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A few months ago, I took some course in Logic and one of the topics was the arity of functions. The arity of a function $f$ is the number of arguments that $f$ needs to works. For example, the function $f$ defined by $f(x,y)=x^2 +y^2$ has arity 2 and the function $g$ defined by $g()=\tanh(\frac{\pi}{3})$ is function with arity $0$.

However, every function that comes to my head can always be written as the composition of functions of arity $0,1$ or $2$ arguments. For example, consider the function $$g(x,y,z,w)=\frac{\sqrt{\sin(xy^2)}}{3+z+\ln(w)}$$ If we let \begin{align} t()&=3\\ p(x)&=x^2\\ r(x)&=\sqrt{x}\\ s(x)&=\sin(x)\\ l(x)&=\ln(x)\\ k(x,y)&=x+y\\ h(x,y)&=xy\\ f(x,y)&=\frac{x}{y} \end{align} Then, $g$ can be rewritten as follows: $$g(x,y,w,z)=f\Bigg(r\bigg(s\Big(h\big(x,p(y)\big)\Big)\bigg),k\bigg(k\Big(z,l(w)\Big),t()\bigg)\Bigg)$$

My question until this point is if there is any function that don't follows this rule, or actually every $n$-ary function could be written as the composition of nullary, unary and binary functions.

I guess my question is because mostly all the function that I know are binary.

  • Your question is a duplicate of the question cited by Bill unless you impose some restrictions. If you look at continuous functions $\Bbb{R}^n \to \Bbb{R}$, then I suspect the answer would be that binary continuous functions $\Bbb{R}^2 \to \Bbb{R}$ are inadequate to generate them all (though I don't claim to know how to prove that). – Rob Arthan Mar 28 '19 at 21:39

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