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How to prove one-to-one and onto?
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up vote
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I am trying to prove $(-1, 1)$ is equinumerous to $mathbbR$ by finding a bijection function from $(-1, 1)$ to $mathbbR$.
I've found a function $f(x) = fracxx^2-1$, and it's a bijection obviously from its graph, but I don't know how to prove it is bijection rigorously. Any help, thanks!
linear-algebra
edited Jul 26 at 19:50
packetpacket
249112
asked Jul 26 at 19:45
Cathy
1067
add a comment |Â
up vote
4
down vote
favorite
I am trying to prove $(-1, 1)$ is equinumerous to $mathbbR$ by finding a bijection function from $(-1, 1)$ to $mathbbR$.
I've found a function $f(x) = fracxx^2-1$, and it's a bijection obviously from its graph, but I don't know how to prove it is bijection rigorously. Any help, thanks!
linear-algebra
edited Jul 26 at 19:50
packetpacket
249112
asked Jul 26 at 19:45
Cathy
1067
You have to show $f$ is one-to-one and onto. Do you know how to prove a function is one-to-one? Do you know how to go about proving a functions is onto?
– saulspatz
Jul 26 at 19:51
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to prove $(-1, 1)$ is equinumerous to $mathbbR$ by finding a bijection function from $(-1, 1)$ to $mathbbR$.
I've found a function $f(x) = fracxx^2-1$, and it's a bijection obviously from its graph, but I don't know how to prove it is bijection rigorously. Any help, thanks!
linear-algebra
edited Jul 26 at 19:50
packetpacket
249112
asked Jul 26 at 19:45
Cathy
1067
I am trying to prove $(-1, 1)$ is equinumerous to $mathbbR$ by finding a bijection function from $(-1, 1)$ to $mathbbR$.
I've found a function $f(x) = fracxx^2-1$, and it's a bijection obviously from its graph, but I don't know how to prove it is bijection rigorously. Any help, thanks!
linear-algebra
edited Jul 26 at 19:50
packetpacket
249112
asked Jul 26 at 19:45
Cathy
1067
edited Jul 26 at 19:50
packetpacket
249112
edited Jul 26 at 19:50
packetpacket
249112
edited Jul 26 at 19:50
packetpacket
249112
249112
asked Jul 26 at 19:45
Cathy
1067
asked Jul 26 at 19:45
Cathy
1067
asked Jul 26 at 19:45
Cathy
1067
1067
You have to show $f$ is one-to-one and onto. Do you know how to prove a function is one-to-one? Do you know how to go about proving a functions is onto?
– saulspatz
Jul 26 at 19:51
add a comment |Â
You have to show $f$ is one-to-one and onto. Do you know how to prove a function is one-to-one? Do you know how to go about proving a functions is onto?
– saulspatz
Jul 26 at 19:51
You have to show $f$ is one-to-one and onto. Do you know how to prove a function is one-to-one? Do you know how to go about proving a functions is onto?
– saulspatz
Jul 26 at 19:51
You have to show $f$ is one-to-one and onto. Do you know how to prove a function is one-to-one? Do you know how to go about proving a functions is onto?
– saulspatz
Jul 26 at 19:51
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
The given function is continous and because of $$(fracxx^2-1)'=fracx^2-1-2x^2(x^2-1)^2=frac-1-x^2(x^2-1)^2<0$$ strictly decreasing in the interval $(-1,1)$ Additionally, we have $$lim_xrightarrow -1+0 fracxx^2-1=infty$$ and $$lim_xrightarrow 1-0 fracxx^2-1=-infty$$
So the range is $mathbb R$ , hence $f^-1(x)$ exists and is defined everywhere.
answered Jul 26 at 19:55
Peter
45k938119
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
add a comment |Â
up vote
7
down vote
Here is a simple proof that $f$ is injective without calculus:
$$
f(x) = f(y)
iff fracxx^2-1 = fracyy^2-1
iff (xy+1)(x-y)=0
iff xy=-1 text or x=y
$$
But $|x|<1$, $|y|<1$ implies $|xy|=|x| |y| <1$ and so $xy$ is never $-1$.
answered Jul 26 at 20:04
lhf
155k9160365
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
add a comment |Â
up vote
4
down vote
Injective: compute the derivative. The function is derivable in the whole domain.
It is $fracx^2-1-xcdot 2x(x^2-1)^2= frac-1-x^2(x^2-1)^2<0$ for all $x$ in the domain. So the function is in fact strictly monotone decreasing.
Surjective: Compute the limits at the endpoints of the interval.
$limlimits_xrightarrow -1 fracxx^2-1= infty$, $limlimits_xrightarrow 1 fracxx^2-1= -infty$.
As the function is continuous, every real value is attained somewhere by the Bolzano theorem.
answered Jul 26 at 19:55
A. Pongrácz
1,574115
add a comment |Â
up vote
4
down vote
To show that a function is a bijection you can simply show that there is an inverse function $f^-1:mathbbRrightarrow(-1,1)$ which can be found by setting
beginalign
x & =fracyy^2-1
\ y&=frac1pmsqrt1+4x^22x
endalign
We take the negative sign to stay within the domain, so we obtain $f^-1(x)=frac1-sqrt1+4x^22x$.
Lastly, we can confirm that $f^-1circ f(x)=x$ for all $xin(-1,1)$ and similarly that $fcirc f^-1(x)=x$ for all $xinmathbbR$.
edited Jul 28 at 0:50
an4s
2,0382417
answered Jul 26 at 19:56
高田航
1,116318
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The given function is continous and because of $$(fracxx^2-1)'=fracx^2-1-2x^2(x^2-1)^2=frac-1-x^2(x^2-1)^2<0$$ strictly decreasing in the interval $(-1,1)$ Additionally, we have $$lim_xrightarrow -1+0 fracxx^2-1=infty$$ and $$lim_xrightarrow 1-0 fracxx^2-1=-infty$$
So the range is $mathbb R$ , hence $f^-1(x)$ exists and is defined everywhere.
answered Jul 26 at 19:55
Peter
45k938119
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
add a comment |Â
up vote
3
down vote
accepted
The given function is continous and because of $$(fracxx^2-1)'=fracx^2-1-2x^2(x^2-1)^2=frac-1-x^2(x^2-1)^2<0$$ strictly decreasing in the interval $(-1,1)$ Additionally, we have $$lim_xrightarrow -1+0 fracxx^2-1=infty$$ and $$lim_xrightarrow 1-0 fracxx^2-1=-infty$$
So the range is $mathbb R$ , hence $f^-1(x)$ exists and is defined everywhere.
answered Jul 26 at 19:55
Peter
45k938119
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The given function is continous and because of $$(fracxx^2-1)'=fracx^2-1-2x^2(x^2-1)^2=frac-1-x^2(x^2-1)^2<0$$ strictly decreasing in the interval $(-1,1)$ Additionally, we have $$lim_xrightarrow -1+0 fracxx^2-1=infty$$ and $$lim_xrightarrow 1-0 fracxx^2-1=-infty$$
So the range is $mathbb R$ , hence $f^-1(x)$ exists and is defined everywhere.
answered Jul 26 at 19:55
Peter
45k938119
The given function is continous and because of $$(fracxx^2-1)'=fracx^2-1-2x^2(x^2-1)^2=frac-1-x^2(x^2-1)^2<0$$ strictly decreasing in the interval $(-1,1)$ Additionally, we have $$lim_xrightarrow -1+0 fracxx^2-1=infty$$ and $$lim_xrightarrow 1-0 fracxx^2-1=-infty$$
So the range is $mathbb R$ , hence $f^-1(x)$ exists and is defined everywhere.
answered Jul 26 at 19:55
Peter
45k938119
answered Jul 26 at 19:55
Peter
45k938119
answered Jul 26 at 19:55
Peter
45k938119
answered Jul 26 at 19:55
Peter
45k938119
45k938119
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
add a comment |Â
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
So we could prove a function is bijection by showing it has inverse rather than to prove step by step according to the definition of bijection? Such as proving injection is to show if f(a) = f(b) then a = b?
– Cathy
Jul 26 at 20:04
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
Yes, but domain and range need not be the same. Here the bijection is between the interval $(-1,1)$ and $mathbb R$. If a function is continous , you can use the intermediate-value-theorem that guarantees that every value in the range is a possible value of the function.
– Peter
Jul 26 at 20:05
add a comment |Â
up vote
7
down vote
Here is a simple proof that $f$ is injective without calculus:
$$
f(x) = f(y)
iff fracxx^2-1 = fracyy^2-1
iff (xy+1)(x-y)=0
iff xy=-1 text or x=y
$$
But $|x|<1$, $|y|<1$ implies $|xy|=|x| |y| <1$ and so $xy$ is never $-1$.
answered Jul 26 at 20:04
lhf
155k9160365
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
add a comment |Â
up vote
7
down vote
Here is a simple proof that $f$ is injective without calculus:
$$
f(x) = f(y)
iff fracxx^2-1 = fracyy^2-1
iff (xy+1)(x-y)=0
iff xy=-1 text or x=y
$$
But $|x|<1$, $|y|<1$ implies $|xy|=|x| |y| <1$ and so $xy$ is never $-1$.
answered Jul 26 at 20:04
lhf
155k9160365
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Here is a simple proof that $f$ is injective without calculus:
$$
f(x) = f(y)
iff fracxx^2-1 = fracyy^2-1
iff (xy+1)(x-y)=0
iff xy=-1 text or x=y
$$
But $|x|<1$, $|y|<1$ implies $|xy|=|x| |y| <1$ and so $xy$ is never $-1$.
answered Jul 26 at 20:04
lhf
155k9160365
Here is a simple proof that $f$ is injective without calculus:
$$
f(x) = f(y)
iff fracxx^2-1 = fracyy^2-1
iff (xy+1)(x-y)=0
iff xy=-1 text or x=y
$$
But $|x|<1$, $|y|<1$ implies $|xy|=|x| |y| <1$ and so $xy$ is never $-1$.
answered Jul 26 at 20:04
lhf
155k9160365
edited Jul 26 at 23:18
answered Jul 26 at 20:04
lhf
155k9160365
answered Jul 26 at 20:04
lhf
155k9160365
answered Jul 26 at 20:04
lhf
155k9160365
155k9160365
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
add a comment |Â
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
Thanks! I've never thought the question can be proved without continuous. Can surjection be proved in a similar way?
– Cathy
Jul 26 at 20:12
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
@Cathy, the answer by 高田航 does that.
– lhf
Jul 26 at 23:17
add a comment |Â
up vote
4
down vote
Injective: compute the derivative. The function is derivable in the whole domain.
It is $fracx^2-1-xcdot 2x(x^2-1)^2= frac-1-x^2(x^2-1)^2<0$ for all $x$ in the domain. So the function is in fact strictly monotone decreasing.
Surjective: Compute the limits at the endpoints of the interval.
$limlimits_xrightarrow -1 fracxx^2-1= infty$, $limlimits_xrightarrow 1 fracxx^2-1= -infty$.
As the function is continuous, every real value is attained somewhere by the Bolzano theorem.
answered Jul 26 at 19:55
A. Pongrácz
1,574115
add a comment |Â
up vote
4
down vote
Injective: compute the derivative. The function is derivable in the whole domain.
It is $fracx^2-1-xcdot 2x(x^2-1)^2= frac-1-x^2(x^2-1)^2<0$ for all $x$ in the domain. So the function is in fact strictly monotone decreasing.
Surjective: Compute the limits at the endpoints of the interval.
$limlimits_xrightarrow -1 fracxx^2-1= infty$, $limlimits_xrightarrow 1 fracxx^2-1= -infty$.
As the function is continuous, every real value is attained somewhere by the Bolzano theorem.
answered Jul 26 at 19:55
A. Pongrácz
1,574115
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Injective: compute the derivative. The function is derivable in the whole domain.
It is $fracx^2-1-xcdot 2x(x^2-1)^2= frac-1-x^2(x^2-1)^2<0$ for all $x$ in the domain. So the function is in fact strictly monotone decreasing.
Surjective: Compute the limits at the endpoints of the interval.
$limlimits_xrightarrow -1 fracxx^2-1= infty$, $limlimits_xrightarrow 1 fracxx^2-1= -infty$.
As the function is continuous, every real value is attained somewhere by the Bolzano theorem.
answered Jul 26 at 19:55
A. Pongrácz
1,574115
Injective: compute the derivative. The function is derivable in the whole domain.
It is $fracx^2-1-xcdot 2x(x^2-1)^2= frac-1-x^2(x^2-1)^2<0$ for all $x$ in the domain. So the function is in fact strictly monotone decreasing.
Surjective: Compute the limits at the endpoints of the interval.
$limlimits_xrightarrow -1 fracxx^2-1= infty$, $limlimits_xrightarrow 1 fracxx^2-1= -infty$.
As the function is continuous, every real value is attained somewhere by the Bolzano theorem.
answered Jul 26 at 19:55
A. Pongrácz
1,574115
answered Jul 26 at 19:55
A. Pongrácz
1,574115
answered Jul 26 at 19:55
A. Pongrácz
1,574115
answered Jul 26 at 19:55
A. Pongrácz
1,574115
1,574115
add a comment |Â
add a comment |Â
up vote
4
down vote
To show that a function is a bijection you can simply show that there is an inverse function $f^-1:mathbbRrightarrow(-1,1)$ which can be found by setting
beginalign
x & =fracyy^2-1
\ y&=frac1pmsqrt1+4x^22x
endalign
We take the negative sign to stay within the domain, so we obtain $f^-1(x)=frac1-sqrt1+4x^22x$.
Lastly, we can confirm that $f^-1circ f(x)=x$ for all $xin(-1,1)$ and similarly that $fcirc f^-1(x)=x$ for all $xinmathbbR$.
edited Jul 28 at 0:50
an4s
2,0382417
answered Jul 26 at 19:56
高田航
1,116318
add a comment |Â
up vote
4
down vote
To show that a function is a bijection you can simply show that there is an inverse function $f^-1:mathbbRrightarrow(-1,1)$ which can be found by setting
beginalign
x & =fracyy^2-1
\ y&=frac1pmsqrt1+4x^22x
endalign
We take the negative sign to stay within the domain, so we obtain $f^-1(x)=frac1-sqrt1+4x^22x$.
Lastly, we can confirm that $f^-1circ f(x)=x$ for all $xin(-1,1)$ and similarly that $fcirc f^-1(x)=x$ for all $xinmathbbR$.
edited Jul 28 at 0:50
an4s
2,0382417
answered Jul 26 at 19:56
高田航
1,116318
add a comment |Â
up vote
4
down vote
up vote
4
down vote
To show that a function is a bijection you can simply show that there is an inverse function $f^-1:mathbbRrightarrow(-1,1)$ which can be found by setting
beginalign
x & =fracyy^2-1
\ y&=frac1pmsqrt1+4x^22x
endalign
We take the negative sign to stay within the domain, so we obtain $f^-1(x)=frac1-sqrt1+4x^22x$.
Lastly, we can confirm that $f^-1circ f(x)=x$ for all $xin(-1,1)$ and similarly that $fcirc f^-1(x)=x$ for all $xinmathbbR$.
edited Jul 28 at 0:50
an4s
2,0382417
answered Jul 26 at 19:56
高田航
1,116318
To show that a function is a bijection you can simply show that there is an inverse function $f^-1:mathbbRrightarrow(-1,1)$ which can be found by setting
beginalign
x & =fracyy^2-1
\ y&=frac1pmsqrt1+4x^22x
endalign
We take the negative sign to stay within the domain, so we obtain $f^-1(x)=frac1-sqrt1+4x^22x$.
Lastly, we can confirm that $f^-1circ f(x)=x$ for all $xin(-1,1)$ and similarly that $fcirc f^-1(x)=x$ for all $xinmathbbR$.
edited Jul 28 at 0:50
an4s
2,0382417
answered Jul 26 at 19:56
高田航
1,116318
edited Jul 28 at 0:50
an4s
2,0382417
edited Jul 28 at 0:50
an4s
2,0382417
edited Jul 28 at 0:50
an4s
2,0382417
2,0382417
answered Jul 26 at 19:56
高田航
1,116318
answered Jul 26 at 19:56
高田航
1,116318
answered Jul 26 at 19:56
高田航
1,116318
1,116318
add a comment |Â
add a comment |Â
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You have to show $f$ is one-to-one and onto. Do you know how to prove a function is one-to-one? Do you know how to go about proving a functions is onto?
– saulspatz
Jul 26 at 19:51