Proving that a function with two variables is bijective
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In a study of cardinality and infinity, I have to prove that the following function is a bijection (first of all that it's injective). $v : mathbb{N} times mathbb{N} to mathbb{N}$ defined by
$$v(g,b)=1/2·b^2+1/2·b·(2·g-1)+1/2·g^2-3/2·g+1$$
Can anybody help me, please?
functions elementary-set-theory
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add a comment |
$begingroup$
In a study of cardinality and infinity, I have to prove that the following function is a bijection (first of all that it's injective). $v : mathbb{N} times mathbb{N} to mathbb{N}$ defined by
$$v(g,b)=1/2·b^2+1/2·b·(2·g-1)+1/2·g^2-3/2·g+1$$
Can anybody help me, please?
functions elementary-set-theory
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What is $N$ here?
$endgroup$
– Guido A.
Dec 10 '18 at 21:04
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The fact that it has "two variables" doesn't change the logic. For injective, show that if $v(a,b) = v(c,d)$ then $a=c$ and $b=d$. For surjective, given $n$ find $a$ and $b$ such that $v(a,b)=n$.
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– Ethan Bolker
Dec 10 '18 at 21:05
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Such a function is usually much easier to understand if you try to visualize it. Take a grid (draw one, or use graph pepper), let that be $Bbb Ntimes Bbb N$ and write the function value at each point. Do you see a pattern?
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– Arthur
Dec 10 '18 at 21:05
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N is the natural numbers.
$endgroup$
– user625033
Dec 10 '18 at 21:08
add a comment |
$begingroup$
In a study of cardinality and infinity, I have to prove that the following function is a bijection (first of all that it's injective). $v : mathbb{N} times mathbb{N} to mathbb{N}$ defined by
$$v(g,b)=1/2·b^2+1/2·b·(2·g-1)+1/2·g^2-3/2·g+1$$
Can anybody help me, please?
functions elementary-set-theory
$endgroup$
In a study of cardinality and infinity, I have to prove that the following function is a bijection (first of all that it's injective). $v : mathbb{N} times mathbb{N} to mathbb{N}$ defined by
$$v(g,b)=1/2·b^2+1/2·b·(2·g-1)+1/2·g^2-3/2·g+1$$
Can anybody help me, please?
functions elementary-set-theory
functions elementary-set-theory
edited Dec 11 '18 at 7:35
asked Dec 10 '18 at 20:58
user625033
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What is $N$ here?
$endgroup$
– Guido A.
Dec 10 '18 at 21:04
$begingroup$
The fact that it has "two variables" doesn't change the logic. For injective, show that if $v(a,b) = v(c,d)$ then $a=c$ and $b=d$. For surjective, given $n$ find $a$ and $b$ such that $v(a,b)=n$.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 21:05
$begingroup$
Such a function is usually much easier to understand if you try to visualize it. Take a grid (draw one, or use graph pepper), let that be $Bbb Ntimes Bbb N$ and write the function value at each point. Do you see a pattern?
$endgroup$
– Arthur
Dec 10 '18 at 21:05
$begingroup$
N is the natural numbers.
$endgroup$
– user625033
Dec 10 '18 at 21:08
add a comment |
$begingroup$
What is $N$ here?
$endgroup$
– Guido A.
Dec 10 '18 at 21:04
$begingroup$
The fact that it has "two variables" doesn't change the logic. For injective, show that if $v(a,b) = v(c,d)$ then $a=c$ and $b=d$. For surjective, given $n$ find $a$ and $b$ such that $v(a,b)=n$.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 21:05
$begingroup$
Such a function is usually much easier to understand if you try to visualize it. Take a grid (draw one, or use graph pepper), let that be $Bbb Ntimes Bbb N$ and write the function value at each point. Do you see a pattern?
$endgroup$
– Arthur
Dec 10 '18 at 21:05
$begingroup$
N is the natural numbers.
$endgroup$
– user625033
Dec 10 '18 at 21:08
$begingroup$
What is $N$ here?
$endgroup$
– Guido A.
Dec 10 '18 at 21:04
$begingroup$
What is $N$ here?
$endgroup$
– Guido A.
Dec 10 '18 at 21:04
$begingroup$
The fact that it has "two variables" doesn't change the logic. For injective, show that if $v(a,b) = v(c,d)$ then $a=c$ and $b=d$. For surjective, given $n$ find $a$ and $b$ such that $v(a,b)=n$.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 21:05
$begingroup$
The fact that it has "two variables" doesn't change the logic. For injective, show that if $v(a,b) = v(c,d)$ then $a=c$ and $b=d$. For surjective, given $n$ find $a$ and $b$ such that $v(a,b)=n$.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 21:05
$begingroup$
Such a function is usually much easier to understand if you try to visualize it. Take a grid (draw one, or use graph pepper), let that be $Bbb Ntimes Bbb N$ and write the function value at each point. Do you see a pattern?
$endgroup$
– Arthur
Dec 10 '18 at 21:05
$begingroup$
Such a function is usually much easier to understand if you try to visualize it. Take a grid (draw one, or use graph pepper), let that be $Bbb Ntimes Bbb N$ and write the function value at each point. Do you see a pattern?
$endgroup$
– Arthur
Dec 10 '18 at 21:05
$begingroup$
N is the natural numbers.
$endgroup$
– user625033
Dec 10 '18 at 21:08
$begingroup$
N is the natural numbers.
$endgroup$
– user625033
Dec 10 '18 at 21:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, try this out with a few examples.
Note, that the proposition is only true if we are using a set of natural numbers without 0.
$v(1,1) = 1\
v(1,2) = 2\
v(2,1) = 3\
v(1,3) = 4\
v(2,2) = 5$
Mark the lattice points on the coordinate grid. (1,1) maps to 1. Now we move in diagonal lines filling in the grid.
Let's play with the algebra:
$frac 12 b^2 + frac 12 b(2g - 1) + frac 12 g^2 -frac 32 g + 1\
frac 12 b^2 + bg + frac 12 g^2 - frac 12 b - frac 32 g +1\
frac 12 (b + g)^2 - frac 12 (b+g) - g +1\
v(g,b) = frac 12 (b + g)(b+g-1) - g + 1$
$frac 12 (b + g)(b+g-1)$ is the sum of all natural numbers less than $b+g$
$v(g,b)$ is injective if $v(b,g) = v(x,y) implies (b,g) =(x,y)$
$frac 12 (b + g)(b+g-1) - g + 1 = frac 12 (x + y)(x+y-1) - y + 1$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) = g-y$
if $b+gne x+y$
WLOG let $b+g > x+y$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) ge g > g-y$
And if $b+g = x+y$
then if follows that $g = y$ and $b=x$
The function is invective.
I will let you prove that it is subjective.
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
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$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
First of all, try this out with a few examples.
Note, that the proposition is only true if we are using a set of natural numbers without 0.
$v(1,1) = 1\
v(1,2) = 2\
v(2,1) = 3\
v(1,3) = 4\
v(2,2) = 5$
Mark the lattice points on the coordinate grid. (1,1) maps to 1. Now we move in diagonal lines filling in the grid.
Let's play with the algebra:
$frac 12 b^2 + frac 12 b(2g - 1) + frac 12 g^2 -frac 32 g + 1\
frac 12 b^2 + bg + frac 12 g^2 - frac 12 b - frac 32 g +1\
frac 12 (b + g)^2 - frac 12 (b+g) - g +1\
v(g,b) = frac 12 (b + g)(b+g-1) - g + 1$
$frac 12 (b + g)(b+g-1)$ is the sum of all natural numbers less than $b+g$
$v(g,b)$ is injective if $v(b,g) = v(x,y) implies (b,g) =(x,y)$
$frac 12 (b + g)(b+g-1) - g + 1 = frac 12 (x + y)(x+y-1) - y + 1$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) = g-y$
if $b+gne x+y$
WLOG let $b+g > x+y$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) ge g > g-y$
And if $b+g = x+y$
then if follows that $g = y$ and $b=x$
The function is invective.
I will let you prove that it is subjective.
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
add a comment |
$begingroup$
First of all, try this out with a few examples.
Note, that the proposition is only true if we are using a set of natural numbers without 0.
$v(1,1) = 1\
v(1,2) = 2\
v(2,1) = 3\
v(1,3) = 4\
v(2,2) = 5$
Mark the lattice points on the coordinate grid. (1,1) maps to 1. Now we move in diagonal lines filling in the grid.
Let's play with the algebra:
$frac 12 b^2 + frac 12 b(2g - 1) + frac 12 g^2 -frac 32 g + 1\
frac 12 b^2 + bg + frac 12 g^2 - frac 12 b - frac 32 g +1\
frac 12 (b + g)^2 - frac 12 (b+g) - g +1\
v(g,b) = frac 12 (b + g)(b+g-1) - g + 1$
$frac 12 (b + g)(b+g-1)$ is the sum of all natural numbers less than $b+g$
$v(g,b)$ is injective if $v(b,g) = v(x,y) implies (b,g) =(x,y)$
$frac 12 (b + g)(b+g-1) - g + 1 = frac 12 (x + y)(x+y-1) - y + 1$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) = g-y$
if $b+gne x+y$
WLOG let $b+g > x+y$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) ge g > g-y$
And if $b+g = x+y$
then if follows that $g = y$ and $b=x$
The function is invective.
I will let you prove that it is subjective.
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
add a comment |
$begingroup$
First of all, try this out with a few examples.
Note, that the proposition is only true if we are using a set of natural numbers without 0.
$v(1,1) = 1\
v(1,2) = 2\
v(2,1) = 3\
v(1,3) = 4\
v(2,2) = 5$
Mark the lattice points on the coordinate grid. (1,1) maps to 1. Now we move in diagonal lines filling in the grid.
Let's play with the algebra:
$frac 12 b^2 + frac 12 b(2g - 1) + frac 12 g^2 -frac 32 g + 1\
frac 12 b^2 + bg + frac 12 g^2 - frac 12 b - frac 32 g +1\
frac 12 (b + g)^2 - frac 12 (b+g) - g +1\
v(g,b) = frac 12 (b + g)(b+g-1) - g + 1$
$frac 12 (b + g)(b+g-1)$ is the sum of all natural numbers less than $b+g$
$v(g,b)$ is injective if $v(b,g) = v(x,y) implies (b,g) =(x,y)$
$frac 12 (b + g)(b+g-1) - g + 1 = frac 12 (x + y)(x+y-1) - y + 1$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) = g-y$
if $b+gne x+y$
WLOG let $b+g > x+y$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) ge g > g-y$
And if $b+g = x+y$
then if follows that $g = y$ and $b=x$
The function is invective.
I will let you prove that it is subjective.
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
$endgroup$
First of all, try this out with a few examples.
Note, that the proposition is only true if we are using a set of natural numbers without 0.
$v(1,1) = 1\
v(1,2) = 2\
v(2,1) = 3\
v(1,3) = 4\
v(2,2) = 5$
Mark the lattice points on the coordinate grid. (1,1) maps to 1. Now we move in diagonal lines filling in the grid.
Let's play with the algebra:
$frac 12 b^2 + frac 12 b(2g - 1) + frac 12 g^2 -frac 32 g + 1\
frac 12 b^2 + bg + frac 12 g^2 - frac 12 b - frac 32 g +1\
frac 12 (b + g)^2 - frac 12 (b+g) - g +1\
v(g,b) = frac 12 (b + g)(b+g-1) - g + 1$
$frac 12 (b + g)(b+g-1)$ is the sum of all natural numbers less than $b+g$
$v(g,b)$ is injective if $v(b,g) = v(x,y) implies (b,g) =(x,y)$
$frac 12 (b + g)(b+g-1) - g + 1 = frac 12 (x + y)(x+y-1) - y + 1$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) = g-y$
if $b+gne x+y$
WLOG let $b+g > x+y$
$frac 12 (b + g)(b+g-1) - frac 12 (x + y)(x+y-1) ge g > g-y$
And if $b+g = x+y$
then if follows that $g = y$ and $b=x$
The function is invective.
I will let you prove that it is subjective.
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
answered Dec 10 '18 at 22:22
Doug MDoug M
45.2k31854
45.2k31854
$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
add a comment |
$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
Thanks a lot! But what exactly do you mean by the sum of all natural numbers "less than b+g"?
$endgroup$
– user625033
Dec 10 '18 at 22:52
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
$begingroup$
$sum_limits{i=1}^{n-1} i = frac 12 (n)(n-1)$ is the sum of all natural numbers less than $n.$ Now we substitute $n$ with $b+g$
$endgroup$
– Doug M
Dec 10 '18 at 23:10
add a comment |
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$begingroup$
What is $N$ here?
$endgroup$
– Guido A.
Dec 10 '18 at 21:04
$begingroup$
The fact that it has "two variables" doesn't change the logic. For injective, show that if $v(a,b) = v(c,d)$ then $a=c$ and $b=d$. For surjective, given $n$ find $a$ and $b$ such that $v(a,b)=n$.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 21:05
$begingroup$
Such a function is usually much easier to understand if you try to visualize it. Take a grid (draw one, or use graph pepper), let that be $Bbb Ntimes Bbb N$ and write the function value at each point. Do you see a pattern?
$endgroup$
– Arthur
Dec 10 '18 at 21:05
$begingroup$
N is the natural numbers.
$endgroup$
– user625033
Dec 10 '18 at 21:08