An operation is defined on $\mathbb{R}$ such that for every $x,y \in \mathbb{R}$, $x \ast y=\sqrt{x^2+y^2}$.
I checked some of the basic properties like commutativity, associativity and whether there exists identity element.
I already checked and confirmed the first two but I wonder about the identity element.
I need to find an element $e \in \mathbb{R}$ such that $ \forall x \in \mathbb{R}$, $\ e \ast x= x \ast e = x$ . Since commutativity applies it's enough to check for just left or right identity element:
$$ x \ast e = \sqrt{x^2+e^2} \iff \sqrt{x^2+e^2}=x \iff x^2+e^2=x^2 \iff e=0$$
If $e=0$ this works out nice for the positive $x$. But what about the negative elements? Take $x=-2$ then $-2 \ast 0 = \sqrt{(-2)^2+0^2}=\sqrt{4} = 2$.
Does this mean that there exists no identity element?
Bonus:
I also need to check if $x \ast y=z \ast y \implies x=z$ and $ y \ast x = y \ast x \implies x=z$ (Is there a term for this property?)
This is equivalent to $\sqrt{x^2+y^2}=\sqrt{z^2+y^2}$ or $x^2=z^2$. But this also doesn't guarantee that $x=z.$
I'm sort of new to this type of math so I would like to know if what I wrote here is correct.
EDIT: Is it true that if the operation is defined like $x \ast y = \sqrt[3]{x^3+y^3}$ then all the properties are true?
I would like to point out that in your proof the equivalence $ \sqrt{x^2+e^2}=x \Leftrightarrow x^2+e^2=x^2 $ is not true if $x$ is a general real number. In fact we have $ \sqrt{x^2+e^2}=|x| \Leftrightarrow x^2+e^2=x^2 $.
As you thought this operation has no identity element.
For your edit: if you consider cubic roots and elevation to the cube then the problem with the sign is not present and $0$ actually defines the identity element.