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Find the identity element, if it exist, where all a, b belongs to R : a*b = a/b + b/a

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This looks like homework. Also please do not make it look like you are giving us homework, show what you have already done, where you got stuck,... – sxd Nov 19 '11 at 14:08
Are you sure it is well defined ? What if a=0 ? – Student Nov 19 '11 at 14:10

First, we must be dealing with $\mathbb{R}_{\not=0}$ (non-zero reals) since $0*b$ and $0*a$ are not defined (for all $a,b$).

Let $a \in \mathbb{R}_{\not=0}$. An identity is an element, call it $e\in\mathbb{R}_{\not=0}$, such that $e*a=a$ and $a*e=a$. Since this operation is commutative (i.e. $a*b=b*a$), we have a single equality to consider.

Suppose we do have an identity $e \in \mathbb{R}_{\not=0}$. Then $a = e*a = a*e = a/e+e/a$ for all $a \in \mathbb{R}_{\not=0}$. This implies that $a = \frac{a^2+e^2}{ae}$. Thus $a^2e=a^2+e^2$ and so $a^2(e-1)=e^2$ and finally $a = \pm \sqrt{\frac{e^2}{e-1}}$. This is non-sense since $a$ can be any non-zero real and $e$ is some fixed non-zero real. Therefore, no identity can exist.

You may want to try to put together a more concrete proof yourself. Consider for example, $a=1$. If there is an identity (for $a$), what would it need to be? Is this possible?

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Thanks a lot for your solution! – Ivan Nov 19 '11 at 15:42

If $a$ were an identity element, then $a*b = b$ for all $b$. So you could just take $b = a$ itself, and you'd have to have $a*a = a$. But your definition implies $a*a = 2$. So $a = 2$ would have to be the identity element. But clearly $2*b = b/2 + 2/b$ is not equal to $b$ for all $b$; choose any random $b$ such as $b = 1$ for example. So there is no identity element.

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What I don't understand is that if in your suggestion, a, b are 2x2 matrices, a is an identity matrix, how can matrix a = identity matrix b in the binary operation a*b = b ? – Ivan Nov 20 '11 at 3:45

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