xyz(x*y)*z=(x+y-xy)*z=(x+y-xy)+z-(x+y-xy)z=x+y-xy+z-xz-yz+xyz.

x∗(y∗z)=x∗(y+z−yz)=x+(y+z−yz)−x(y+z−yz)=x+y+z−yz−xy−xz+xyzx*(y*z)=x*(y+z-yz)=x+(y+z-yz)-x(y+z-yz)=x+y+z-yz-xy-xz+xyz.

Both expand to x+y+z−xy−xz−yz+xyzx+y+z-xy-xz-yz+xyz -- equal. Associative. …

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