The intervals $(2,4)$ and $(-1,17)$ have the same cardinality
$begingroup$
I have to prove that $(2,4)$ and $(-1,17)$ have the same cardinality.
I have the definition of cardinality but my prof words things in the most confusing way possible. Help!
discrete-mathematics elementary-set-theory proof-writing
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
asked Feb 17 '16 at 13:55
gabforsegabforse
363
$endgroup$
add a comment |
$begingroup$
I have to prove that $(2,4)$ and $(-1,17)$ have the same cardinality.
I have the definition of cardinality but my prof words things in the most confusing way possible. Help!
discrete-mathematics elementary-set-theory proof-writing
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
asked Feb 17 '16 at 13:55
gabforsegabforse
363
$endgroup$
1
$begingroup$
Try to find a linear function, i.e. a line from $(2,4)$ to $(-1,17)$.
$endgroup$
– sqtrat
Feb 17 '16 at 13:57
2
$begingroup$
Strictly speaking, an affine function, $xmapsto ax+b$. ($b$ will be nonzero.)
$endgroup$
– BrianO
Feb 17 '16 at 14:02
add a comment |
$begingroup$
I have to prove that $(2,4)$ and $(-1,17)$ have the same cardinality.
I have the definition of cardinality but my prof words things in the most confusing way possible. Help!
discrete-mathematics elementary-set-theory proof-writing
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
asked Feb 17 '16 at 13:55
gabforsegabforse
363
$endgroup$
I have to prove that $(2,4)$ and $(-1,17)$ have the same cardinality.
I have the definition of cardinality but my prof words things in the most confusing way possible. Help!
discrete-mathematics elementary-set-theory proof-writing
discrete-mathematics elementary-set-theory proof-writing
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
asked Feb 17 '16 at 13:55
gabforsegabforse
363
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
asked Feb 17 '16 at 13:55
gabforsegabforse
363
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
edited Feb 17 '16 at 14:42
Mankind
10.3k72444
10.3k72444
asked Feb 17 '16 at 13:55
gabforsegabforse
363
asked Feb 17 '16 at 13:55
gabforsegabforse
363
asked Feb 17 '16 at 13:55
gabforsegabforse
363
363
1
$begingroup$
Try to find a linear function, i.e. a line from $(2,4)$ to $(-1,17)$.
$endgroup$
– sqtrat
Feb 17 '16 at 13:57
2
$begingroup$
Strictly speaking, an affine function, $xmapsto ax+b$. ($b$ will be nonzero.)
$endgroup$
– BrianO
Feb 17 '16 at 14:02
add a comment |
1
$begingroup$
Try to find a linear function, i.e. a line from $(2,4)$ to $(-1,17)$.
$endgroup$
– sqtrat
Feb 17 '16 at 13:57
2
$begingroup$
Strictly speaking, an affine function, $xmapsto ax+b$. ($b$ will be nonzero.)
$endgroup$
– BrianO
Feb 17 '16 at 14:02
1
1
$begingroup$
Try to find a linear function, i.e. a line from $(2,4)$ to $(-1,17)$.
$endgroup$
– sqtrat
Feb 17 '16 at 13:57
$begingroup$
Try to find a linear function, i.e. a line from $(2,4)$ to $(-1,17)$.
$endgroup$
– sqtrat
Feb 17 '16 at 13:57
2
2
$begingroup$
Strictly speaking, an affine function, $xmapsto ax+b$. ($b$ will be nonzero.)
$endgroup$
– BrianO
Feb 17 '16 at 14:02
$begingroup$
Strictly speaking, an affine function, $xmapsto ax+b$. ($b$ will be nonzero.)
$endgroup$
– BrianO
Feb 17 '16 at 14:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You need to find a bijective function $fcolon (2,4)rightarrow (-1,17)$. As suggested in the comments, try with a function of the form $f(x) = ax+b$.
You know that the interval $(2,4)$ of length $2$ should be stretched out to fit onto the interval $(-1,17)$, which has length $18$. In other words,
$$a = 18/2 = 9.$$
But $f(x) = 9x$ doesn't work. It maps $(2,4)$ to $(18,36)$, so you need to adjust $f$ by adding some number $b$, i.e. you need to find $b$, such that $f(x) = 9x+b$ maps $(2,4)$ to the correct interval.
When you have found a function that maps to the right place, you need to check that it is indeed a bijection.
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
$endgroup$
add a comment |
$begingroup$
The general way to show two sets $X,Y$ have the same cardinality is to show that there is a function $f:Xrightarrow Y$ that is both 1) injective and 2) surjective. That is 1) for all $aneq bin X$ we must have $f(a)neq f(b)$ and 2) for all $yin Y$ there must exist $xin X$ such that $f(x)=y$.
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
You need to find a bijective function $fcolon (2,4)rightarrow (-1,17)$. As suggested in the comments, try with a function of the form $f(x) = ax+b$.
You know that the interval $(2,4)$ of length $2$ should be stretched out to fit onto the interval $(-1,17)$, which has length $18$. In other words,
$$a = 18/2 = 9.$$
But $f(x) = 9x$ doesn't work. It maps $(2,4)$ to $(18,36)$, so you need to adjust $f$ by adding some number $b$, i.e. you need to find $b$, such that $f(x) = 9x+b$ maps $(2,4)$ to the correct interval.
When you have found a function that maps to the right place, you need to check that it is indeed a bijection.
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
$endgroup$
add a comment |
$begingroup$
You need to find a bijective function $fcolon (2,4)rightarrow (-1,17)$. As suggested in the comments, try with a function of the form $f(x) = ax+b$.
You know that the interval $(2,4)$ of length $2$ should be stretched out to fit onto the interval $(-1,17)$, which has length $18$. In other words,
$$a = 18/2 = 9.$$
But $f(x) = 9x$ doesn't work. It maps $(2,4)$ to $(18,36)$, so you need to adjust $f$ by adding some number $b$, i.e. you need to find $b$, such that $f(x) = 9x+b$ maps $(2,4)$ to the correct interval.
When you have found a function that maps to the right place, you need to check that it is indeed a bijection.
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
$endgroup$
add a comment |
$begingroup$
You need to find a bijective function $fcolon (2,4)rightarrow (-1,17)$. As suggested in the comments, try with a function of the form $f(x) = ax+b$.
You know that the interval $(2,4)$ of length $2$ should be stretched out to fit onto the interval $(-1,17)$, which has length $18$. In other words,
$$a = 18/2 = 9.$$
But $f(x) = 9x$ doesn't work. It maps $(2,4)$ to $(18,36)$, so you need to adjust $f$ by adding some number $b$, i.e. you need to find $b$, such that $f(x) = 9x+b$ maps $(2,4)$ to the correct interval.
When you have found a function that maps to the right place, you need to check that it is indeed a bijection.
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
$endgroup$
You need to find a bijective function $fcolon (2,4)rightarrow (-1,17)$. As suggested in the comments, try with a function of the form $f(x) = ax+b$.
You know that the interval $(2,4)$ of length $2$ should be stretched out to fit onto the interval $(-1,17)$, which has length $18$. In other words,
$$a = 18/2 = 9.$$
But $f(x) = 9x$ doesn't work. It maps $(2,4)$ to $(18,36)$, so you need to adjust $f$ by adding some number $b$, i.e. you need to find $b$, such that $f(x) = 9x+b$ maps $(2,4)$ to the correct interval.
When you have found a function that maps to the right place, you need to check that it is indeed a bijection.
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
answered Feb 17 '16 at 14:10
MankindMankind
10.3k72444
10.3k72444
add a comment |
add a comment |
$begingroup$
The general way to show two sets $X,Y$ have the same cardinality is to show that there is a function $f:Xrightarrow Y$ that is both 1) injective and 2) surjective. That is 1) for all $aneq bin X$ we must have $f(a)neq f(b)$ and 2) for all $yin Y$ there must exist $xin X$ such that $f(x)=y$.
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
$endgroup$
add a comment |
$begingroup$
The general way to show two sets $X,Y$ have the same cardinality is to show that there is a function $f:Xrightarrow Y$ that is both 1) injective and 2) surjective. That is 1) for all $aneq bin X$ we must have $f(a)neq f(b)$ and 2) for all $yin Y$ there must exist $xin X$ such that $f(x)=y$.
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
$endgroup$
add a comment |
$begingroup$
The general way to show two sets $X,Y$ have the same cardinality is to show that there is a function $f:Xrightarrow Y$ that is both 1) injective and 2) surjective. That is 1) for all $aneq bin X$ we must have $f(a)neq f(b)$ and 2) for all $yin Y$ there must exist $xin X$ such that $f(x)=y$.
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
$endgroup$
The general way to show two sets $X,Y$ have the same cardinality is to show that there is a function $f:Xrightarrow Y$ that is both 1) injective and 2) surjective. That is 1) for all $aneq bin X$ we must have $f(a)neq f(b)$ and 2) for all $yin Y$ there must exist $xin X$ such that $f(x)=y$.
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
answered Feb 17 '16 at 13:57
MarcMarc
5,63511022
5,63511022
add a comment |
add a comment |
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1
$begingroup$
Try to find a linear function, i.e. a line from $(2,4)$ to $(-1,17)$.
$endgroup$
– sqtrat
Feb 17 '16 at 13:57
2
$begingroup$
Strictly speaking, an affine function, $xmapsto ax+b$. ($b$ will be nonzero.)
$endgroup$
– BrianO
Feb 17 '16 at 14:02