# [SOLVED]Associativity puzzle

#### Klaas van Aarsen

##### MHB Seeker
Staff member
There was a thread on another forum that I'd like to share.

Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?

#### Amer

##### Active member
what do you mean by associative function ? I know the associative property

#### Bacterius

##### Well-known member
MHB Math Helper
what do you mean by associative function ? I know the associative property
I think in this case the associative property is $f(f(a, b), c) = f(a, f(b, c))$? My best guess

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yep. That's the one.

Does $f(x,f(y,z)) = f(f(x,y),z)$ hold for all $x, y, z \in \mathbb R$?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
As the famous math joke goes, "Yes, it IS obvious!" (Notices of the AMS, 52:1, PDF).

Use hyperbolic functions.

Staff member
That's right!
But how?

#### Amer

##### Active member
There was a thread on another forum that I'd like to share.

Is $f(x,y) = x \sqrt{1+y^2} + y \sqrt{1+x^2}$ associative?
$$f(f(x,y),z) = f \sqrt{ 1 + z^2} + z \sqrt{1 + f^2 }$$

$$f(f(x,y),z) = \left( x \sqrt{1+y^2} + y \sqrt{1+x^2}\right) \sqrt{ 1 + z^2} + z \sqrt{1 + \left( x \sqrt{1+y^2} + y \sqrt{1+x^2}\right)^2 }$$
Let
$$x = \sinh u , y = \sinh v , z = \sinh w$$
and note
$$\sinh^2 a + 1 = \cosh^2 a$$

$$f(f(u,v),w) = \left( \sinh u \mid \cosh v \mid + \sinh v \mid \cosh u \mid\right) \mid \cosh w \mid + \sinh w \sqrt{1 + \left( \sinh u \mid \cosh v \mid + \sinh v \mid \cosh u \mid \right)^2 }$$

the other one
$$f(x, f(y,z)) = x \sqrt{ 1 + f^2} + f \sqrt{1 + x^2 }$$
as the previous
$$f(y,z) = y \sqrt{ 1 + z^2} + z\sqrt{1+ y^2}$$
$$f(v,w) = \sinh u \mid \cosh w \mid + \sinh w \mid \cosh v \mid$$

$$f(u,f(v,w)) = \sinh u \sqrt{ 1 + \left( \sinh v \mid \cosh w \mid + \sinh w \mid \cosh v\mid \right)^2 } + \left(\sinh u \mid \cosh w \mid + \sinh w \mid \cosh v \mid \right) \mid \cosh u \mid$$

i cant see how they are the same, if i did it right at the first place

#### Opalg

##### MHB Oldtimer
Staff member
$$f(f(x,y),z) = f \sqrt{ 1 + z^2} + z \sqrt{1 + f^2 }$$

$$f(f(x,y),z) = \left( x \sqrt{1+y^2} + y \sqrt{1+x^2}\right) \sqrt{ 1 + z^2} + z \sqrt{1 + \left( x \sqrt{1+y^2} + y \sqrt{1+x^2}\right)^2 }$$
Let
$$x = \sinh u , y = \sinh v , z = \sinh w$$
and note
$$\sinh^2 a + 1 = \cosh^2 a$$
Also, $\cosh a$ is always positive, so we can write $\sqrt{\sinh^2 a + 1} = \cosh a$. Thus $$f(x,y) = \sinh u\cosh v + \cosh u\sinh v = \sinh(u+v).$$ It follows that $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Yes, in other words, $\sinh$ is a surjective homomorphism (in fact, an isomorphism) from $\mathbb{R}$ with $+$ to $\mathbb{R}$ with $f$. For this problem, it is only important that $\sinh$ is surjective and respects the operations, but it can be viewed as an isomorphism of abelian groups $\langle\mathbb{R},{+},0,{-}\rangle$ and $\langle\mathbb{R},f,0,{-}\rangle$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Also, $\cosh a$ is always positive, so we can write $\sqrt{\sinh^2 a + 1} = \cosh a$. Thus $$f(x,y) = \sinh u\cosh v + \cosh u\sinh v = \sinh(u+v).$$ It follows that $f(f(x,y),z) = \sinh((u+v)+w) = \ldots$.
Yep!
That takes care of it.

Follow-up extended solution:
$$f(x,y) = \sinh(\sinh^{-1}x+\sinh^{-1}y)$$
So:
\begin{aligned}
f\big(x,f(y,z)\big) &= \sinh \Big(\sinh^{-1}x+\sinh^{-1}\big(\sinh(\sinh^{-1}y+\sinh^{-1}z)\big)\Big) \\
&= \sinh\big(\sinh^{-1}x+(\sinh^{-1}y+\sinh^{-1}z)\big) \\
&= \sinh\big((\sinh^{-1}x+\sinh^{-1}y)+\sinh^{-1}z)\big) \\
&= f\big(f(x,y),z\big) & \blacksquare
\end{aligned}

More generally, any $f(x,y)$ of the form
$$f(x,y) = g\big(g^{-1}(x) + g^{-1}(y)\big)$$
is associative.

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Yes, in other words, $\sinh$ is a surjective homomorphism (in fact, an isomorphism) from $\mathbb{R}$ with $+$ to $\mathbb{R}$ with $f$. For this problem, it is only important that $\sinh$ is surjective and respects the operations, but it can be viewed as an isomorphism of abelian groups $\langle\mathbb{R},{+},0,{-}\rangle$ and $\langle\mathbb{R},f,0,{-}\rangle$.
Nice insight!

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