Injective/Surjective Of Functions Help
$begingroup$
So I've created functions based off questions I have and I have to prove the functions are bijections.
$A ∪ B$ and $[n + m]$ where $A$ has $n$ elements, $B$ has $m$ elements, and $A ∩ B = ∅$.
Function: $$h(m)= begin{cases}a_k hspace{1.4 cm}{k le n} \ b_{k-n} hspace{1 cm}k > n end{cases}$$
Pf:
First we will prove injectivity. Let $m_1,m_2in mathbb{N}$ s.t. $m_1 neq m_2$. We will show $h(m_1) neq h(m_2)$. Case 1: Let $m_1leq n$ and $m_2leq n$. Then $h(m_1)=a_{m_1}neq a_{m_2}=h(m_2)$. Case 2: Let $m_1>n$ and $m_2leq n$. Then $h(m_1)=b_{m_1-n}neq a_{m_2}=h(m_2)$. I have two more cases that follow the format of the first two cases. So this shows the function is injective. Now we will prove this function is surjective. Let $nin (A cup B)$. We will find an $minmathbb{N}$ such that $h(m)=n$. Case 1: Let $mleq n$. Let $n=a_1$. Then $mleq n$ so $h(m)=a_1=n$. Case 2: Let $m>n$. Let $n=b_1$. Then $m>n$ so $h(m)=b_1=n$. We conclude h is surjective so the function is bijective.
- Let $[n]times[m]$ be the usual cartesian product of sets and $[nm]$ Function: $f((i,j))=i+(j-1)n$
Pf: We will first prove the function is injective. Let $i_1, i_2, j_1, j_2$ exist in $f$ s.t. $f(i_1,j_1)=f(i_2,j_2)$. Then $i_1+(j_1-1)n=i_2+(j_2-1)n$. Then $i_1+j_1n=i_2+j_2n$. So $i_1=i_2$ and $j_1=j_2$. As a result this function is an injection. The surjective part I am confused. I think I should be using the divison algorithm but I am not sure.
functions proof-verification elementary-set-theory proof-writing
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
$endgroup$
|
show 7 more comments
$begingroup$
So I've created functions based off questions I have and I have to prove the functions are bijections.
$A ∪ B$ and $[n + m]$ where $A$ has $n$ elements, $B$ has $m$ elements, and $A ∩ B = ∅$.
Function: $$h(m)= begin{cases}a_k hspace{1.4 cm}{k le n} \ b_{k-n} hspace{1 cm}k > n end{cases}$$
Pf:
First we will prove injectivity. Let $m_1,m_2in mathbb{N}$ s.t. $m_1 neq m_2$. We will show $h(m_1) neq h(m_2)$. Case 1: Let $m_1leq n$ and $m_2leq n$. Then $h(m_1)=a_{m_1}neq a_{m_2}=h(m_2)$. Case 2: Let $m_1>n$ and $m_2leq n$. Then $h(m_1)=b_{m_1-n}neq a_{m_2}=h(m_2)$. I have two more cases that follow the format of the first two cases. So this shows the function is injective. Now we will prove this function is surjective. Let $nin (A cup B)$. We will find an $minmathbb{N}$ such that $h(m)=n$. Case 1: Let $mleq n$. Let $n=a_1$. Then $mleq n$ so $h(m)=a_1=n$. Case 2: Let $m>n$. Let $n=b_1$. Then $m>n$ so $h(m)=b_1=n$. We conclude h is surjective so the function is bijective.
- Let $[n]times[m]$ be the usual cartesian product of sets and $[nm]$ Function: $f((i,j))=i+(j-1)n$
Pf: We will first prove the function is injective. Let $i_1, i_2, j_1, j_2$ exist in $f$ s.t. $f(i_1,j_1)=f(i_2,j_2)$. Then $i_1+(j_1-1)n=i_2+(j_2-1)n$. Then $i_1+j_1n=i_2+j_2n$. So $i_1=i_2$ and $j_1=j_2$. As a result this function is an injection. The surjective part I am confused. I think I should be using the divison algorithm but I am not sure.
functions proof-verification elementary-set-theory proof-writing
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
$endgroup$
$begingroup$
Please add mathjax to make this readable. I did this for the last paragraph so that should help you see how this done.
$endgroup$
– Mason
Dec 5 '18 at 0:16
$begingroup$
I just made an account. I do not know what mathjax is sorry.
$endgroup$
– Marcus Dillions
Dec 5 '18 at 0:21
$begingroup$
Yup. There is a link I can give you to help you learn it. I can find it. Or someone else will shortly post it for you. For now: see my edit and you will pick it up. It's not hard. Look for the $ signs
$endgroup$
– Mason
Dec 5 '18 at 0:23
$begingroup$
To make it easier to see that your function is a bijection, rename the element as follows $A={a_1,ldots,a_n}$ and $B={a_{n+1},ldots, a_{n+m}}$. Now just definie $f(a_i)=i$ and it is easy to show that this function is bijective.
$endgroup$
– user9077
Dec 5 '18 at 0:25
$begingroup$
Mathjax
$endgroup$
– Mason
Dec 5 '18 at 0:26
|
show 7 more comments
$begingroup$
So I've created functions based off questions I have and I have to prove the functions are bijections.
$A ∪ B$ and $[n + m]$ where $A$ has $n$ elements, $B$ has $m$ elements, and $A ∩ B = ∅$.
Function: $$h(m)= begin{cases}a_k hspace{1.4 cm}{k le n} \ b_{k-n} hspace{1 cm}k > n end{cases}$$
Pf:
First we will prove injectivity. Let $m_1,m_2in mathbb{N}$ s.t. $m_1 neq m_2$. We will show $h(m_1) neq h(m_2)$. Case 1: Let $m_1leq n$ and $m_2leq n$. Then $h(m_1)=a_{m_1}neq a_{m_2}=h(m_2)$. Case 2: Let $m_1>n$ and $m_2leq n$. Then $h(m_1)=b_{m_1-n}neq a_{m_2}=h(m_2)$. I have two more cases that follow the format of the first two cases. So this shows the function is injective. Now we will prove this function is surjective. Let $nin (A cup B)$. We will find an $minmathbb{N}$ such that $h(m)=n$. Case 1: Let $mleq n$. Let $n=a_1$. Then $mleq n$ so $h(m)=a_1=n$. Case 2: Let $m>n$. Let $n=b_1$. Then $m>n$ so $h(m)=b_1=n$. We conclude h is surjective so the function is bijective.
- Let $[n]times[m]$ be the usual cartesian product of sets and $[nm]$ Function: $f((i,j))=i+(j-1)n$
Pf: We will first prove the function is injective. Let $i_1, i_2, j_1, j_2$ exist in $f$ s.t. $f(i_1,j_1)=f(i_2,j_2)$. Then $i_1+(j_1-1)n=i_2+(j_2-1)n$. Then $i_1+j_1n=i_2+j_2n$. So $i_1=i_2$ and $j_1=j_2$. As a result this function is an injection. The surjective part I am confused. I think I should be using the divison algorithm but I am not sure.
functions proof-verification elementary-set-theory proof-writing
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
$endgroup$
So I've created functions based off questions I have and I have to prove the functions are bijections.
$A ∪ B$ and $[n + m]$ where $A$ has $n$ elements, $B$ has $m$ elements, and $A ∩ B = ∅$.
Function: $$h(m)= begin{cases}a_k hspace{1.4 cm}{k le n} \ b_{k-n} hspace{1 cm}k > n end{cases}$$
Pf:
First we will prove injectivity. Let $m_1,m_2in mathbb{N}$ s.t. $m_1 neq m_2$. We will show $h(m_1) neq h(m_2)$. Case 1: Let $m_1leq n$ and $m_2leq n$. Then $h(m_1)=a_{m_1}neq a_{m_2}=h(m_2)$. Case 2: Let $m_1>n$ and $m_2leq n$. Then $h(m_1)=b_{m_1-n}neq a_{m_2}=h(m_2)$. I have two more cases that follow the format of the first two cases. So this shows the function is injective. Now we will prove this function is surjective. Let $nin (A cup B)$. We will find an $minmathbb{N}$ such that $h(m)=n$. Case 1: Let $mleq n$. Let $n=a_1$. Then $mleq n$ so $h(m)=a_1=n$. Case 2: Let $m>n$. Let $n=b_1$. Then $m>n$ so $h(m)=b_1=n$. We conclude h is surjective so the function is bijective.
- Let $[n]times[m]$ be the usual cartesian product of sets and $[nm]$ Function: $f((i,j))=i+(j-1)n$
Pf: We will first prove the function is injective. Let $i_1, i_2, j_1, j_2$ exist in $f$ s.t. $f(i_1,j_1)=f(i_2,j_2)$. Then $i_1+(j_1-1)n=i_2+(j_2-1)n$. Then $i_1+j_1n=i_2+j_2n$. So $i_1=i_2$ and $j_1=j_2$. As a result this function is an injection. The surjective part I am confused. I think I should be using the divison algorithm but I am not sure.
functions proof-verification elementary-set-theory proof-writing
functions proof-verification elementary-set-theory proof-writing
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
edited Dec 5 '18 at 14:05
Martin Sleziak
44.8k9118272
44.8k9118272
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
asked Dec 5 '18 at 0:11
Marcus DillionsMarcus Dillions
63
63
$begingroup$
Please add mathjax to make this readable. I did this for the last paragraph so that should help you see how this done.
$endgroup$
– Mason
Dec 5 '18 at 0:16
$begingroup$
I just made an account. I do not know what mathjax is sorry.
$endgroup$
– Marcus Dillions
Dec 5 '18 at 0:21
$begingroup$
Yup. There is a link I can give you to help you learn it. I can find it. Or someone else will shortly post it for you. For now: see my edit and you will pick it up. It's not hard. Look for the $ signs
$endgroup$
– Mason
Dec 5 '18 at 0:23
$begingroup$
To make it easier to see that your function is a bijection, rename the element as follows $A={a_1,ldots,a_n}$ and $B={a_{n+1},ldots, a_{n+m}}$. Now just definie $f(a_i)=i$ and it is easy to show that this function is bijective.
$endgroup$
– user9077
Dec 5 '18 at 0:25
$begingroup$
Mathjax
$endgroup$
– Mason
Dec 5 '18 at 0:26
|
show 7 more comments
$begingroup$
Please add mathjax to make this readable. I did this for the last paragraph so that should help you see how this done.
$endgroup$
– Mason
Dec 5 '18 at 0:16
$begingroup$
I just made an account. I do not know what mathjax is sorry.
$endgroup$
– Marcus Dillions
Dec 5 '18 at 0:21
$begingroup$
Yup. There is a link I can give you to help you learn it. I can find it. Or someone else will shortly post it for you. For now: see my edit and you will pick it up. It's not hard. Look for the $ signs
$endgroup$
– Mason
Dec 5 '18 at 0:23
$begingroup$
To make it easier to see that your function is a bijection, rename the element as follows $A={a_1,ldots,a_n}$ and $B={a_{n+1},ldots, a_{n+m}}$. Now just definie $f(a_i)=i$ and it is easy to show that this function is bijective.
$endgroup$
– user9077
Dec 5 '18 at 0:25
$begingroup$
Mathjax
$endgroup$
– Mason
Dec 5 '18 at 0:26
$begingroup$
Please add mathjax to make this readable. I did this for the last paragraph so that should help you see how this done.
$endgroup$
– Mason
Dec 5 '18 at 0:16
$begingroup$
Please add mathjax to make this readable. I did this for the last paragraph so that should help you see how this done.
$endgroup$
– Mason
Dec 5 '18 at 0:16
$begingroup$
I just made an account. I do not know what mathjax is sorry.
$endgroup$
– Marcus Dillions
Dec 5 '18 at 0:21
$begingroup$
I just made an account. I do not know what mathjax is sorry.
$endgroup$
– Marcus Dillions
Dec 5 '18 at 0:21
$begingroup$
Yup. There is a link I can give you to help you learn it. I can find it. Or someone else will shortly post it for you. For now: see my edit and you will pick it up. It's not hard. Look for the $ signs
$endgroup$
– Mason
Dec 5 '18 at 0:23
$begingroup$
Yup. There is a link I can give you to help you learn it. I can find it. Or someone else will shortly post it for you. For now: see my edit and you will pick it up. It's not hard. Look for the $ signs
$endgroup$
– Mason
Dec 5 '18 at 0:23
$begingroup$
To make it easier to see that your function is a bijection, rename the element as follows $A={a_1,ldots,a_n}$ and $B={a_{n+1},ldots, a_{n+m}}$. Now just definie $f(a_i)=i$ and it is easy to show that this function is bijective.
$endgroup$
– user9077
Dec 5 '18 at 0:25
$begingroup$
To make it easier to see that your function is a bijection, rename the element as follows $A={a_1,ldots,a_n}$ and $B={a_{n+1},ldots, a_{n+m}}$. Now just definie $f(a_i)=i$ and it is easy to show that this function is bijective.
$endgroup$
– user9077
Dec 5 '18 at 0:25
$begingroup$
Mathjax
$endgroup$
– Mason
Dec 5 '18 at 0:26
$begingroup$
Mathjax
$endgroup$
– Mason
Dec 5 '18 at 0:26
|
show 7 more comments
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$begingroup$
Please add mathjax to make this readable. I did this for the last paragraph so that should help you see how this done.
$endgroup$
– Mason
Dec 5 '18 at 0:16
$begingroup$
I just made an account. I do not know what mathjax is sorry.
$endgroup$
– Marcus Dillions
Dec 5 '18 at 0:21
$begingroup$
Yup. There is a link I can give you to help you learn it. I can find it. Or someone else will shortly post it for you. For now: see my edit and you will pick it up. It's not hard. Look for the $ signs
$endgroup$
– Mason
Dec 5 '18 at 0:23
$begingroup$
To make it easier to see that your function is a bijection, rename the element as follows $A={a_1,ldots,a_n}$ and $B={a_{n+1},ldots, a_{n+m}}$. Now just definie $f(a_i)=i$ and it is easy to show that this function is bijective.
$endgroup$
– user9077
Dec 5 '18 at 0:25
$begingroup$
Mathjax
$endgroup$
– Mason
Dec 5 '18 at 0:26