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$\lfloor\cdot\rfloor: Q \rightarrow\mathbb Z$ with $\lfloor x\rfloor :=$ floor of $x$.

I know a function is injective by using $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ and a function is surjective if each element of the codomain, $y\in Y$, is the image of some element in the domain $x\in X$, and bijective if the function is both injective and surjective.

I don't know what floor of $x$ is.

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    $\begingroup$ The floor of $x$ is the largest integer $\le x$: it’s the unique integer $n$ such that $n\le x<n+1$. For instance, the floor of $2$ is $2$, the floor of $2.5$ is $2$, the floor of $\pi$ is $3$, the floor of $-2$ is $-2$, and the floor of $-2.5$ is $-3$. (You may have to think a moment about that last one.) $\endgroup$ – Brian M. Scott Dec 2 '15 at 13:12
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    $\begingroup$ The floor function returns the largestr integer $\le x$. $\endgroup$ – Hagen von Eitzen Dec 2 '15 at 13:13
  • $\begingroup$ Following the above comments, you need to consider: Is any integer $y\in\mathbb{Z}$ the floor of some $x\in\mathbb{Q}$ (surjectivity)? Can one integer $y\in\mathbb{Z}$ be the floor of more than one rational number (non-injectivity)? $\endgroup$ – Mankind Dec 2 '15 at 13:16
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The floor, $\lfloor x\rfloor,$ of a real number $x$ is defined as the largest integer less than or equal to $x$. For example;

  • $\lfloor 3.1\rfloor = 3$,
  • $\lfloor 3\rfloor = 3$ and
  • $\lfloor \pi \rfloor = 3$.

These examples demonstrate that the function $f:\mathbb{Q}\to \mathbb{Z}$ given by $f(x) = \lfloor x \rfloor$ is not injective since, for example, $\lfloor \pi \rfloor = \lfloor 3 \rfloor = \lfloor 3.1 \rfloor = 3$. This immediately implies that $f$ cannot be bijective (since a bijection is a function which is both injective and surjective).

However the function is surjective because any integer $k\in \mathbb{Z}$ is the image of itself. That is $f(k) = \lfloor k\rfloor = k$ for all $k\in \mathbb{Z}$.

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The function is surjective, but not injective.

(1). Injectivity.

Take $x=2$ and $y=2.5$. We have $\lfloor x \rfloor = \lfloor y \rfloor = 2$, but we don't have $x=y$.

(2). Surjectivity.

Take $x=n$. Then, we have $\lfloor x \rfloor = n$.

Therefore, for all $n \in \mathbb{Z}$, we can find an $x$ (for example, $n$) in $\mathbb{Q}$ such that $\lfloor x \rfloor = n$.

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Definition:Floor of a number $x$, $\lfloor x\rfloor=$ largest positive integer $a\in \Bbb Z$ such that $a\le x$

For example: $\lfloor 3\rfloor=3$

$\lfloor 3.456\rfloor=3$

$\lfloor e\rfloor=2$

$\lfloor -\frac32\rfloor=-2$

For proof, try answering these questions:

  1. $\frac12, \frac13\in\Bbb Q$. Using above definition, what is $\lfloor \frac12\rfloor$ and $\lfloor \frac13\rfloor$? What does that suggest about injectivity?

  2. For any $x\in\Bbb Z$, what is $\lfloor x\rfloor$? What does that suggest about surjectivity?

  3. Using conclusions from above questions, what can you conclude about bijectivity?

Extra information: Since OP seems new to floor function: What happens if $\lfloor .\rfloor$ was defined from $\Bbb Z\to\Bbb Z$ or $\Bbb N \to \Bbb Z$?

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