Mixing
Checking if a function is injective and surjective
$begingroup$
I am doing past paper question and came across the following question:
For each of the following functions, decide whether it is injective
and surjective. Justify your answer.
$f: $ {$-1, 0, 1$} $to$ {$-1, 0, 1$}
$f(x) = x^3$
$g: $ {$0, 1$} $to$ {$0, 1, 2, 3, 4, 5$}
$g(x) = 3x + 1$
I have only recently started studying functions, so hoped to check my answers here, because I do not have access to a marking scheme.
My answers and reasoning:
$f$ is not injective, because $pm x neq pm x$
$f$ is surjective because the co-domain {$-1, 0 ,1$} $=$ the range {$-1, 0 ,1$}
$g$ is injective, because $x = x$
$g$ is not surjective, because the co-domain {$0, 1, 2, 3, 4, 5$} $neq$ the range {$1, 4$}
Please let me know if I have made any errors in my answers or reasoning. Thank you.
functions
asked Jan 5 '18 at 13:29
ShannonShannon
567
$endgroup$
add a comment |
$begingroup$
I am doing past paper question and came across the following question:
For each of the following functions, decide whether it is injective
and surjective. Justify your answer.
$f: $ {$-1, 0, 1$} $to$ {$-1, 0, 1$}
$f(x) = x^3$
$g: $ {$0, 1$} $to$ {$0, 1, 2, 3, 4, 5$}
$g(x) = 3x + 1$
I have only recently started studying functions, so hoped to check my answers here, because I do not have access to a marking scheme.
My answers and reasoning:
$f$ is not injective, because $pm x neq pm x$
$f$ is surjective because the co-domain {$-1, 0 ,1$} $=$ the range {$-1, 0 ,1$}
$g$ is injective, because $x = x$
$g$ is not surjective, because the co-domain {$0, 1, 2, 3, 4, 5$} $neq$ the range {$1, 4$}
Please let me know if I have made any errors in my answers or reasoning. Thank you.
functions
asked Jan 5 '18 at 13:29
ShannonShannon
567
$endgroup$
$begingroup$
@LinkingPark but because we must $sqrt[3]{x^3}$ does that not mean that $x$ can be either positive or negative, and therefore $pm x$ does not imply one another?
$endgroup$
– Shannon
Jan 5 '18 at 13:37
1
$begingroup$
You can easily see that $f(-1)=-1 ,f(1)=1,f(0)=0$ and then you can see that this is bijective mapping (surjective and injective).I dont understand you way of thinking.
$endgroup$
– user508685
Jan 5 '18 at 13:41
$begingroup$
I can see that, but I am confused because I followed this youtu.be/bZred_Ksz2k?t=220
$endgroup$
– Shannon
Jan 5 '18 at 13:46
1
$begingroup$
If you use that you must know that $({x^{3}})^{1/3}=x$
$endgroup$
– user508685
Jan 5 '18 at 13:56
add a comment |
$begingroup$
I am doing past paper question and came across the following question:
For each of the following functions, decide whether it is injective
and surjective. Justify your answer.
$f: $ {$-1, 0, 1$} $to$ {$-1, 0, 1$}
$f(x) = x^3$
$g: $ {$0, 1$} $to$ {$0, 1, 2, 3, 4, 5$}
$g(x) = 3x + 1$
I have only recently started studying functions, so hoped to check my answers here, because I do not have access to a marking scheme.
My answers and reasoning:
$f$ is not injective, because $pm x neq pm x$
$f$ is surjective because the co-domain {$-1, 0 ,1$} $=$ the range {$-1, 0 ,1$}
$g$ is injective, because $x = x$
$g$ is not surjective, because the co-domain {$0, 1, 2, 3, 4, 5$} $neq$ the range {$1, 4$}
Please let me know if I have made any errors in my answers or reasoning. Thank you.
functions
asked Jan 5 '18 at 13:29
ShannonShannon
567
$endgroup$
I am doing past paper question and came across the following question:
For each of the following functions, decide whether it is injective
and surjective. Justify your answer.
$f: $ {$-1, 0, 1$} $to$ {$-1, 0, 1$}
$f(x) = x^3$
$g: $ {$0, 1$} $to$ {$0, 1, 2, 3, 4, 5$}
$g(x) = 3x + 1$
I have only recently started studying functions, so hoped to check my answers here, because I do not have access to a marking scheme.
My answers and reasoning:
$f$ is not injective, because $pm x neq pm x$
$f$ is surjective because the co-domain {$-1, 0 ,1$} $=$ the range {$-1, 0 ,1$}
$g$ is injective, because $x = x$
$g$ is not surjective, because the co-domain {$0, 1, 2, 3, 4, 5$} $neq$ the range {$1, 4$}
Please let me know if I have made any errors in my answers or reasoning. Thank you.
functions
functions
asked Jan 5 '18 at 13:29
ShannonShannon
567
asked Jan 5 '18 at 13:29
ShannonShannon
567
asked Jan 5 '18 at 13:29
ShannonShannon
567
asked Jan 5 '18 at 13:29
ShannonShannon
567
asked Jan 5 '18 at 13:29
ShannonShannon
567
567
$begingroup$
@LinkingPark but because we must $sqrt[3]{x^3}$ does that not mean that $x$ can be either positive or negative, and therefore $pm x$ does not imply one another?
$endgroup$
– Shannon
Jan 5 '18 at 13:37
1
$begingroup$
You can easily see that $f(-1)=-1 ,f(1)=1,f(0)=0$ and then you can see that this is bijective mapping (surjective and injective).I dont understand you way of thinking.
$endgroup$
– user508685
Jan 5 '18 at 13:41
$begingroup$
I can see that, but I am confused because I followed this youtu.be/bZred_Ksz2k?t=220
$endgroup$
– Shannon
Jan 5 '18 at 13:46
1
$begingroup$
If you use that you must know that $({x^{3}})^{1/3}=x$
$endgroup$
– user508685
Jan 5 '18 at 13:56
add a comment |
$begingroup$
@LinkingPark but because we must $sqrt[3]{x^3}$ does that not mean that $x$ can be either positive or negative, and therefore $pm x$ does not imply one another?
$endgroup$
– Shannon
Jan 5 '18 at 13:37
1
$begingroup$
You can easily see that $f(-1)=-1 ,f(1)=1,f(0)=0$ and then you can see that this is bijective mapping (surjective and injective).I dont understand you way of thinking.
$endgroup$
– user508685
Jan 5 '18 at 13:41
$begingroup$
I can see that, but I am confused because I followed this youtu.be/bZred_Ksz2k?t=220
$endgroup$
– Shannon
Jan 5 '18 at 13:46
1
$begingroup$
If you use that you must know that $({x^{3}})^{1/3}=x$
$endgroup$
– user508685
Jan 5 '18 at 13:56
$begingroup$
@LinkingPark but because we must $sqrt[3]{x^3}$ does that not mean that $x$ can be either positive or negative, and therefore $pm x$ does not imply one another?
$endgroup$
– Shannon
Jan 5 '18 at 13:37
$begingroup$
@LinkingPark but because we must $sqrt[3]{x^3}$ does that not mean that $x$ can be either positive or negative, and therefore $pm x$ does not imply one another?
$endgroup$
– Shannon
Jan 5 '18 at 13:37
1
1
$begingroup$
You can easily see that $f(-1)=-1 ,f(1)=1,f(0)=0$ and then you can see that this is bijective mapping (surjective and injective).I dont understand you way of thinking.
$endgroup$
– user508685
Jan 5 '18 at 13:41
$begingroup$
You can easily see that $f(-1)=-1 ,f(1)=1,f(0)=0$ and then you can see that this is bijective mapping (surjective and injective).I dont understand you way of thinking.
$endgroup$
– user508685
Jan 5 '18 at 13:41
$begingroup$
I can see that, but I am confused because I followed this youtu.be/bZred_Ksz2k?t=220
$endgroup$
– Shannon
Jan 5 '18 at 13:46
$begingroup$
I can see that, but I am confused because I followed this youtu.be/bZred_Ksz2k?t=220
$endgroup$
– Shannon
Jan 5 '18 at 13:46
1
1
$begingroup$
If you use that you must know that $({x^{3}})^{1/3}=x$
$endgroup$
– user508685
Jan 5 '18 at 13:56
$begingroup$
If you use that you must know that $({x^{3}})^{1/3}=x$
$endgroup$
– user508685
Jan 5 '18 at 13:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Another way to think about it.
If $f:Xto Y$ is a function then for every $yin Y$ we have the set $f^{-1}({y}):={xin Xmid f(x)=y}$.
(non-empty subsets of this form are the so-called fibers of $f$ and form a partition of $X$)
Based on that you can say:
- $f$ is injective iff $f^{-1}({y})$ has at most one element for every $yin Y$.
- $f$ is surjective iff $f^{-1}({y})$ has at least one element for every $yin Y$.
So checking a function on injectivity and/or surjectivity boils down to checking how the sets $f^{-1}({y})$ behave.
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
I think you understood the concept of surjection but not injection.
A function $h$ is injective if $h(a)=h(b)$ implies that $a=b$. Equivalently, if $a neq b$ then $h(a) neq h(b)$.
$f$ is injective because $f(-1)=-1, f(0)=0, f(1)=1$, we can see that the images are distinct. Hence it is injective.
$g(0)=1$ and $g(1)=4$, again, it is injective.
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 14:12
$begingroup$
But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:55
1
$begingroup$
then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 15:17
$begingroup$
I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
$endgroup$
– Shannon
Jan 5 '18 at 21:20
|
show 1 more comment
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Another way to think about it.
If $f:Xto Y$ is a function then for every $yin Y$ we have the set $f^{-1}({y}):={xin Xmid f(x)=y}$.
(non-empty subsets of this form are the so-called fibers of $f$ and form a partition of $X$)
Based on that you can say:
- $f$ is injective iff $f^{-1}({y})$ has at most one element for every $yin Y$.
- $f$ is surjective iff $f^{-1}({y})$ has at least one element for every $yin Y$.
So checking a function on injectivity and/or surjectivity boils down to checking how the sets $f^{-1}({y})$ behave.
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Another way to think about it.
If $f:Xto Y$ is a function then for every $yin Y$ we have the set $f^{-1}({y}):={xin Xmid f(x)=y}$.
(non-empty subsets of this form are the so-called fibers of $f$ and form a partition of $X$)
Based on that you can say:
- $f$ is injective iff $f^{-1}({y})$ has at most one element for every $yin Y$.
- $f$ is surjective iff $f^{-1}({y})$ has at least one element for every $yin Y$.
So checking a function on injectivity and/or surjectivity boils down to checking how the sets $f^{-1}({y})$ behave.
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Another way to think about it.
If $f:Xto Y$ is a function then for every $yin Y$ we have the set $f^{-1}({y}):={xin Xmid f(x)=y}$.
(non-empty subsets of this form are the so-called fibers of $f$ and form a partition of $X$)
Based on that you can say:
- $f$ is injective iff $f^{-1}({y})$ has at most one element for every $yin Y$.
- $f$ is surjective iff $f^{-1}({y})$ has at least one element for every $yin Y$.
So checking a function on injectivity and/or surjectivity boils down to checking how the sets $f^{-1}({y})$ behave.
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
$endgroup$
Another way to think about it.
If $f:Xto Y$ is a function then for every $yin Y$ we have the set $f^{-1}({y}):={xin Xmid f(x)=y}$.
(non-empty subsets of this form are the so-called fibers of $f$ and form a partition of $X$)
Based on that you can say:
- $f$ is injective iff $f^{-1}({y})$ has at most one element for every $yin Y$.
- $f$ is surjective iff $f^{-1}({y})$ has at least one element for every $yin Y$.
So checking a function on injectivity and/or surjectivity boils down to checking how the sets $f^{-1}({y})$ behave.
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
answered Jan 5 '18 at 14:21
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
I think you understood the concept of surjection but not injection.
A function $h$ is injective if $h(a)=h(b)$ implies that $a=b$. Equivalently, if $a neq b$ then $h(a) neq h(b)$.
$f$ is injective because $f(-1)=-1, f(0)=0, f(1)=1$, we can see that the images are distinct. Hence it is injective.
$g(0)=1$ and $g(1)=4$, again, it is injective.
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 14:12
$begingroup$
But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:55
1
$begingroup$
then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 15:17
$begingroup$
I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
$endgroup$
– Shannon
Jan 5 '18 at 21:20
|
show 1 more comment
$begingroup$
I think you understood the concept of surjection but not injection.
A function $h$ is injective if $h(a)=h(b)$ implies that $a=b$. Equivalently, if $a neq b$ then $h(a) neq h(b)$.
$f$ is injective because $f(-1)=-1, f(0)=0, f(1)=1$, we can see that the images are distinct. Hence it is injective.
$g(0)=1$ and $g(1)=4$, again, it is injective.
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 14:12
$begingroup$
But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:55
1
$begingroup$
then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 15:17
$begingroup$
I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
$endgroup$
– Shannon
Jan 5 '18 at 21:20
|
show 1 more comment
$begingroup$
I think you understood the concept of surjection but not injection.
A function $h$ is injective if $h(a)=h(b)$ implies that $a=b$. Equivalently, if $a neq b$ then $h(a) neq h(b)$.
$f$ is injective because $f(-1)=-1, f(0)=0, f(1)=1$, we can see that the images are distinct. Hence it is injective.
$g(0)=1$ and $g(1)=4$, again, it is injective.
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
I think you understood the concept of surjection but not injection.
A function $h$ is injective if $h(a)=h(b)$ implies that $a=b$. Equivalently, if $a neq b$ then $h(a) neq h(b)$.
$f$ is injective because $f(-1)=-1, f(0)=0, f(1)=1$, we can see that the images are distinct. Hence it is injective.
$g(0)=1$ and $g(1)=4$, again, it is injective.
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 5 '18 at 13:56
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 14:12
$begingroup$
But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:55
1
$begingroup$
then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 15:17
$begingroup$
I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
$endgroup$
– Shannon
Jan 5 '18 at 21:20
|
show 1 more comment
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 14:12
$begingroup$
But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:55
1
$begingroup$
then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 15:17
$begingroup$
I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
$endgroup$
– Shannon
Jan 5 '18 at 21:20
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
So to check if it is injective, I should put every value from the domain into the function, and then check that all the outputs are both unique from one-another, and, that all the outputs match up to a value in the co-domain?
$endgroup$
– Shannon
Jan 5 '18 at 14:09
$begingroup$
I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
$endgroup$
– Siong Thye Goh
Jan 5 '18 at 14:12
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I prefer using $h(a)=h(b)$ implies $a=b$ actually. Also, for this question, $f$ and $g$ are increasing function, hence they are injective.
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– Siong Thye Goh
Jan 5 '18 at 14:12
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But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
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– Shannon
Jan 5 '18 at 14:55
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But lets say from the example "$g(0) = 1$ and $g(1) = 4$", but instead imagine it was $g(0) = 1$ and $g(1) = 6$, then would it still be injective even though $6$ isn't in the co-domain?
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– Shannon
Jan 5 '18 at 14:55
1
1
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then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
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– Siong Thye Goh
Jan 5 '18 at 15:17
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then we have an even bigger issue... the function wasn't defined properly. First, we define the function properly, state the domain, the codomain, and the rule. After the function is well defined, then we can talk about whether it is injective or surjective.
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– Siong Thye Goh
Jan 5 '18 at 15:17
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I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
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– Shannon
Jan 5 '18 at 21:20
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I understand that if $g(1) = 6$ then the function would not be properly formatted. I am just asking, hypothetically, would the function still be injective if $g(1) = 6$, but $6$ is outside of the codomain (i.e. the codomain is specified as being {$1,2,3,4,5$}?
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– Shannon
Jan 5 '18 at 21:20
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$begingroup$
@LinkingPark but because we must $sqrt[3]{x^3}$ does that not mean that $x$ can be either positive or negative, and therefore $pm x$ does not imply one another?
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– Shannon
Jan 5 '18 at 13:37
1
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You can easily see that $f(-1)=-1 ,f(1)=1,f(0)=0$ and then you can see that this is bijective mapping (surjective and injective).I dont understand you way of thinking.
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– user508685
Jan 5 '18 at 13:41
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I can see that, but I am confused because I followed this youtu.be/bZred_Ksz2k?t=220
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– Shannon
Jan 5 '18 at 13:46
1
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If you use that you must know that $({x^{3}})^{1/3}=x$
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– user508685
Jan 5 '18 at 13:56