Can you solve this function problem with a hint and the given condition?

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The discussion revolves around solving a functional equation where f is defined from real numbers to real numbers, with the conditions f(1) not equal to 0, f(x+y)=f(x)+f(y), and f(xy)=f(x)(y). Participants suggest that knowing f(1) is non-zero allows for division in equations, leading to the conclusion that f(0) must equal 0. They also indicate that proving f(n) = n for integers through induction is a logical step, and there is a need to extend this proof to all real numbers. Clarification is provided regarding a potential typo in the original problem statement, which could simplify the solution. The conversation emphasizes the importance of using the given conditions effectively to derive the function's properties.
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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
 
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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).

Some things to start with are
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.
 
Mark44 said:
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).

Some things to start with are
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.

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:
Gib Z said:
The problem we have left is to show it for all real numbers other than just the integers.
Next on to the rationals! Alas, we should wait for Barathiviji to provide some work in this endeavor.
 
Mark44 said:
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).

Some things to start with are
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.

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

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