**$f(n) =\pm n$****$f(n) = \sqrt {n^2 + 1}$****$f(n) = \dfrac{1}{n^2 -4}$**

The aim of this question is to find out if the given equations are **functions** from **Z ****to ****R**.

The basic concept behind solving this problem is to have sound knowledge of all **sets** and the conditions for which a given equation is a **function** from Z **to **R.

Here we have:

\[\mathbb{R}= Real\ Numbers\]

Which means it contains all other set such as, **Rational numbers ** {$…,-2.5, -2, -1.5, 1, 0.5, 0, 0.5, 1, 1.5,…$}, **Integers** {$…,-3, -2, -1, 0, 1, 2, 3,…$}, **Whole numbers** {$0,1,2,3,4,5,6,7,….$}, **Natural numbers** {$1,2,3,4,5,6,7…….$}, **Irrational numbers** {$\pi$, $\sqrt 2$, $\sqrt 3$, $…$}.

\[\mathbb{Z} = Integers\]

\[ \mathbb{Z}\ = {…..,-3,\ -2, -1,\ 0,\ 1,\ 2,\ 3,…..} \]

## Expert Answer

**(a)** To solve this problem first we have to evaluate the given equation $f(n) =\pm (n)$ as a **function** in the **domain** and **range** set.

\[n_1 \times n_2 \in \mathbb{Z}\]

Such that:

\[n_1 =n_2 \]

As the given function is:

\[f(n) = \pm n\]

We can write it with both **positive** and **negative values** as:

\[f(n)=n \]

\[ f(n_1) = n_1\]

Which will also be equal to:

\[f(n_2) = n_2\]

Now it can also be written as:

\[f(n)= – n \]

\[ f(n_1) = – n_1\]

Which will also be equal to:

\[f(n_2) = – n_2\]

For both** positive and negative** values the **function **$f$ is **defined** but as it gives $2$ different values instead of $1$ single value, therefore $f(n) =\pm n$ is **not a function** from **$\mathbb{Z}$ to $\mathbb{R}$**.

**(b) ** Given function is $f(n) = \sqrt {n^2 + 1}$

\[n_1 \times n_2 \in \mathbb{Z}\]

Such that:

\[{n_1}^2 = {n_2}^2 \]

As there is square on $n$ so what ever value we will put it be positive.

\[{n_1}^2 + 1 = {n_2}^2 + 1 \]

\[\sqrt{{n_1}^2 + 1} = \sqrt{{n_2}^2 + 1} \]

So we can write:

\[ f(n_1) = f( n_2) \]

Thus we conclude that $f(n) = \sqrt {n^2 + 1}$ **is a function** from **$\mathbb{Z}$ to $\mathbb{R}$**.

**(c)** Given function $f(n) = \dfrac{1}{n^2 -4}$

\[n_1 \times n_2 \in \mathbb{Z}\]

Such that:

\[{n_1}^2 = {n_2}^2 \]

\[{n_1}^2 – 4 = {n_2}^2 -\ 4 \]

But now if $n=2$ or $n= -2$, we have:

\[f(2)= \frac{1}{ {2}^2 –\ 4} ; f(-2)= \frac{1}{ {-2}^2\ –\ 4}\]

\[f(2)= \frac{1}{ 4 – 4} ; f(-2)= \frac{1}{ 4 – 4}\]

\[f(2)= \frac{1}{ 0} ; f(-2)= \frac{1}{ 0}\]

Here we can see that the **function** $f$ is now equal to $\infty $ and therefore it **cannot be defined** so $f(n) = \dfrac{1}{n^2 -4}$ is **not a function** from **$\mathbb{Z}$ to $\mathbb{R}$**.

## Numerical Results

$f(n) =\pm n$ is **not a function** from $\mathbb{Z}$ to $\mathbb{R}$.

$f(n) = \sqrt {n^2 + 1}$ is** a function** from $\mathbb{Z}$ to $\mathbb{R}$.

$f(n) = \dfrac{1}{n^2 -4}$ is **not a function** from $\mathbb{Z}$ to $\mathbb{R}$.

## Example

Find if $f(n) = \sqrt {n^2 + 8}$ is a function from $\mathbb{Z}$ to $\mathbb{R}$.

**Solution**

\[n_1 \times n_2 \in \mathbb{Z}\]

\[{n_1}^2={n_2}^2\]

\[{n_1}^2+8={n_2}^2+8\]

\[\sqrt{{n_1}^2+8}=\sqrt{{n_2}^2+8} \]

\[f(n_1)=f( n_2)\]

Is** a function** from **$\mathbb{Z}$ to $\mathbb{R}$**.