1. ## Continuity

Suppose f,g:R->R are continuous functions such that f(r)=f(r) for all r in Q, that is, f and g are equal on the rational numbers. Prove that f(x)=g(x) for all x in R.

Not sure what to do. Any guidance would be appreciated.

2. ## Re: Continuity

Hey lovesmath.

Continuity implies the that the limit for each point is equal to the function value evaluated at that point.

You might want to show that if the limit exists then the epsilon delta definition of continuity makes sure that the function equals the definition at all numbers.

There is a result that relates the existence of rational numbers and the construction of real numbers between rationals, but I'm guessing that as these become dense enough, then you will have a situation where the limit to each rational number from either side has to have the right properties.

One may be able to show this specifically by saying that if you have a rational number r, then no matter how different you choose a number q where q > r then there exists a smaller number q* where q* > r but q* < q.

Then using this you can use continuity definitions on the function f and conclude that if it is continuous then all real points correspond to the function value.

I forget the whole delta epsilon thingy formulation and I'm not a pure mathematician, but hopefully these give you ideas to bounce off.

3. ## Re: Continuity

Originally Posted by lovesmath
Suppose f,g:R->R are continuous functions such that f(r)=f(r) for all r in Q, that is, f and g are equal on the rational numbers. Prove that f(x)=g(x) for all x in R.
Suppose that $\displaystyle f(a)\ne g(a)$ for some real number $\displaystyle a$.
There is a sequence of rational numbers $\displaystyle r_n$ such $\displaystyle (r_n)\to a$.
But $\displaystyle f(r_n)=g(r_n)\to f(a)$. That is a contradiction.