What does $f(2n) = n + 3$ mean? [closed]
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Here is the full question. Consider a function $f:mathbb{Z}longrightarrowmathbb{Z}$ such that $f(2n) = n + 3$ for $ninmathbb{N}$. Prove that $f$ is not an injective function.
At first I thought I could solve it by replace the n on the right with 2n, so f(n) = (2n) + 3. But that doesn't seem right. I also tried looking for this form of functions, but I don't know what to search for.
functions notation
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
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closed as unclear what you're asking by José Carlos Santos, 5xum, MisterRiemann, Siong Thye Goh, Masacroso Dec 4 '18 at 10:47
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 2 more comments
$begingroup$
Here is the full question. Consider a function $f:mathbb{Z}longrightarrowmathbb{Z}$ such that $f(2n) = n + 3$ for $ninmathbb{N}$. Prove that $f$ is not an injective function.
At first I thought I could solve it by replace the n on the right with 2n, so f(n) = (2n) + 3. But that doesn't seem right. I also tried looking for this form of functions, but I don't know what to search for.
functions notation
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
$endgroup$
closed as unclear what you're asking by José Carlos Santos, 5xum, MisterRiemann, Siong Thye Goh, Masacroso Dec 4 '18 at 10:47
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Is that all the information there is about the function? In that case, it seems only to be defined for even integers, and then it is indeed injective.
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– Teddan the Terran
Dec 4 '18 at 10:36
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You have to prove it is not injective...?
$endgroup$
– StackTD
Dec 4 '18 at 10:38
$begingroup$
The function seems injective, but it's also quite unclear what exactly the function is. Without further details, this question is ripe for closing as it is unclear.
$endgroup$
– 5xum
Dec 4 '18 at 10:38
$begingroup$
You could always turn it into $g(m)$ where $m=2n$ then convert it back when solved. ie turn what you're unsure how to handle into something you are sure how to handle.
$endgroup$
– timtfj
Dec 4 '18 at 10:47
$begingroup$
$f: mathbb{Z} to mathbb{Z}$ maps all even integers $2n$ to $n+3$, therefore it cannot be injective on $mathbb{Z}$: $f(1) = f(m)$ for $m = 2(f(1)-3)$.
$endgroup$
– Martin R
Dec 4 '18 at 10:51
|
show 2 more comments
$begingroup$
Here is the full question. Consider a function $f:mathbb{Z}longrightarrowmathbb{Z}$ such that $f(2n) = n + 3$ for $ninmathbb{N}$. Prove that $f$ is not an injective function.
At first I thought I could solve it by replace the n on the right with 2n, so f(n) = (2n) + 3. But that doesn't seem right. I also tried looking for this form of functions, but I don't know what to search for.
functions notation
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
$endgroup$
Here is the full question. Consider a function $f:mathbb{Z}longrightarrowmathbb{Z}$ such that $f(2n) = n + 3$ for $ninmathbb{N}$. Prove that $f$ is not an injective function.
At first I thought I could solve it by replace the n on the right with 2n, so f(n) = (2n) + 3. But that doesn't seem right. I also tried looking for this form of functions, but I don't know what to search for.
functions notation
functions notation
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
edited Dec 4 '18 at 11:17
soyfroggy
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
asked Dec 4 '18 at 10:32
soyfroggysoyfroggy
133
133
closed as unclear what you're asking by José Carlos Santos, 5xum, MisterRiemann, Siong Thye Goh, Masacroso Dec 4 '18 at 10:47
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by José Carlos Santos, 5xum, MisterRiemann, Siong Thye Goh, Masacroso Dec 4 '18 at 10:47
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Is that all the information there is about the function? In that case, it seems only to be defined for even integers, and then it is indeed injective.
$endgroup$
– Teddan the Terran
Dec 4 '18 at 10:36
$begingroup$
You have to prove it is not injective...?
$endgroup$
– StackTD
Dec 4 '18 at 10:38
$begingroup$
The function seems injective, but it's also quite unclear what exactly the function is. Without further details, this question is ripe for closing as it is unclear.
$endgroup$
– 5xum
Dec 4 '18 at 10:38
$begingroup$
You could always turn it into $g(m)$ where $m=2n$ then convert it back when solved. ie turn what you're unsure how to handle into something you are sure how to handle.
$endgroup$
– timtfj
Dec 4 '18 at 10:47
$begingroup$
$f: mathbb{Z} to mathbb{Z}$ maps all even integers $2n$ to $n+3$, therefore it cannot be injective on $mathbb{Z}$: $f(1) = f(m)$ for $m = 2(f(1)-3)$.
$endgroup$
– Martin R
Dec 4 '18 at 10:51
|
show 2 more comments
$begingroup$
Is that all the information there is about the function? In that case, it seems only to be defined for even integers, and then it is indeed injective.
$endgroup$
– Teddan the Terran
Dec 4 '18 at 10:36
$begingroup$
You have to prove it is not injective...?
$endgroup$
– StackTD
Dec 4 '18 at 10:38
$begingroup$
The function seems injective, but it's also quite unclear what exactly the function is. Without further details, this question is ripe for closing as it is unclear.
$endgroup$
– 5xum
Dec 4 '18 at 10:38
$begingroup$
You could always turn it into $g(m)$ where $m=2n$ then convert it back when solved. ie turn what you're unsure how to handle into something you are sure how to handle.
$endgroup$
– timtfj
Dec 4 '18 at 10:47
$begingroup$
$f: mathbb{Z} to mathbb{Z}$ maps all even integers $2n$ to $n+3$, therefore it cannot be injective on $mathbb{Z}$: $f(1) = f(m)$ for $m = 2(f(1)-3)$.
$endgroup$
– Martin R
Dec 4 '18 at 10:51
$begingroup$
Is that all the information there is about the function? In that case, it seems only to be defined for even integers, and then it is indeed injective.
$endgroup$
– Teddan the Terran
Dec 4 '18 at 10:36
$begingroup$
Is that all the information there is about the function? In that case, it seems only to be defined for even integers, and then it is indeed injective.
$endgroup$
– Teddan the Terran
Dec 4 '18 at 10:36
$begingroup$
You have to prove it is not injective...?
$endgroup$
– StackTD
Dec 4 '18 at 10:38
$begingroup$
You have to prove it is not injective...?
$endgroup$
– StackTD
Dec 4 '18 at 10:38
$begingroup$
The function seems injective, but it's also quite unclear what exactly the function is. Without further details, this question is ripe for closing as it is unclear.
$endgroup$
– 5xum
Dec 4 '18 at 10:38
$begingroup$
The function seems injective, but it's also quite unclear what exactly the function is. Without further details, this question is ripe for closing as it is unclear.
$endgroup$
– 5xum
Dec 4 '18 at 10:38
$begingroup$
You could always turn it into $g(m)$ where $m=2n$ then convert it back when solved. ie turn what you're unsure how to handle into something you are sure how to handle.
$endgroup$
– timtfj
Dec 4 '18 at 10:47
$begingroup$
You could always turn it into $g(m)$ where $m=2n$ then convert it back when solved. ie turn what you're unsure how to handle into something you are sure how to handle.
$endgroup$
– timtfj
Dec 4 '18 at 10:47
$begingroup$
$f: mathbb{Z} to mathbb{Z}$ maps all even integers $2n$ to $n+3$, therefore it cannot be injective on $mathbb{Z}$: $f(1) = f(m)$ for $m = 2(f(1)-3)$.
$endgroup$
– Martin R
Dec 4 '18 at 10:51
$begingroup$
$f: mathbb{Z} to mathbb{Z}$ maps all even integers $2n$ to $n+3$, therefore it cannot be injective on $mathbb{Z}$: $f(1) = f(m)$ for $m = 2(f(1)-3)$.
$endgroup$
– Martin R
Dec 4 '18 at 10:51
|
show 2 more comments
1 Answer
1
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$begingroup$
I suppose that $f$ is defined on $D={2n: n in mathbb N}$
Then we have
$f(2)=f(2 cdot 1)=1+3=4, quad f(4)=f(2 cdot 2)=2+3=5,quad f(6)=f(2 cdot 3)=3+3=6,$ etc ....
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
$endgroup$
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
1
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suppose that $f$ is defined on $D={2n: n in mathbb N}$
Then we have
$f(2)=f(2 cdot 1)=1+3=4, quad f(4)=f(2 cdot 2)=2+3=5,quad f(6)=f(2 cdot 3)=3+3=6,$ etc ....
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
$endgroup$
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
1
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
add a comment |
$begingroup$
I suppose that $f$ is defined on $D={2n: n in mathbb N}$
Then we have
$f(2)=f(2 cdot 1)=1+3=4, quad f(4)=f(2 cdot 2)=2+3=5,quad f(6)=f(2 cdot 3)=3+3=6,$ etc ....
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
$endgroup$
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
1
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
add a comment |
$begingroup$
I suppose that $f$ is defined on $D={2n: n in mathbb N}$
Then we have
$f(2)=f(2 cdot 1)=1+3=4, quad f(4)=f(2 cdot 2)=2+3=5,quad f(6)=f(2 cdot 3)=3+3=6,$ etc ....
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
$endgroup$
I suppose that $f$ is defined on $D={2n: n in mathbb N}$
Then we have
$f(2)=f(2 cdot 1)=1+3=4, quad f(4)=f(2 cdot 2)=2+3=5,quad f(6)=f(2 cdot 3)=3+3=6,$ etc ....
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
answered Dec 4 '18 at 10:36
FredFred
44.7k1846
44.7k1846
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
1
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
add a comment |
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
1
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
$begingroup$
does it mean the function $f$ is only defined in the domain where x is even?
$endgroup$
– soyfroggy
Dec 4 '18 at 10:48
1
1
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
$begingroup$
@soyfroggy Not necessarily. The function simply satisfies the condition $f(2n) = n+3$. It could be, just as a hypothetical example, that for odd numbers it is zero always, i.e. $f(2n+1) = 0.$ You have simply been given how the function is defined for even numbers, that doesn't by itself say anything about how it is defined on odd numbers.
$endgroup$
– Eff
Dec 4 '18 at 11:23
add a comment |
$begingroup$
Is that all the information there is about the function? In that case, it seems only to be defined for even integers, and then it is indeed injective.
$endgroup$
– Teddan the Terran
Dec 4 '18 at 10:36
$begingroup$
You have to prove it is not injective...?
$endgroup$
– StackTD
Dec 4 '18 at 10:38
$begingroup$
The function seems injective, but it's also quite unclear what exactly the function is. Without further details, this question is ripe for closing as it is unclear.
$endgroup$
– 5xum
Dec 4 '18 at 10:38
$begingroup$
You could always turn it into $g(m)$ where $m=2n$ then convert it back when solved. ie turn what you're unsure how to handle into something you are sure how to handle.
$endgroup$
– timtfj
Dec 4 '18 at 10:47
$begingroup$
$f: mathbb{Z} to mathbb{Z}$ maps all even integers $2n$ to $n+3$, therefore it cannot be injective on $mathbb{Z}$: $f(1) = f(m)$ for $m = 2(f(1)-3)$.
$endgroup$
– Martin R
Dec 4 '18 at 10:51