Originally Posted by

**mwanahisab** I had posted this problem before, however the expression didn't appear as expected. The final expression should read f ^2=f

Let the commutative ring F(R) be the set of all functions R-R equipped with the operations of pointwise addition and pointwise multiplication: given f, g are members F(R), define functions f+g and fg by

*f+g: a→ f(a) + g(a)* and *fg: →f(a)g(a)*

(Notice fg is not their composite)

Pointwise addition and multiplication are the operations on functions occurring in calculus for example.

∫*f(x) g(x)dx* =∫*f(x) dx* +∫*g(x) dx*

And D(fg)=D(f)g+f(D(g), where D denotes derivative.

The sum f+g in the first integrand is pointwise addition while the product fg in the derivative formula is pointwise multiplication.

Show that the above commutative ring F(R) contains elements *f*≠0,1 with *f2=f*

The expresiion should read the square of f equals f. What I mean here is that f multiplied by f equals f. Symbolicaly this is f ^2=f