Is the following function form Z to Z one-to-one? 1) f(n)=n−1 2) f(n)=n2+1 3) f(n)=n3 4) f(n)=⌈n∕2⌉

The function is one-to-one if f(a)=f(b), a=b and f is 1−1. 1) f(n)=n−1 Let n1,n2∈Z If f(n1)=f(n2), then n1=n2 n1−1=n2−1 n1=n2 Thus, the function is one-to-one 2) f(n)=n2+1 Let n1,n2∈Z If f(n1)=f(n2), then n1=n2 n12+1=n22+1 n12=n22 ±n1=±n2 n11=−n2and−n1=n2 Thus, the function is not one-to-one 3) f(n)=n3 Let n1,n2∈Z If f(n1)=f(n2), then n1=n2 n13=n23 n1=n2 Thus, the function is one-to-one 4) f(n)=⌈n∕2⌉ Let n1,n2∈Z If f(n1)=f(n2),then n1=n2 n12=n22 n1=n2 Thus, the function is one-to-one

Resolved: "Is the following function form Z to Z..."

High School
Geometry
Trigonometry
Trigonometric Functions

Armorikam

Answered question

2021-09-05

Is the following function form

Z

Z

one-to-one?
1)

f (n) = n - 1

f (n) = n^{2} + 1

f (n) = n^{3}

f (n) = ⌈ n ∕ 2 ⌉

Answer & Explanation

unett

Skilled2021-09-06Added 119 answers

The function is one-to-one if $f (a) = f (b)$ , a=b and f is $1 - 1$ .
1) $f (n) = n - 1$
Let $\underset{1}{n}, \underset{2}{n} \in Z$
If $f (\underset{1}{n}) = f (\underset{2}{n})$ , then $\underset{1}{n} = \underset{2}{n}$
$\underset{1}{n} - 1 = \underset{2}{n} - 1$
$\underset{1}{n} = \underset{2}{n}$
Thus, the function is one-to-one
2) $f (n) = n^{2} + 1$
Let $\underset{1}{n}, \underset{2}{n} \in Z$
If $f (\underset{1}{n}) = f (\underset{2}{n})$ , then $\underset{1}{n} = \underset{2}{n}$
${\underset{1}{n}}^{2} + 1 = {\underset{2}{n}}^{2} + 1$
${\underset{1}{n}}^{2} = {\underset{2}{n}}^{2}$
$\pm \underset{1}{n} = \pm \underset{2}{n}$
$\underset{1}{n} 1 = - \underset{2}{n} and - \underset{1}{n} = \underset{2}{n}$
Thus, the function is not one-to-one
3) $f (n) = n^{3}$
Let $\underset{1}{n}, \underset{2}{n} \in Z$
If $f (\underset{1}{n}) = f (\underset{2}{n})$ , then $\underset{1}{n} = \underset{2}{n}$
${\underset{1}{n}}^{3} = {\underset{2}{n}}^{3}$
$\underset{1}{n} = \underset{2}{n}$
Thus, the function is one-to-one
4) $f (n) = ⌈ n ∕ 2 ⌉$
Let $\underset{1}{n}, \underset{2}{n} \in Z$
If $f (\underset{1}{n}) = f (\underset{2}{n}), t h e n \underset{1}{n} = \underset{2}{n}$
$\frac{\underset{1}{n}}{2} = \frac{\underset{2}{n}}{2}$
$\underset{1}{n} = \underset{2}{n}$
Thus, the function is one-to-one

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