# Continuous extension of XOR

The $$\text{XOR}$$ function is a function from $$\{0,1\}^2$$ to $$\{0,1\}$$, defined as:

$$\text{XOR}(0,0)=\text{XOR}(1,1)=0$$

$$\text{XOR}(1,0)=\text{XOR}(0,1)=1$$

I am interested in finding an extension of this function to $$\mathbb{R}$$. To be more specific, I am looking for a function $$f$$ from $$\mathbb{R}^2$$ to $$\mathbb{R}$$ with the following properties (for all $$x,y,z\in\mathbb{R}$$):

• $$f(x,y)=f(y,x)$$

• $$f(x,0)=x$$

• $$f(x,x)=0$$

• $$f(x,f(y,z))=f(f(x,y),z)$$

• $$f$$ is continuous

Does such a function exist? If yes, how do I construct it? If not, how do I prove so?

• $f(x,y)=\max\{x,y\}-\min\{x,y\}$ Aug 5, 2019 at 12:47
• @SimplyBeautifulArt: but $f(x,y)=|x-y|$ does not fulfill $f(x,0)=x$ for any $x$, just for $x\geq 0$. Aug 5, 2019 at 12:56
• Oops my bad @JackD'Aurizio :P Aug 5, 2019 at 12:58
• It is also not associative for $x=1,y=2,z=3$
– Lior
Aug 5, 2019 at 13:01
• @orlp that follows from associativity and the identities $f(x,x)=0$ and $f(0,y)=y$. Aug 5, 2019 at 13:08

Let's denote $$f(x,y) = x \circ y$$. We have:

1)$$x \circ y = y \circ x$$

2)$$x \circ x = 0$$

3) $$x \circ 0 = x$$

4)$$(x \circ y)\circ z = x \circ (y \circ z)$$

Setting $$x = y$$ in $$(4)$$ we get $$(x \circ x)\circ z = x \circ (x \circ z) \rightarrow 0 \circ z = x \circ (x \circ z) \implies z = x \circ (x \circ z)$$.

$$x \circ (x \circ z) = z$$ $$f(x,f(x,z)) = z$$

This means $$\circ$$ behaves on $$R$$ like the composition for an abelian group where every non-trivial element is of order $$2$$. Our $$f$$ can be defined by pulling back the group composition operation trough a bijection between $$\mathbb{R}$$ and an uncountable product of $$Z_{2}$$ (where the trivial element corresponds to $$0$$). We shall now prove that such an $$f$$ cannot be continuous.

Consider $$g(x) = f(1,x)$$. If $$f$$ is continuous, so is $$g$$. We have $$g(0) = 1$$ and $$g(1) = 0$$. Consider $$h(x) = g(x) - x$$. We have $$h(0) = 1$$ and $$h(1) = -1$$. By the intermediate value theorem, there exists $$y\in (0,1)$$ such that $$h(y) = 0$$ . Therefore, there exists $$y \in (0,1)$$ such that $$g(y) = y$$. So $$f(1,y) = y$$. But $$f(f(1,y),y) = f(y,y)$$. So $$1 = 0$$ . Contradiction. A continuous $$f$$ satisfying $$1-4$$ does not exist.

Edit: Imho $$\text{XOR}$$ behaves like bit addition. We can identify every real number with its remainder mod. 2. That is $$3.31 \rightarrow 1.31$$ and $$8.161616 \rightarrow 0.161616$$, etc. Then $$a$$ $$\text{XOR}$$ $$b$$ becomes $$a + b$$. This is informally defined on $$[0,2)$$. The alternative would be to simply make bit-wise $$\text{XOR}$$ with numbers written as base $$2$$ sequences. Eg. $$0.101101...\:\:\text{XOR}\:\:0.011 = 0.110101...$$.