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Determine whether f is a function from Z to R for given functions

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

The aim of this question is to find out if the given equations are functions from 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 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 $\mathbb{R}$.

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