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Show that the operation x*y=(1/x) + (1/y) on Q, is well defined and commutative, but that it is not associative, nor does it have an identity.

Not really too sure how I would go about proving that the function is well defined on Q as if x and y are equal to zero, then they wouldn't be (at least I don't think they would be).

When I wanted to prove that the binary operation was commutative, I said:

x*y = (1/x) + (1/y) = (1/y) + (1/x) = y*x

I have no idea how to do the rest of the question as well. I tried:

Associativity: (x*y)z = x(y*z)

(1/x + 1/y)z = 1/x(1/y + z)

(1/x + 1/y)z = 1/x(1/y + z)

z/x + z/y = 1/xy + z/x

Untrue and therefore not associative.

Absolutely no idea how to find out if there is an identity or not. All I was able to do was define what an identity actually was.

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Please read about accepting answers here and here. –  Git Gud Nov 30 '13 at 15:52
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Indeed, as it is, it isn't a well defined operation for the exact reason you mention. –  Git Gud Nov 30 '13 at 15:53
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1 Answer

Hint (for lacking an identity): Show that there exists no $e\in\boldsymbol Q$ such that $1\ast e=1$.

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Can 0 be an identity? So then I would have x*e=e*x= 1/x + 0 = 1/x? –  Roy Sheehan Nov 30 '13 at 16:17
    
@RoySheehan $1\ast e=1\iff 1/1+1/e=1\iff 1/e=0$, got it? –  Michael Hoppe Nov 30 '13 at 16:22
    
@RoySheehan Are you seriously not accepting any answers after my comment to your question? –  Git Gud Nov 30 '13 at 16:43
    
@GitGud The wiser head gives in. –  Michael Hoppe Nov 30 '13 at 17:25
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