Is this a valid identity element? A dilemma

**x3bnm** · October 27th 2012, 01:10 PM

There is an operation on a set $\mathbb{R}$ (real numbers) such that:

$x * y = \sqrt{x^2 + y^2}$

Does $\mathbb{R}$ have an identity element with respect to operation $*$ ?

Now I find the identity element like this:

$\begin{align*}x * e =& \sqrt{x^2 + e^2} \\ x =& \sqrt{x^2 + e^2}......................\text{[by definition of identity element]} \\ x^2 =& (x^2 + e^2) \\ e^2 =& 0 \\ e =& 0 \end{align*}$

So now $x * e = x * 0 = \pm x$

Is this $e = 0$ a valid identity element for this operation?

Because an identity element has a property that it doesn't change the original element when used with the operation $*$ on set $\mathbb{R}$ .

$x * e = \sqrt{x^2 + 0^2}$ can be $-x$ The original element is modified.

But it's also possible that $x * e = +x$ . The original element is left untouched. I'm confused by these two situations.

So my question is:
Is this $e = 0$ a valid identity element of $\mathbb{R}$ with respect to $*$ ?

**Plato** · October 27th 2012, 01:18 PM

Originally Posted by x3bnm

There is an operation on a set $\mathbb{R}$ (real numbers) such that:
$x * y = \sqrt{x^2 + y^2}$
Does $\mathbb{R}$ have an identity element with respect to operation $*$ ?

There is no $e$ such that $(-1)*e=-1$ .
So the answer is no.

**x3bnm** · October 27th 2012, 01:37 PM

Originally Posted by Plato

There is no $e$ such that $(-1)*e=-1$ .
So the answer is no.

Thanks Plato for answer. I didn't think of that way. Again thanks.

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