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How to give an injective function?



The 2019 Stack Overflow Developer Survey Results Are InFunction problem (kernel, surjective & injective function)Injective, Surjective functions helpFind the domain, co-domain and range of a function$sqrt{4x -3}$ injective? Bijective? Inverse?Cardinality of the Domain vs Codomain in Surjective (non-injective) & Injective (non-surjective) functionsCardinality of Surjective only & Injective only functionsHow many functions from ${0,1} times {0,1}$ to ${0,1,2}$ are there?Construct a function that is surjective, but not injectiveHow many injective and surjective functions are there from $A$ to $B$?Prove that any function can be written as the sum of an even function and an odd function.












1












$begingroup$


Totally lost with this question :




Give an injective function $f : 2^{{0,1}} to {0,1} times {0,1}$



Can the codomain be written as {(0,0), (0,1), (1,0), (1,1)}?




I'm not sure I understand the $2^{{0,1}}$ part?



Any help would be greatly appreciated! Toally new to functions.






functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes on codomain; $2^{{0,1}}$ is the (four) subsets of ${0,1}.$
    $endgroup$
    – coffeemath
    Mar 21 at 22:14










  • $begingroup$
    I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain.
    $endgroup$
    – Robin
    Mar 21 at 22:20










  • $begingroup$
    They want you to provide an explicit example of an injective function with the given domain and codomain. For instance, could you give an example of an injective function $g:{0,1}to {0,2,7}$? (To practise giving examples of functions.)
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 22:23


















1












$begingroup$


Totally lost with this question :




Give an injective function $f : 2^{{0,1}} to {0,1} times {0,1}$



Can the codomain be written as {(0,0), (0,1), (1,0), (1,1)}?




I'm not sure I understand the $2^{{0,1}}$ part?



Any help would be greatly appreciated! Toally new to functions.






functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes on codomain; $2^{{0,1}}$ is the (four) subsets of ${0,1}.$
    $endgroup$
    – coffeemath
    Mar 21 at 22:14










  • $begingroup$
    I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain.
    $endgroup$
    – Robin
    Mar 21 at 22:20










  • $begingroup$
    They want you to provide an explicit example of an injective function with the given domain and codomain. For instance, could you give an example of an injective function $g:{0,1}to {0,2,7}$? (To practise giving examples of functions.)
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 22:23
















1












1








1


1



$begingroup$


Totally lost with this question :




Give an injective function $f : 2^{{0,1}} to {0,1} times {0,1}$



Can the codomain be written as {(0,0), (0,1), (1,0), (1,1)}?




I'm not sure I understand the $2^{{0,1}}$ part?



Any help would be greatly appreciated! Toally new to functions.






functions







share|cite|improve this question











$endgroup$




Totally lost with this question :




Give an injective function $f : 2^{{0,1}} to {0,1} times {0,1}$



Can the codomain be written as {(0,0), (0,1), (1,0), (1,1)}?




I'm not sure I understand the $2^{{0,1}}$ part?



Any help would be greatly appreciated! Toally new to functions.






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 22:18









Brian

1,508316




1,508316










asked Mar 21 at 22:11









RobinRobin

625




625








  • 1




    $begingroup$
    Yes on codomain; $2^{{0,1}}$ is the (four) subsets of ${0,1}.$
    $endgroup$
    – coffeemath
    Mar 21 at 22:14










  • $begingroup$
    I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain.
    $endgroup$
    – Robin
    Mar 21 at 22:20










  • $begingroup$
    They want you to provide an explicit example of an injective function with the given domain and codomain. For instance, could you give an example of an injective function $g:{0,1}to {0,2,7}$? (To practise giving examples of functions.)
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 22:23
















  • 1




    $begingroup$
    Yes on codomain; $2^{{0,1}}$ is the (four) subsets of ${0,1}.$
    $endgroup$
    – coffeemath
    Mar 21 at 22:14










  • $begingroup$
    I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain.
    $endgroup$
    – Robin
    Mar 21 at 22:20










  • $begingroup$
    They want you to provide an explicit example of an injective function with the given domain and codomain. For instance, could you give an example of an injective function $g:{0,1}to {0,2,7}$? (To practise giving examples of functions.)
    $endgroup$
    – Minus One-Twelfth
    Mar 21 at 22:23










1




1




$begingroup$
Yes on codomain; $2^{{0,1}}$ is the (four) subsets of ${0,1}.$
$endgroup$
– coffeemath
Mar 21 at 22:14




$begingroup$
Yes on codomain; $2^{{0,1}}$ is the (four) subsets of ${0,1}.$
$endgroup$
– coffeemath
Mar 21 at 22:14












$begingroup$
I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain.
$endgroup$
– Robin
Mar 21 at 22:20




$begingroup$
I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain.
$endgroup$
– Robin
Mar 21 at 22:20












$begingroup$
They want you to provide an explicit example of an injective function with the given domain and codomain. For instance, could you give an example of an injective function $g:{0,1}to {0,2,7}$? (To practise giving examples of functions.)
$endgroup$
– Minus One-Twelfth
Mar 21 at 22:23






$begingroup$
They want you to provide an explicit example of an injective function with the given domain and codomain. For instance, could you give an example of an injective function $g:{0,1}to {0,2,7}$? (To practise giving examples of functions.)
$endgroup$
– Minus One-Twelfth
Mar 21 at 22:23












1 Answer
1






active

oldest

votes


















3












$begingroup$

The symbol $2^S$ denotes the power set of the set $S$. In your case,
$$ 2^{{0,1}} =left{ emptyset, {0}, {1}, {0,1} right}$$



Similarly, ${0,1} times {0,1}$ denotes the direct product of the set ${0,1}$ with itself; that is,
begin{align}
{0,1} times {0,1} &= left{(x,y) mid x,y in {0,1} right} \
&= left{(0,0), (0,1), (1,0), (1,1) right}
end{align}




In response to your more recent comments, a function is injective if and only if it maps distinct elements in its domain to distinct element in its codomain. If $f$ is an injective function, then $$f(a) = f(b) iff a = b $$
In your question, you want to create an injective function between the set $2^{{0,1}}$ and the set ${0,1} times {0,1}$. One such function would be begin{align}
emptyset &mapsto (0,0) \
{0} &mapsto (1,0) \
{1} &mapsto (1,1) \
{0, 1} &mapsto (0,1) \
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
    $endgroup$
    – Robin
    Mar 21 at 22:21






  • 1




    $begingroup$
    @Robin See my most recent edit.
    $endgroup$
    – Brian
    Mar 21 at 22:33












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The symbol $2^S$ denotes the power set of the set $S$. In your case,
$$ 2^{{0,1}} =left{ emptyset, {0}, {1}, {0,1} right}$$



Similarly, ${0,1} times {0,1}$ denotes the direct product of the set ${0,1}$ with itself; that is,
begin{align}
{0,1} times {0,1} &= left{(x,y) mid x,y in {0,1} right} \
&= left{(0,0), (0,1), (1,0), (1,1) right}
end{align}




In response to your more recent comments, a function is injective if and only if it maps distinct elements in its domain to distinct element in its codomain. If $f$ is an injective function, then $$f(a) = f(b) iff a = b $$
In your question, you want to create an injective function between the set $2^{{0,1}}$ and the set ${0,1} times {0,1}$. One such function would be begin{align}
emptyset &mapsto (0,0) \
{0} &mapsto (1,0) \
{1} &mapsto (1,1) \
{0, 1} &mapsto (0,1) \
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
    $endgroup$
    – Robin
    Mar 21 at 22:21






  • 1




    $begingroup$
    @Robin See my most recent edit.
    $endgroup$
    – Brian
    Mar 21 at 22:33
















3












$begingroup$

The symbol $2^S$ denotes the power set of the set $S$. In your case,
$$ 2^{{0,1}} =left{ emptyset, {0}, {1}, {0,1} right}$$



Similarly, ${0,1} times {0,1}$ denotes the direct product of the set ${0,1}$ with itself; that is,
begin{align}
{0,1} times {0,1} &= left{(x,y) mid x,y in {0,1} right} \
&= left{(0,0), (0,1), (1,0), (1,1) right}
end{align}




In response to your more recent comments, a function is injective if and only if it maps distinct elements in its domain to distinct element in its codomain. If $f$ is an injective function, then $$f(a) = f(b) iff a = b $$
In your question, you want to create an injective function between the set $2^{{0,1}}$ and the set ${0,1} times {0,1}$. One such function would be begin{align}
emptyset &mapsto (0,0) \
{0} &mapsto (1,0) \
{1} &mapsto (1,1) \
{0, 1} &mapsto (0,1) \
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
    $endgroup$
    – Robin
    Mar 21 at 22:21






  • 1




    $begingroup$
    @Robin See my most recent edit.
    $endgroup$
    – Brian
    Mar 21 at 22:33














3












3








3





$begingroup$

The symbol $2^S$ denotes the power set of the set $S$. In your case,
$$ 2^{{0,1}} =left{ emptyset, {0}, {1}, {0,1} right}$$



Similarly, ${0,1} times {0,1}$ denotes the direct product of the set ${0,1}$ with itself; that is,
begin{align}
{0,1} times {0,1} &= left{(x,y) mid x,y in {0,1} right} \
&= left{(0,0), (0,1), (1,0), (1,1) right}
end{align}




In response to your more recent comments, a function is injective if and only if it maps distinct elements in its domain to distinct element in its codomain. If $f$ is an injective function, then $$f(a) = f(b) iff a = b $$
In your question, you want to create an injective function between the set $2^{{0,1}}$ and the set ${0,1} times {0,1}$. One such function would be begin{align}
emptyset &mapsto (0,0) \
{0} &mapsto (1,0) \
{1} &mapsto (1,1) \
{0, 1} &mapsto (0,1) \
end{align}






share|cite|improve this answer











$endgroup$



The symbol $2^S$ denotes the power set of the set $S$. In your case,
$$ 2^{{0,1}} =left{ emptyset, {0}, {1}, {0,1} right}$$



Similarly, ${0,1} times {0,1}$ denotes the direct product of the set ${0,1}$ with itself; that is,
begin{align}
{0,1} times {0,1} &= left{(x,y) mid x,y in {0,1} right} \
&= left{(0,0), (0,1), (1,0), (1,1) right}
end{align}




In response to your more recent comments, a function is injective if and only if it maps distinct elements in its domain to distinct element in its codomain. If $f$ is an injective function, then $$f(a) = f(b) iff a = b $$
In your question, you want to create an injective function between the set $2^{{0,1}}$ and the set ${0,1} times {0,1}$. One such function would be begin{align}
emptyset &mapsto (0,0) \
{0} &mapsto (1,0) \
{1} &mapsto (1,1) \
{0, 1} &mapsto (0,1) \
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 21 at 22:33

























answered Mar 21 at 22:20









BrianBrian

1,508316




1,508316












  • $begingroup$
    Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
    $endgroup$
    – Robin
    Mar 21 at 22:21






  • 1




    $begingroup$
    @Robin See my most recent edit.
    $endgroup$
    – Brian
    Mar 21 at 22:33


















  • $begingroup$
    Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
    $endgroup$
    – Robin
    Mar 21 at 22:21






  • 1




    $begingroup$
    @Robin See my most recent edit.
    $endgroup$
    – Brian
    Mar 21 at 22:33
















$begingroup$
Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
$endgroup$
– Robin
Mar 21 at 22:21




$begingroup$
Thanks Brian. I guess I'm confused what "Give an injective function means". I know a function is injective when there are more that 1 elements in the domain that map to 1 codomain
$endgroup$
– Robin
Mar 21 at 22:21




1




1




$begingroup$
@Robin See my most recent edit.
$endgroup$
– Brian
Mar 21 at 22:33




$begingroup$
@Robin See my most recent edit.
$endgroup$
– Brian
Mar 21 at 22:33


















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