LoomiTymnk63x

2023-03-29

Whether f is a function from Z to R if
a) $f\left(n\right)=±n$
b) $f\left(n\right)=\sqrt{{n}^{2}+1}$
c) $f\left(n\right)=\frac{1}{{n}^{2}-4}$.

?

rangiranitlh9

Let's examine each function to determine if it is a function from Z (integers) to R (real numbers).
a) $f\left(n\right)=±n$:
This function is not well-defined since it does not specify whether it should output $+n$ or $-n$ for each input $n$. For example, if we take $n=1$, it is unclear whether $f\left(1\right)$ should be $+1$ or $-1$. Therefore, function $f$ is not a function from Z to R.
b) $f\left(n\right)=\sqrt{{n}^{2}+1}$:
For any integer $n$, the expression ${n}^{2}+1$ is always a non-negative number. The square root of a non-negative real number is a real number. Therefore, the function $f\left(n\right)$ is well-defined for all integers $n$, and it is a function from Z to R.
c) $f\left(n\right)=\frac{1}{{n}^{2}-4}$:
This function is not well-defined for certain integers. Specifically, when the denominator ${n}^{2}-4$ becomes zero, the function is undefined. In this case, when $n=2$ or $n=-2$, the denominator becomes zero, resulting in a division by zero error. Therefore, the function $f\left(n\right)$ is not a function from Z to R.
In summary:
a) The function $f\left(n\right)=±n$ is not a function from Z to R.
b) The function $f\left(n\right)=\sqrt{{n}^{2}+1}$ is a function from Z to R.
c) The function $f\left(n\right)=\frac{1}{{n}^{2}-4}$ is not a function from Z to R.

Do you have a similar question?