Determine if $y = x^2$ is injective
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1
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I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).
However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.
Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.
This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.
However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.
Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
functions
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show 2 more comments
up vote
1
down vote
favorite
I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).
However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.
Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.
This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.
However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.
Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
functions
You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48
The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58
It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59
Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11
1
You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).
However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.
Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.
This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.
However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.
Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
functions
I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example).
However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.
Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.
This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.
However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.
Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
functions
functions
edited Feb 10 at 22:36
N. F. Taussig
43.2k93254
43.2k93254
asked Sep 9 '16 at 2:45
Ham Sandwich
226316
226316
You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48
The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58
It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59
Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11
1
You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17
|
show 2 more comments
You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48
The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58
It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59
Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11
1
You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17
You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48
You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48
The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58
The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58
It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59
It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59
Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11
Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11
1
1
You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17
You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17
|
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
add a comment |
up vote
1
down vote
to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.
First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
add a comment |
up vote
1
down vote
Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.
If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.
The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.
It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.
Let me take an example. Lets show that $f(x) = x^3$ is injective.
We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.
The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
add a comment |
up vote
1
down vote
accepted
The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x neq y$. The contrapositive fails as well because you have $x neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
answered Sep 9 '16 at 2:53
Ross Millikan
290k23195368
290k23195368
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add a comment |
up vote
1
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to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.
First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
add a comment |
up vote
1
down vote
to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.
First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
add a comment |
up vote
1
down vote
up vote
1
down vote
to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.
to prove that if $f(x)=x^2$ is injective you have to check that if $x_1=x_2 Rightarrow f(x_1)=f(x_2)$ but this isn't the case because if $x_1=1$ and $x_2=-1 Rightarrow f(x_1)=f(x_2)$ yet $x_1 neq x_2$, making $f(x)$ not injective.
answered Sep 9 '16 at 3:17
R.Mor
728
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First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
add a comment |
First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
First sentence: "no". You need to prove that $f(x_1)=f(x_2)implies x_1=x_2$
– user228113
Sep 9 '16 at 3:36
add a comment |
up vote
1
down vote
Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.
If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.
The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.
It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.
Let me take an example. Lets show that $f(x) = x^3$ is injective.
We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.
The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.
add a comment |
up vote
1
down vote
Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.
If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.
The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.
It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.
Let me take an example. Lets show that $f(x) = x^3$ is injective.
We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.
The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.
add a comment |
up vote
1
down vote
up vote
1
down vote
Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.
If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.
The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.
It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.
Let me take an example. Lets show that $f(x) = x^3$ is injective.
We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.
The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.
Thats right. As you say $-1,1$ both map on $1$ under the function of $x^2$. This means that $f$ cant be injective. The definition you had in class pretty much does the same. If you have two values like $x=-1$ and $y=1$ with property of $f(x) = f(y) = 1$ them $f$ cant be injective because two different values are mapping onto the same value.
If you take general unknown $x$ and $y$ and say that theyre having the property of $f(x) = f(y)$ then it has to follow that $x = y$. This means that the general unknown $x,y$ you have picked are actually the same. So you did not find any two values with the same value under $f$. This means that $f$ is injective.
The other definition is just the other way around. Like $A Rightarrow B$ is equal to $neg B Rightarrow neg A$.
It does not matter which way you are going. Both will work. After time you will get a feeling which one works the best to prove.
Let me take an example. Lets show that $f(x) = x^3$ is injective.
We take general $x,y in mathbb{R}$. For them we say that $f(x) = f(y)$. Then we know what $f$ does to them and we will get $x^3 = y^3$. Applying the third-root will give us $x=y$. That means that $f(x) = x^3$ is injective.
The same argument wont work with $f(x)=x^2$. You have to careful applying the square root ok both sides. What you did is $sqrt{y} = x$ but that not enough. You also get $sqrt{y} = - x$. For example $sqrt{y = 1} = pm 1$ which makes perfectly sense because both $x = -1$ and $x = 1$ are mapping onto $y = 1$. This is where you might messed up something.
answered Nov 23 at 9:59
Arjihad
378111
378111
add a comment |
add a comment |
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You want to find value of $x,y$ such that $f(x)=f(y)$. Then you have already shown that it is not injective by counterexample.
– lEm
Sep 9 '16 at 2:48
The meaning of "$y$" changes throughout the text you've written. $f$ has not really been defined, and it gets consistently used with two different meanings in the second and third paragraph. If you want my opinion, you'll have a hard time reconciling what your book says with whatever you're trying to say, since your book and you use two different notations.
– user228113
Sep 9 '16 at 2:58
It is however true that the function $$g : [0,infty)to [0,infty)$$$$g(h)=h^2$$ is bijective. Its inverse function is called $sqrt{bullet}$.
– user228113
Sep 9 '16 at 2:59
Okay, so "y" is simply another point, so if y = -1, that means that (-1)^2 = 1, so f(y) = 1. It's the same as f(x1), f(x2). Right?
– Ham Sandwich
Sep 9 '16 at 3:11
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You consistently write sentences where $f(2)=2^2$ is immediately followed by $f(4)=sqrt4$.
– user228113
Sep 9 '16 at 3:17