Bijection discrete math
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$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isnâÂÂt. If yes, find an expression computing $f^âÂÂ1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
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up vote
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$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isnâÂÂt. If yes, find an expression computing $f^âÂÂ1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
New contributor
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?
â JMoravitz
1 hour ago
Just try to solve $f(x)=1$.
â Michael Hoppe
57 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 isnâÂÂt. If yes, find an expression computing $f^âÂÂ1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
New contributor
$f: BbbZ to BbbZ, f(x) = 3x + 6$
Is $f$ a bijection? If no, explain why it isnâÂÂt. If yes, find an expression computing $f^âÂÂ1(y)$ for $y in BbbZ$.
How to approach that question?
functions discrete-mathematics elementary-set-theory
functions discrete-mathematics elementary-set-theory
New contributor
New contributor
edited 58 mins ago
JMoravitz
45.4k33684
45.4k33684
New contributor
asked 1 hour ago
FreshmanUCSD
111
111
New contributor
New contributor
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?
â JMoravitz
1 hour ago
Just try to solve $f(x)=1$.
â Michael Hoppe
57 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?
â JMoravitz
1 hour ago
Just try to solve $f(x)=1$.
â Michael Hoppe
57 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?
â JMoravitz
1 hour 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?
â JMoravitz
1 hour ago
Just try to solve $f(x)=1$.
â Michael Hoppe
57 mins ago
Just try to solve $f(x)=1$.
â Michael Hoppe
57 mins ago
add a comment |Â
3 Answers
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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?
add a comment |Â
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1
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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.
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.
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?
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?
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?
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 53 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.
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.
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.
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 50 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.
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.
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.
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 48 mins ago
ArsenBerk
7,35021234
7,35021234
add a comment |Â
add a comment |Â
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?
â JMoravitz
1 hour ago
Just try to solve $f(x)=1$.
â Michael Hoppe
57 mins ago