Do you know its property ? -5

Algebra Level 5

Concerning the function:

$f(x)=\begin{cases} 2 ,\ \ \text{if}\ \space x\in\mathbb{Q}\\-2,\ \ \text{if}\ \space x\notin\mathbb{Q}\ \end{cases},$

which of the following statement(s) is/are correct?

It is periodic with period 1
It is periodic with period 2
It is non-periodic
It is periodic but fundamental period is not defined.

Note: $\mathbb{Q}$ represents the set of rational numbers.

You can try more such problems of the set Do you know its property ?

4 only 3 only 1 and 2 1,2, and 4

2 solutions

Sandeep Bhardwaj
Oct 16, 2014

Given:

$f(x)=$ $\begin{cases} 2 ,if \space x\in{Q}\\-2, if \space x\notin{Q}\ \end{cases}$

Let $a$ be a $Rational Number$ . $\implies f(a)=2$

Just next number after every Rational Number is an Irrational Number. & Just next number after every Irrational Number is a Rational Number.

Suppose $b$ is just next number after $a$ and Just next after $b$ is $c$

$\implies f(b)=-2$

$\implies f(c)=2$

and so on........

Hence f(x) has same value at every two distinct numbers which have exactly one number between them.

So it will behave like a $\large constant$ function.

So $f(x)$ is periodic But fundamental period is not defined. It will also have period 1 & 2.

So statements $(1),(2) and (4)$ are correct.

I'd be more careful with phrases like 'the next number'. The way we define the real numbers ensures that there is no such thing as 'the next real number'.

No $b$ can 'the next number just after' $a$ since $\frac{a+b}{2}$ is always strictly between $a$ and $b$ if they are distinct.

However, it is true that there is an irrational number between every two (distinct) rational numbers.

Mursalin Habib - 6 years, 7 months ago

By saying,

It is periodic with period 1

doesnt it mean that you are implying that the fundemental period is 1.

shouldnt it be f(x+1)=f(x) .....?

A Former Brilliant Member - 6 years, 7 months ago

Yeah, but we have $f(x) = f(x+y)$ for all $y \in \mathbb{Q}$ .

Since the fundamental period is always a positive real number that can defined as

$\min \lbrace y \in \mathbb{R}^{+} | x \in \mathbb{R} \wedge f(x) = f(x+y) \rbrace$

we will consider the set of positive rational numbers $\mathbb{Q}^{+}$ .

The infimum of $\mathbb{Q}^{+}$ exists:

$\inf \mathbb{Q}^{+} = 0$

However, the minimum of $\mathbb{Q}^{+}$ is not defined ; if it is, it cannot be the period of $f$ since it is not a real number but rather an infinitesimal number .

You may wish you read on continuity and the Dirichlet function .

Jake Lai - 6 years, 5 months ago

How can we say the next number to any rational number number is irrational?

Could you please explain?

Soumo Mukherjee - 6 years, 5 months ago

I got trolled again.. (No offense)

Nishant Sharma - 6 years, 7 months ago

I do not think that this question constitutes "trolling".

It is a valid question about fundamental periods.

Calvin Lin Staff - 6 years, 7 months ago

Tanay Wakhare
Dec 18, 2014

This is all application of the definition of a period. If x is rational, then x+n and x+2n where n is an integer will also be rational and the function will still be 2. If x is irrational, then x+n and x+2n will also be irrational for any integer n, and so answers 1 and 2 are true.

Since at least one period exists, 3 is false.

Now a fundamental period is the smallest positive constant K such than the function is periodic with period K. However suppose that this K exists. K/2 will also be a period since it preserves rationality and irrationality under addition. Therefore there can be no smallest K and 4 is true.

Do you know its property ? -5

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