What are the properties of the following functions?

Question

I've been doing some exercises to learn how to identify surjective, bijective, injective and normal functions. I was wondering if someone can critique my work and see if there are any places where i went wrong.

1. $f : \mathbb N \to ℕ$ defined as $f(n) = n^2$

   Bijection, because every input entered had a different answer due to the domain being natural numbers we can only enter numbers greater than or equal to zero this makes it injective. For every input entered there is an output, which satisfies the rule of surjection.

2. $f : \mathbb Z \to ℕ$ defined as $f(n) = n^2$

   Surjective, for every input entered there is an output. Because we can also deal with negatives inputs here, the function is not injective.

3. $f : \mathbb N \to \mathbb Z$ defined as $f(n) = n^2$

   Bijection, because every natural number entered produces a different output unless they're the same and every input has some output.

4. $f: \mathbb Z \to \mathbb Z$ defined as $f(n) = n^2$

   Surjective, for every input entered there some output.

5. $f: \mathbb R \to \mathbb N$ defined as $f(n) = n^2$

   Not a function, because every input entered doesn't match to a natural number.

6. $f: \mathbb N \to \mathbb R$ defined as $f(n) = n^2$

   Bijection, every natural number entered maps to a natural number. The set of natural numbers is in the set of real numbers. Every number entered produces a different output therefore the function is injective. Every input entered has some output, there the function is also surjective.

7. $f: \mathbb N \to \mathbb N$ defined as $f(n) = \sqrt{n}$

   Not a function, because every natural number doesn't output a natural number.

8. $f: \mathbb R \to \mathbb R$ defined as $f(n) = \sqrt{n}$

   Not a function, every real number doesn't output a real number. Because we can enter in negative numbers and it won't result in a real number.

9. $f: \mathbb R \to $ { x $\in \mathbb R$ | x $\ge$ 0 } defined as $f(n)= \sqrt{n}$

   Not a function, every real number doesn't output a real number. Because we can enter in negative numbers and it won't result in a real number.

10. $f:$ { x $\in \mathbb R$ | x $\ge$ 0 } $\to$ { x $\in \mathbb R$ | x $\ge$ 0 }, $f(n) = \sqrt{n}$

    Bijection, Every Real Number greater than 0 entered results in a real number greater than 0. Every input entered returns a result unique to that input unless they are the same, therefore the function is injective. For every input there is some output, therefore the function is surjective.

11. $f: \mathbb N \to \mathcal P(\mathbb N)$, where f is some injective function.

    Bijection. I sort of guessed this one without any real explanation.

12. $f:$ {0,1,2} $\to$ {3,4}, where $f$ is some surjective function

    surjective, because there is an output for every input. But 1 number maybe mapped to another number so it can't be injective.

13. $f:$ {breakfast, lunch, dinner} $\to$ {shakshuka, soondubu, maafe}, where $f$ is some injection

    Bijection. Every input has it's own unique output, therefore it's injective. For every input there is some output, there it's surjective.

Answer 1

Thank you for editing your post to include explanations. Please recall that a function $f: A\to B$ is

• injective iff $f$ sends different inputs to different outputs, or mathematically: $x\neq y \implies f(x) \neq f(y)$.
• surjective iff all things in $B$ are reached by $f$, or mathematically $\forall b\in B, \exists a\in A$ s.t. $f(a)=b$.
• bijective iff $f$ is BOTH injective and surjective

Examples:

$f: \mathbb N \to \mathbb N$, $f(n) = n^2$ is NOT surjective because not all $n \in \mathbb N$ are reached by $f$; e.g. $3$ is not a square of a natural number. Thus, functions like $f: \mathbb N \to \mathbb Z$, $f(n) = n^2$ and $f: \mathbb N \to \mathbb R$, $f(n) = n^2$ are also NOT surjective.

$f: \mathbb Z \to \mathbb N$, $f(n) = n^2$ is NOT injective because $f(-1)=f(1)=1$. Please try again and I will check over your solutions.

For number 11, please read the proof here: https://math.stackexchange.com/a/2453/405572

Answer 2

1: $f:\mathbb N \to \mathbb N$ defined as $f(n)= n^2$.

"because every input entered had a different $\require{cancel}$$\cancel{\text{answer}}\text{output}$ due to the domain being natural numbers".

It is true that if $k \ne j$ then $f(k) \ne f(j)$ and therefore the function is injective but was does "due to the domain being natural numbers" have anything to do with it.

consider $g: \mathbb N\to \mathbb N$ defined by $g(n) = n^2 - 6n + 10$. Then $g(2)=2$ and $g(4)=2$ so $g$ is not injective. Why does the domain being natural numbers mean every input having a different output.

" For every input entered there is an output, which satisfies the rule of surjection."

That's true for all functions. It's part of the definition of functions. Not all functions are surjective and that is not the definition of surjection.

Surjective means for every potential output, $y$ there is a least one input $x$ so that $f(x) = y$.

Is this true? If $f(n) =17$ so that $n^2 = 17$ what is $n$? Is there any? If not it is not surjective.

2) $f: \mathbb Z \to \mathbb N$.

Surjective, for every input entered there is an output.

That's not what surjective means. What is $n$ if $f(n) = 17$.

Because we can also deal with negatives inputs here, the function is not injective.

What does negative inputs have to with it? Consider $g: \mathbb Z\to \mathbb Z$ defined by $f(n) = n^3$. And takes negative values and is injective.

3. $f:\mathbb N \to \mathbb Z$.

because every natural number entered produces a different output unless they're the same

Prove this, or explain it in greater detail.

and every input has some output.

That's not what surjective means. This is true of every function.

....

keep trying.