Is f(a+b) = f(a)f(b) true for all real numbers a and b?

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SUMMARY

The discussion centers on proving that if the functional equation f(a+b) = f(a)f(b) holds for all real numbers a and b, then the function f is continuous at every real number. The proof begins with the assumption that f is continuous at 0, leading to the conclusion that f(0) is defined and equal to the limit of f(x) as x approaches 0. The argument extends to show that continuity at 0 implies continuity for all real numbers, confirming the assertion definitively.

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teme6@yahoo.c
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there is one problem. the problem is related with contuinity of afunction and i tried like as shown below.so if anyone who is intersted to help me i like ..
the problem is
prove that if f(a+b)=f(a)f(b) for all a and b ,then f is cntiniuous at every real number.here there is given information that is the domain of f is the set of all real number and f is continuous at 0.
I tried to approache this problem like this
since f is continuous at 0, f(0) must defined and it must be equal to the value of limf(x) =f(0)
x-->0
and a and b are real numbers then limf(a+b)=limf(a).limf(b)
x--->0 x-->0 x-->0
=f(a).f(b) (since f(a) and f(b) are constnt fu.)
this implies f is continuous at every real number.
am i write? i want to be sure so please help
 
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Hi, welcome to PF.

Try looking at
[tex]\lim_{x \to a} f(x) = \lim_{x \to a} f( (x - a) + a )[/tex].
 

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