Can Two Discontinuous Functions Become Continuous When Added Together?

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SUMMARY

The discussion centers on the mathematical concept of combining two discontinuous functions to create a continuous function. Participants suggest using functions such as F(x) = 1 for x ≤ 0 and F(x) = 0 for x > 1, and G(x) = 1 for x ≤ 0 and G(x) = 0 for x = 1. However, it is established that simply adding these functions does not yield continuity due to domain issues. A valid approach is to define G(x) as G(x) = 1 - F(x), which can lead to a continuous result when properly defined.

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semidevil
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just a basic question, so if I'm asked to find 2 functions that are discontinus, but when added together, becomes continuous, how do I approach that?

can I say like, let

F(x) = 1 for x =< 0, and f(x) = 0 for x > 1.
G(x) = 1 for x =<0 and g(x) = 0 for x = 1.

can I just somehow "add" f + g and say that is continuous? I don't know...tips?
 
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Let f be your favourite discontinuous function, then let g=1-f.

Prove g is discontinuous, and hence find a continuous function that is the sum of two continuous ones.

Your f and g, did you mean to capitalize them? Note g doesn't have the same domain as f; g isn't defined for any positive real numbers, and hence neither is f+g.
 
semidevil said:
just a basic question, so if I'm asked to find 2 functions that are discontinus, but when added together, becomes continuous, how do I approach that?

can I say like, let

F(x) = 1 for x =< 0, and f(x) = 0 for x > 1.
G(x) = 1 for x =<0 and g(x) = 0 for x = 1.

can I just somehow "add" f + g and say that is continuous? I don't know...tips?

Well, you can't just say it's continuous- because it isn't!

(f+ g(x)= 2 for x<0, 1 if x= 0 and 0 if x>0
(I've switched the last "1" to "0". The f and g you give are not defined between 0 and 1. Unless that's a typo, I have the uncomfortable feeling that you don't know what is meant by "defining" a function.

Taking f(x)= 1 for x<= 0, f(x)= 0 for x> 0, which is not continuous at x= 0,
try matt grimes' suggestion. What is g(x)= 1- f(x)?
 

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