F(x) =
So what i am really confused about is because all professors are telling me its periodic whose period is not defined, i mean it doesn't make any sense to me. So they countered me by telling in constant function periodic i said yes but again they asked what is its period. I mean the problem is what is exact definition of periodic function they are not drawing the line. So please help me with understanding it.
On
The function is indeed periodic because $\mathbb{Q}$ is closed for the addition so for example $\forall x \in \mathbb{R}, x \in \mathbb{Q} \iff x+1\in \mathbb{Q}$. This means that $\forall x \in \mathbb{R}, F(x+1) = F(x)$.
We can conclude that $F$ is periodic and $1$ is one period.
Moreover, $\forall p\in\mathbb Q,\forall x \in \mathbb{R}, x \in \mathbb{Q} \iff x+p\in \mathbb{Q}$, which is enough to prove that: $$\forall p\in\mathbb Q,\forall x \in \mathbb{R}, F(x+p) = F(x)$$
Said differently any rational number is a period for $F$.
But we cannot give the minimum period because $\inf(\mathbb{Q_+}) = 0$.
That is the reason why F is periodic (we could exhibit its periods) but we cannot define the period (commonly the minimal positive period) because the inferior bound of the positive periods is $0$.
The period of a function is the smallest positive $r$ so that $f(x+r) = f(x)$ for all $x$.
But some functions will have an infinite number of $r$s but none of them being a smallest positive $r$ so that $f(x+r) =f(x)$ for all $x$.
These functions are periodic because there do exist $r$ so that $f(x+r) = f(x)$ for all $x$. But the do not have known periods because there is not smallest such (positive) number that do that.
In this case if $r\in \mathbb Q$ then $x+r\in \mathbb Q\iff x\in \mathbb Q$ and so $f(x+r) = f(x)$. But there is no smallest positive $r$ that does this. So $f$ is periodic without a defined specific period.
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Note the restriction is that $r \ne 0$. Obviously all functions have $f(x+0) = f(x)$ and obviously that doesn't count.