# Give hint to this problem

1. Jan 24, 2009

### barathiviji

give me solution to this problem........

Let f be a function from the set of real numbers R to R such that f(1) is not equal to 0 , and f(x+y)=f(x)+f(y), f(xy)=f(x)(y) for all x,y belongs to R. Then show that f(x)=x for all x in R

2. Jan 24, 2009

### Staff: Mentor

I don't have a solution, but I'm pretty certain you need to use the information that f(1) is not zero. This suggests to me that you would be able to divide both sides of an equation by f(1).

f(1 + 0) = f(1) + f(0) <I don't know what to do with this.>
Now f(0) = f(1*0), so f(0) = f(1)*f(0) so f(0)/f(0) = f(1).
That might be a start for you.

3. Jan 25, 2009

### Gib Z

Well the first equation in fact tells you f(0) = 0, so it probably wasn't a good idea to divide by that in the second line =].

It should be fairly straight forward to show by induction that f(n) = n f(1), integral n. Now if we use that fact in f(xy) = f(x)f(y), we can see that f(1) = 1, so now we can conclude f(n) = n for integer n.

EDIT: Throughout this thread we have assumed the OP made a typo meaning f(xy) = f(x) f(y), though he wrote f(xy) = yf(x). If that is the actual question, then this is quite trivial.
The problem we have left is to show it for all real numbers other than just the integers.

Last edited: Jan 25, 2009
4. Jan 25, 2009

### Unco

Next on to the rationals! Alas, we should wait for Barathiviji to provide some work in this endeavor.

5. Jan 25, 2009

### Staff: Mentor

Sorry, I must have gotten in a hurry. I first said to divide by f(1), but later divided by f(0).