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