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Bijective function excluding 0
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Construct a bijective function $f:mathbb Rtomathbb Rsetminus0$. Prove that the function is bijective.
Im having trouble with this... A few concepts that I get so far
is that function essentially should be representing a set of real numbers, R, which in turn maps to the same set of real numbers excluding 0. Im thinking about proving its bijective by proving the function is both injective and surjective. Im thinking about doing x^2 + 1 but am uncertain if this is the right sort of track to take. Would really appreciate guidance
discrete-mathematics
edited 27 mins ago
Parcly Taxel
34.5k136991
asked 29 mins ago
sphynx888
624
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sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
Construct a bijective function $f:mathbb Rtomathbb Rsetminus0$. Prove that the function is bijective.
Im having trouble with this... A few concepts that I get so far
is that function essentially should be representing a set of real numbers, R, which in turn maps to the same set of real numbers excluding 0. Im thinking about proving its bijective by proving the function is both injective and surjective. Im thinking about doing x^2 + 1 but am uncertain if this is the right sort of track to take. Would really appreciate guidance
discrete-mathematics
edited 27 mins ago
Parcly Taxel
34.5k136991
asked 29 mins ago
sphynx888
624
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
When you see a cardinality question like this you should suspect that you won't find a nice formula for the function you need. You will probably need something weird, defined with cases. There are several answers here that use that strategy - it's worth learning.
– Ethan Bolker
16 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Construct a bijective function $f:mathbb Rtomathbb Rsetminus0$. Prove that the function is bijective.
Im having trouble with this... A few concepts that I get so far
is that function essentially should be representing a set of real numbers, R, which in turn maps to the same set of real numbers excluding 0. Im thinking about proving its bijective by proving the function is both injective and surjective. Im thinking about doing x^2 + 1 but am uncertain if this is the right sort of track to take. Would really appreciate guidance
discrete-mathematics
edited 27 mins ago
Parcly Taxel
34.5k136991
asked 29 mins ago
sphynx888
624
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Construct a bijective function $f:mathbb Rtomathbb Rsetminus0$. Prove that the function is bijective.
Im having trouble with this... A few concepts that I get so far
is that function essentially should be representing a set of real numbers, R, which in turn maps to the same set of real numbers excluding 0. Im thinking about proving its bijective by proving the function is both injective and surjective. Im thinking about doing x^2 + 1 but am uncertain if this is the right sort of track to take. Would really appreciate guidance
discrete-mathematics
discrete-mathematics
edited 27 mins ago
Parcly Taxel
34.5k136991
asked 29 mins ago
sphynx888
624
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 27 mins ago
Parcly Taxel
34.5k136991
asked 29 mins ago
sphynx888
624
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 27 mins ago
Parcly Taxel
34.5k136991
edited 27 mins ago
Parcly Taxel
34.5k136991
edited 27 mins ago
Parcly Taxel
34.5k136991
34.5k136991
asked 29 mins ago
sphynx888
624
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 29 mins ago
sphynx888
624
asked 29 mins ago
sphynx888
624
624
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sphynx888 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
When you see a cardinality question like this you should suspect that you won't find a nice formula for the function you need. You will probably need something weird, defined with cases. There are several answers here that use that strategy - it's worth learning.
– Ethan Bolker
16 mins ago
add a comment |Â
1
When you see a cardinality question like this you should suspect that you won't find a nice formula for the function you need. You will probably need something weird, defined with cases. There are several answers here that use that strategy - it's worth learning.
– Ethan Bolker
16 mins ago
1
1
When you see a cardinality question like this you should suspect that you won't find a nice formula for the function you need. You will probably need something weird, defined with cases. There are several answers here that use that strategy - it's worth learning.
– Ethan Bolker
16 mins ago
When you see a cardinality question like this you should suspect that you won't find a nice formula for the function you need. You will probably need something weird, defined with cases. There are several answers here that use that strategy - it's worth learning.
– Ethan Bolker
16 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
The classic natural number shifting argument works:
$$f(x)=begincasesx+1&xinmathbb N\x&textelseendcases$$
(Here $0inmathbb N$.)
Suppose $f(a)=f(b)$. Then either this value is not a positive integer, in which case $a=b$ (since the first case only produces positive integers), or it is a positive integer, whence $a=b=f(a)-1=f(b)-1$ again. Hence $f$ is injective.
Real numbers that are not positive integers are the images of themselves, while the positive integers have as image their respective predecessors. Thus $f$ is also surjective, hence bijective.
The inverse function is
$$f^-1=begincasesx-1&xinmathbb N,xne0\x&textelseendcases$$
answered 17 mins ago
Parcly Taxel
34.5k136991
add a comment |Â
up vote
2
down vote
Hint
Consider $A=ninmathbbNcup 0.$ Define:
$xin mathbbRsetminus Aimplies f(x)=x;$
$f(0)=1;$- $f(1)=1/2;$
- $f(1/2)=1/3;$
answered 21 mins ago
mfl
25.4k12141
add a comment |Â
up vote
2
down vote
Here's a function I would use. Let $mathbbN$ be the set of natural numbers ($0,1,2,3...$) including $0$
$$f(x) = begincases x+1&xinmathbbN\x&xnotinmathbb Nendcases$$
It's quite trivial to see that $f:;mathbbRtomathbbRsetminus0$. Showing injectivity and surjectivity with this piecewise function is quite easy, so give it a shot!
answered 20 mins ago
Rushabh Mehta
3,120225
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
The classic natural number shifting argument works:
$$f(x)=begincasesx+1&xinmathbb N\x&textelseendcases$$
(Here $0inmathbb N$.)
Suppose $f(a)=f(b)$. Then either this value is not a positive integer, in which case $a=b$ (since the first case only produces positive integers), or it is a positive integer, whence $a=b=f(a)-1=f(b)-1$ again. Hence $f$ is injective.
Real numbers that are not positive integers are the images of themselves, while the positive integers have as image their respective predecessors. Thus $f$ is also surjective, hence bijective.
The inverse function is
$$f^-1=begincasesx-1&xinmathbb N,xne0\x&textelseendcases$$
answered 17 mins ago
Parcly Taxel
34.5k136991
add a comment |Â
up vote
3
down vote
The classic natural number shifting argument works:
$$f(x)=begincasesx+1&xinmathbb N\x&textelseendcases$$
(Here $0inmathbb N$.)
Suppose $f(a)=f(b)$. Then either this value is not a positive integer, in which case $a=b$ (since the first case only produces positive integers), or it is a positive integer, whence $a=b=f(a)-1=f(b)-1$ again. Hence $f$ is injective.
Real numbers that are not positive integers are the images of themselves, while the positive integers have as image their respective predecessors. Thus $f$ is also surjective, hence bijective.
The inverse function is
$$f^-1=begincasesx-1&xinmathbb N,xne0\x&textelseendcases$$
answered 17 mins ago
Parcly Taxel
34.5k136991
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The classic natural number shifting argument works:
$$f(x)=begincasesx+1&xinmathbb N\x&textelseendcases$$
(Here $0inmathbb N$.)
Suppose $f(a)=f(b)$. Then either this value is not a positive integer, in which case $a=b$ (since the first case only produces positive integers), or it is a positive integer, whence $a=b=f(a)-1=f(b)-1$ again. Hence $f$ is injective.
Real numbers that are not positive integers are the images of themselves, while the positive integers have as image their respective predecessors. Thus $f$ is also surjective, hence bijective.
The inverse function is
$$f^-1=begincasesx-1&xinmathbb N,xne0\x&textelseendcases$$
answered 17 mins ago
Parcly Taxel
34.5k136991
The classic natural number shifting argument works:
$$f(x)=begincasesx+1&xinmathbb N\x&textelseendcases$$
(Here $0inmathbb N$.)
Suppose $f(a)=f(b)$. Then either this value is not a positive integer, in which case $a=b$ (since the first case only produces positive integers), or it is a positive integer, whence $a=b=f(a)-1=f(b)-1$ again. Hence $f$ is injective.
Real numbers that are not positive integers are the images of themselves, while the positive integers have as image their respective predecessors. Thus $f$ is also surjective, hence bijective.
The inverse function is
$$f^-1=begincasesx-1&xinmathbb N,xne0\x&textelseendcases$$
answered 17 mins ago
Parcly Taxel
34.5k136991
answered 17 mins ago
Parcly Taxel
34.5k136991
answered 17 mins ago
Parcly Taxel
34.5k136991
answered 17 mins ago
Parcly Taxel
34.5k136991
34.5k136991
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint
Consider $A=ninmathbbNcup 0.$ Define:
$xin mathbbRsetminus Aimplies f(x)=x;$
$f(0)=1;$- $f(1)=1/2;$
- $f(1/2)=1/3;$
answered 21 mins ago
mfl
25.4k12141
add a comment |Â
up vote
2
down vote
Hint
Consider $A=ninmathbbNcup 0.$ Define:
$xin mathbbRsetminus Aimplies f(x)=x;$
$f(0)=1;$- $f(1)=1/2;$
- $f(1/2)=1/3;$
answered 21 mins ago
mfl
25.4k12141
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint
Consider $A=ninmathbbNcup 0.$ Define:
$xin mathbbRsetminus Aimplies f(x)=x;$
$f(0)=1;$- $f(1)=1/2;$
- $f(1/2)=1/3;$
answered 21 mins ago
mfl
25.4k12141
Hint
Consider $A=ninmathbbNcup 0.$ Define:
$xin mathbbRsetminus Aimplies f(x)=x;$
$f(0)=1;$- $f(1)=1/2;$
- $f(1/2)=1/3;$
answered 21 mins ago
mfl
25.4k12141
answered 21 mins ago
mfl
25.4k12141
answered 21 mins ago
mfl
25.4k12141
answered 21 mins ago
mfl
25.4k12141
25.4k12141
add a comment |Â
add a comment |Â
up vote
2
down vote
Here's a function I would use. Let $mathbbN$ be the set of natural numbers ($0,1,2,3...$) including $0$
$$f(x) = begincases x+1&xinmathbbN\x&xnotinmathbb Nendcases$$
It's quite trivial to see that $f:;mathbbRtomathbbRsetminus0$. Showing injectivity and surjectivity with this piecewise function is quite easy, so give it a shot!
answered 20 mins ago
Rushabh Mehta
3,120225
add a comment |Â
up vote
2
down vote
Here's a function I would use. Let $mathbbN$ be the set of natural numbers ($0,1,2,3...$) including $0$
$$f(x) = begincases x+1&xinmathbbN\x&xnotinmathbb Nendcases$$
It's quite trivial to see that $f:;mathbbRtomathbbRsetminus0$. Showing injectivity and surjectivity with this piecewise function is quite easy, so give it a shot!
answered 20 mins ago
Rushabh Mehta
3,120225
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's a function I would use. Let $mathbbN$ be the set of natural numbers ($0,1,2,3...$) including $0$
$$f(x) = begincases x+1&xinmathbbN\x&xnotinmathbb Nendcases$$
It's quite trivial to see that $f:;mathbbRtomathbbRsetminus0$. Showing injectivity and surjectivity with this piecewise function is quite easy, so give it a shot!
answered 20 mins ago
Rushabh Mehta
3,120225
Here's a function I would use. Let $mathbbN$ be the set of natural numbers ($0,1,2,3...$) including $0$
$$f(x) = begincases x+1&xinmathbbN\x&xnotinmathbb Nendcases$$
It's quite trivial to see that $f:;mathbbRtomathbbRsetminus0$. Showing injectivity and surjectivity with this piecewise function is quite easy, so give it a shot!
answered 20 mins ago
Rushabh Mehta
3,120225
answered 20 mins ago
Rushabh Mehta
3,120225
answered 20 mins ago
Rushabh Mehta
3,120225
answered 20 mins ago
Rushabh Mehta
3,120225
3,120225
add a comment |Â
add a comment |Â
sphynx888 is a new contributor. Be nice, and check out our Code of Conduct.
sphynx888 is a new contributor. Be nice, and check out our Code of Conduct.
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1
When you see a cardinality question like this you should suspect that you won't find a nice formula for the function you need. You will probably need something weird, defined with cases. There are several answers here that use that strategy - it's worth learning.
– Ethan Bolker
16 mins ago