Mixing
Bijection discrete math
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$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
edited 54 mins ago
JMoravitz
45.4k33684
asked 1 hour ago
FreshmanUCSD
111
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FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
favorite
$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
edited 54 mins ago
JMoravitz
45.4k33684
asked 1 hour ago
FreshmanUCSD
111
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
59 mins ago
Just try to solve $f(x)=1$.
– Michael Hoppe
53 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
edited 54 mins ago
JMoravitz
45.4k33684
asked 1 hour ago
FreshmanUCSD
111
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isn’t. If yes, find an expression computing $f^−1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
functions discrete-mathematics elementary-set-theory
edited 54 mins ago
JMoravitz
45.4k33684
asked 1 hour ago
FreshmanUCSD
111
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
JMoravitz
45.4k33684
asked 1 hour ago
FreshmanUCSD
111
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 54 mins ago
JMoravitz
45.4k33684
edited 54 mins ago
JMoravitz
45.4k33684
edited 54 mins ago
JMoravitz
45.4k33684
45.4k33684
asked 1 hour ago
FreshmanUCSD
111
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
FreshmanUCSD
111
asked 1 hour ago
FreshmanUCSD
111
111
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
FreshmanUCSD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
59 mins ago
Just try to solve $f(x)=1$.
– Michael Hoppe
53 mins ago
add a comment |Â
2
Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
59 mins ago
Just try to solve $f(x)=1$.
– Michael Hoppe
53 mins ago
2
2
Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
59 mins ago
Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
59 mins ago
Just try to solve $f(x)=1$.
– Michael Hoppe
53 mins ago
Just try to solve $f(x)=1$.
– Michael Hoppe
53 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?
answered 49 mins ago
Matt Samuel
35k43060
add a comment |Â
up vote
1
down vote
Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.
answered 46 mins ago
coffeemath
1,6841313
add a comment |Â
up vote
0
down vote
We can also use a different approach since it is easy to invert the function.
Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
$$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.
answered 44 mins ago
ArsenBerk
7,35021234
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?
answered 49 mins ago
Matt Samuel
35k43060
add a comment |Â
up vote
3
down vote
It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?
answered 49 mins ago
Matt Samuel
35k43060
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?
answered 49 mins ago
Matt Samuel
35k43060
It's easy to see that the function's value is a multiple of 3. Is every integer a multiple of 3? No. So which property does that violate?
answered 49 mins ago
Matt Samuel
35k43060
answered 49 mins ago
Matt Samuel
35k43060
answered 49 mins ago
Matt Samuel
35k43060
answered 49 mins ago
Matt Samuel
35k43060
35k43060
add a comment |Â
add a comment |Â
up vote
1
down vote
Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.
answered 46 mins ago
coffeemath
1,6841313
add a comment |Â
up vote
1
down vote
Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.
answered 46 mins ago
coffeemath
1,6841313
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.
answered 46 mins ago
coffeemath
1,6841313
Since $f(x)=3(x+2),$ the image of $f$ is not all of $mathbbZ$ but only the multiples of $3$. So there can be no inverse map from $mathbbZ$ to $mathbbZ.$ But $f$ does happen to be an injection.
answered 46 mins ago
coffeemath
1,6841313
answered 46 mins ago
coffeemath
1,6841313
answered 46 mins ago
coffeemath
1,6841313
answered 46 mins ago
coffeemath
1,6841313
1,6841313
add a comment |Â
add a comment |Â
up vote
0
down vote
We can also use a different approach since it is easy to invert the function.
Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
$$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.
answered 44 mins ago
ArsenBerk
7,35021234
add a comment |Â
up vote
0
down vote
We can also use a different approach since it is easy to invert the function.
Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
$$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.
answered 44 mins ago
ArsenBerk
7,35021234
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We can also use a different approach since it is easy to invert the function.
Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
$$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.
answered 44 mins ago
ArsenBerk
7,35021234
We can also use a different approach since it is easy to invert the function.
Suppose for a contradiction that $f$ is a bijection, which implies $f$ is invertible. Then inverse function $f^-1$ of $f$ is
$$f^-1(x) = fracx-63, xinmathbbZ, f^-1:mathbbZtomathbbZ$$
But this is not even a function since $x$ values that are not multiple of $3$ are not even mapped to an element in codomain. So we have a contradiction as required.
answered 44 mins ago
ArsenBerk
7,35021234
answered 44 mins ago
ArsenBerk
7,35021234
answered 44 mins ago
ArsenBerk
7,35021234
answered 44 mins ago
ArsenBerk
7,35021234
7,35021234
add a comment |Â
add a comment |Â
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FreshmanUCSD is a new contributor. Be nice, and check out our Code of Conduct.
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2
Ask yourself whether or not the function is both injective and surjective. That is to say, will you always get different outputs if you give it different inputs or are there some inputs that give the same output? Is every element in the codomain actually an output that happens?
– JMoravitz
59 mins ago
Just try to solve $f(x)=1$.
– Michael Hoppe
53 mins ago