Associativity and neutral element of the composition $x \circ y=x\sqrt{y^2+1}+y\sqrt{x^2+1}$

Question

I have to demonstrate that this composition $x \circ y=x\sqrt{y^2+1}+y\sqrt{x^2+1}$ is a commutative group, but I can't demonstrate that it is associative, and I can't find the neutral element $e$.

Answer 1

You may show associativity with hyperbolic functions. Let $x=\sinh(a), \ y=\sinh(b)$. We have no restriction on $x, y \in \mathbb{R}$ as the hyperbolic sine is bijective.

$$\begin{align} x \circ y &= x\sqrt{y^2+1}+y\sqrt{x^2+1} \\ &= \sinh(a)\cosh(b)+\sinh(b)\cosh(a)\\ &=\sinh(a+b). \end{align}$$

Now let $z=\sinh(c)$, so

$$\begin{align} (x\circ y) \circ z &= \sinh(a+b) \circ z\\ &= \sinh(a+b+c)\\ &= x \circ \sinh(b+c)\\ &= x\circ (y\circ z). \end{align}$$

A similar method was followed here.

Answer 2

You must show $(x\circ y)\circ z = x\circ (y\circ z)$.

So we must show $(x\sqrt{y^2+1}+y\sqrt{x^2+1})\sqrt{z^2 + 1} + z\sqrt{(x\sqrt{y^2+1}+y\sqrt{x^2+1})^2+1}=$

$x\sqrt{(y\sqrt{z^2+1}+z\sqrt{y^2+1})^2+1}+(y\sqrt{z^2+1}+z\sqrt{y^2+1})\sqrt{x^2 + 1}$

And to find the identity $e$ you must find the value where

$x\sqrt{e^2+1}+e\sqrt{x^2+1}=x$

so solve for $e$.

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To do the first:

$(x\sqrt{y^2+1}+y\sqrt{x^2+1})\sqrt{z^2 + 1} + z\sqrt{(x\sqrt{y^2+1}+y\sqrt{x^2+1})^2+1}=$

$x\sqrt{y^2 + 1}\sqrt{z^2 + 1} + y\sqrt{x^2+1}\sqrt{z^2 + 1}+ z\sqrt{(x\sqrt{y^2+1}+y\sqrt{x^2+1})^2+1}=$

$x\sqrt{y^2z^2 + y^2 + z^2 + 1} +y\sqrt{x^2z^2 + x^2 + z^2 + 1} + z\sqrt{x^2(y^2+1)+y^2(x^2+ 1) +1+ 2xy\sqrt{x^2y^2+ x^2 + y^2 + 1}}=$

$x\sqrt{y^2z^2 + y^2 + z^2 + 1} +y\sqrt{x^2z^2 + x^2 + z^2 + 1} + z\sqrt{x^2y^2+(x^2y^2+x^2+y^2 +1)+ 2xy\sqrt{x^2y^2+ x^2 + y^2 + 1}}=$

$x\sqrt{y^2z^2 + y^2 + z^2 + 1} +y\sqrt{x^2z^2 + x^2 + z^2 + 1} + z\sqrt{(xy + \sqrt{x^2y^2 + x^2 + y^2+1})^2 }=$

$x\sqrt{y^2z^2 + y^2 + z^2 + 1} +y\sqrt{x^2z^2 + x^2 + z^2 + 1} + z\sqrt{(xy + \sqrt{x^2y^2 + x^2 + y^2+1})^2 }=$

(Note $\sqrt{x^2y^2 + x^2 +y^2 + 1} >\sqrt{x^2y^2} = |xy|$ so even if $xy < 0$ we have $xy + \sqrt{x^2y^2 + x^2 + y^2+1} > 0$ so)

$x\sqrt{y^2z^2 + y^2 + z^2 + 1} +y\sqrt{x^2z^2 + x^2 + z^2 + 1} + z(xy + \sqrt{x^2y^2 + x^2 + y^2+1})=$

$x\sqrt{y^2z^2 + y^2 + z^2 + 1} +y\sqrt{x^2z^2 + x^2 + z^2 + 1} + z\sqrt{x^2y^2 + x^2 + y^2+1}+ xyz$

And be symmetry (or repeated calculations) we can see $x\sqrt{(y\sqrt{z^2+1}+z\sqrt{y^2+1})^2+1}+(y\sqrt{z^2+1}+z\sqrt{y^2+1})\sqrt{x^2 + 1}$ is equal to the same thing.

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To do the second.

$x\sqrt{e^2+1}+e\sqrt{x^2+1}=x$

Well, if $e=0$ we get $x\sqrt{e^2+1} + e\sqrt{x^2+1} = x\sqrt 1 + 0*\sqrt{x^2 +1} = x$. So that is a solution. So that is an identity. We havent proven it is a unique solution but we don't need to yet.

All that remains to prove it is a group is to find the inverse:

If $x\sqrt{y^2 + 1}+y{x^2 + 1} =0$ we can solve for $y$ in terms of $x$. We can see that if $y=-x$ this is true. So $y=-x$ is one solution; we haven't shown it is unique but we don't have to.

that's enough to show it is a group and therefore the solutions must be unique.

Then $x(\sqrt{e^2+1}-1)= -e\sqrt{x^2 +1}$