Prove Continuity at a: f(x+y)=f(x)+f(y) & f(0) Continuous

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Homework Statement


Suppose that "f" satisfies "f(x+y)=f(x)+f(y)", and that "f" is continuous at 0. Prove that "f" is continuous at a for all a.

Homework Equations


In class we were given 3 main ways to solve continuity proofs.

A function "f" is continuous at x=a if:

a.)
Limit of f(x) as x->a = f(a), (a can be all real numbers)

b.)
Limit of f(a+h) as h->0 = f(a), (Let x=a+h)

c.)
(delta-epsilon proof)
For all epsilon greater than 0, there exists some delta greater than 0, such that for all x, if |x-a|< delta then |f(x)-f(a)|< epsilon.

The Attempt at a Solution



So far, by working with my teacher I was able to get this much as being correct:

Theres not enough information for a delta-epsilon proof so

[
Were told that f(x+y)=f(x)+f(y), for all x,y belonging to real numbers then
In partiality:
f(0+0)=f(0)+f(0)
f(0)=2f(0)
Only way this can be possible is if f(0)=0
hence f(0)=0

Now consider:

Limit of f(a+h)-f(a) as h->0 is equal to 0.
]

Im not as to what i should do now and what the second part that my teacher wants me to consider will prove.

Any ideas would be greatly appreciated!
 
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Use the information given. You can rewrite f(a + h) - f(a) to reduce the problem around a to a problem around zero.
 
How can I reduce the problem to be around zero?
Also what does this prove?
 
OK, firstly: do you see what \lim_{h \to 0} f(a + h) - f(a) = 0 is good for?

Then, note that you have been given that f is continuous at zero and you want to prove continuity at some arbitrary a, about which you don't know anything. So you can try to use the properties you have to make the problem of continuity at a into a problem of continuity at 0, because you have more information of f around 0 than around a.
 
You have to use the fact that f(x+ y)= f(x)+ f(y). So f(a+ h)= ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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