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Mathematics · Ch 1 — Sets, Relations and Functions

Algebra of Functions

1.6.6

Algebra of Functions

A function whose co-domain is RR (or a subset of RR) is a real-valued function. When f,gf,g share the same domain XX, we can combine their output values using ordinary real-number arithmetic, defining new functions on XX:

(f+g)(x)=f(x)+g(x),(f−g)(x)=f(x)−g(x),(fg)(x)=f(x)g(x),(f+g)(x)=f(x)+g(x),\quad (f-g)(x)=f(x)-g(x),\quad (fg)(x)=f(x)g(x),

(fg)(x)=f(x)g(x) (g(x)≠0),(cf)(x)=c f(x) (c∈R),(−f)(x)=−f(x).\left(\frac fg\right)(x)=\frac{f(x)}{g(x)}\ (g(x)\ne0),\quad (cf)(x)=c\,f(x)\ (c\in R),\quad (-f)(x)=-f(x).

The domain need not be a set of numbers at all -- e.g. if XX is a class of students and f,gf,g are their marks in two different tests, f+gf+g is exactly their combined (total) marks function.

Properties (mirroring the field properties of RR itself): (f+g)+h=f+(g+h)(f+g)+h=f+(g+h); f+g=g+ff+g=g+f; 0+f=f+0=f0+f=f+0=f (zero function as identity); f+(−f)=(−f)+f=0f+(-f)=(-f)+f=0; f(g+h)=fg+fhf(g+h)=fg+fh (distributivity); (c1+c2)f=c1f+c2f(c_1+c_2)f=c_1f+c_2f.

Proof of distributivity (representative proof). To show f(g+h)=fg+fhf(g+h)=fg+fh, check both sides agree at every xx in the common domain:

(f(g+h))(x)=f(x) (g+h)(x)=f(x)[g(x)+h(x)]=f(x)g(x)+f(x)h(x)=(fg)(x)+(fh)(x)=(fg+fh)(x),\big(f(g+h)\big)(x)=f(x)\,(g+h)(x)=f(x)\big[g(x)+h(x)\big]=f(x)g(x)+f(x)h(x)=(fg)(x)+(fh)(x)=(fg+fh)(x), …