Let f: \mathbb{R} \to \mathbb{R} be a continuous function such that for every rational number q, f(x+q) = f(x) for all x \in \mathbb{R}. Prove that f is a constant function?
Answer (2)
The function f is shown to be constant because it maintains the same output for any input modified by a rational number. Utilizing the density of rational numbers and the property of continuity, we can extend the function's value across all real numbers. Hence, the conclusion is that f ( x ) = c for all x , where c is a constant value.
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**Answer:Proving that a continuous function **
** with the property **
** for all rational **
** and **
** is a constant function.**
What's given in the problem
** is a continuous function.**
**For every rational number **
**, **
** for all **
**. **
Helpful information
**The set of rational numbers **
** is dense in the set of real numbers **
. This means that for any real number, there is a sequence of rational numbers that converges to it.
**If a function **
** is continuous at a point **
**, and **
** is a sequence converging to **
**, then **
** converges to **
.
How to solve
**Show that for any real number **
**, there exists a sequence of rational numbers converging to **
**. Then use the continuity of **
** and the given property to prove that **
** is constant.**
1. Step 1
**Consider an arbitrary real number **
.
**Since **
** is dense in **
**, there exists a sequence of rational numbers **
** such that **
.
2. Step 2
**Use the continuity of **
.
**Since **
** is continuous, if **
**, then **
.
**We know that **
** for all **
** because **
** is rational and **
.
**Therefore, **
.
3. Step 3
**Show that **
** is a constant function.**
**Since **
** was arbitrary, this holds for any **
;
