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We have $n$ real numbers around the circle and among any consecutive 3 one is AM of the other two. Then all the numbers are the same or $3mid n$.
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There are $n$ real numbers around the circle and among any consecutive 3 one is arithmetic mean of the other two. Prove that all the numbers are the same or $3mid n$.
Hint was to use a linear algebra.
It is obviously that if among some three consecutive numbers some two are the same then all are three the same: Say we have $$(a,a,b)implies b =a+aover 2= a;;;rm or;;;a =a+bover 2 implies a=b$$
But then all the numbers are the same. So we can assume that among any consecutive 3 there are all different.
Any way, if all the number are $a_1,a_2,....,a_n$ then for any three consecutive (inidices are modulo $n$) we have $$a_i-1+a_i+a_i+1 equiv_3 0$$
combinatorics vector-spaces algebraic-combinatorics hamel-basis
asked Jul 25 at 14:07
greedoid
26.1k93473
add a comment |Â
up vote
6
down vote
favorite
There are $n$ real numbers around the circle and among any consecutive 3 one is arithmetic mean of the other two. Prove that all the numbers are the same or $3mid n$.
Hint was to use a linear algebra.
It is obviously that if among some three consecutive numbers some two are the same then all are three the same: Say we have $$(a,a,b)implies b =a+aover 2= a;;;rm or;;;a =a+bover 2 implies a=b$$
But then all the numbers are the same. So we can assume that among any consecutive 3 there are all different.
Any way, if all the number are $a_1,a_2,....,a_n$ then for any three consecutive (inidices are modulo $n$) we have $$a_i-1+a_i+a_i+1 equiv_3 0$$
combinatorics vector-spaces algebraic-combinatorics hamel-basis
asked Jul 25 at 14:07
greedoid
26.1k93473
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
There are $n$ real numbers around the circle and among any consecutive 3 one is arithmetic mean of the other two. Prove that all the numbers are the same or $3mid n$.
Hint was to use a linear algebra.
It is obviously that if among some three consecutive numbers some two are the same then all are three the same: Say we have $$(a,a,b)implies b =a+aover 2= a;;;rm or;;;a =a+bover 2 implies a=b$$
But then all the numbers are the same. So we can assume that among any consecutive 3 there are all different.
Any way, if all the number are $a_1,a_2,....,a_n$ then for any three consecutive (inidices are modulo $n$) we have $$a_i-1+a_i+a_i+1 equiv_3 0$$
combinatorics vector-spaces algebraic-combinatorics hamel-basis
asked Jul 25 at 14:07
greedoid
26.1k93473
There are $n$ real numbers around the circle and among any consecutive 3 one is arithmetic mean of the other two. Prove that all the numbers are the same or $3mid n$.
Hint was to use a linear algebra.
It is obviously that if among some three consecutive numbers some two are the same then all are three the same: Say we have $$(a,a,b)implies b =a+aover 2= a;;;rm or;;;a =a+bover 2 implies a=b$$
But then all the numbers are the same. So we can assume that among any consecutive 3 there are all different.
Any way, if all the number are $a_1,a_2,....,a_n$ then for any three consecutive (inidices are modulo $n$) we have $$a_i-1+a_i+a_i+1 equiv_3 0$$
combinatorics vector-spaces algebraic-combinatorics hamel-basis
asked Jul 25 at 14:07
greedoid
26.1k93473
edited Jul 25 at 18:37
asked Jul 25 at 14:07
greedoid
26.1k93473
asked Jul 25 at 14:07
greedoid
26.1k93473
asked Jul 25 at 14:07
greedoid
26.1k93473
26.1k93473
add a comment |Â
add a comment |Â
3 Answers
3
active
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up vote
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The problem can be solved in an elementary way.
Let $bf x=(x_1,x_2,ldots,x_n)$ be the given cyclic sequence of numbers and $bf d$ the cyclic sequence of their first differences $d_k:=x_k+1-x_k$. The basic condition on the sequence $bf x$ then implies
$$d_k=hquadRightarrow quad d_k+1inlefth,-hover2,-2hrightqquadforall> kin[n] .$$
It follows that there is an $hinmathbb R$ such that $$d_k=pm 2^j_kh,quad j_kinmathbb Z,qquadforall,kin[n] .$$ If $h=0$ all $x_k$ are equal. If $hne0$ then after multiplying $bf x$, hence $bf d$, with a suitable constant we may assume that the $d_k$ of smallest absolute value is $=1$. We then know that
$$d_kinbigl1,-2,4,-8,16,-32,ldotsbigr .$$
Now $sum_k=1^n d_k=0$, hence this sum is divisible by $3$. But all $d_k$ have remainder $1$ mod $3$, hence $n$ has to be divisible by $3$.
answered Jul 26 at 6:47
Christian Blatter
163k7107306
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
add a comment |Â
up vote
4
down vote
This question has a lot in common with Prove that the elements of X all have the same weight:
- It was originally meant to be solved using linear algebra.
- Someone mistakenly assumed the numbers were integers.
- Christian Blatter pointed out the mistake.
- The answer can be salvaged by reducing to the integer case.
You can essentially apply the same reasoning as in my answer to that question:
Assume the elements are not all the same. Scale them by a factor large enough that their nearest integers are not all the same. Now apply the simultaneous version of Dirichlet's approximation theorem to find an integer $q$ to multiply them by, such that the resulting numbers all differ by less than $frac14$ from the nearest integer. These rescalings preserve the premise. Since $4$ differences from the nearest integers add up to less than $1$, the arithmetic-mean condition $a+b=2c$ in the premise must also hold for the nearest integers, which are by construction not all the same. Thus it suffices to prove the claim for integers, as Ross has done (in a deleted answer which I hope he'll undelete upon seeing this).
answered Jul 25 at 22:01
joriki
164k10180328
add a comment |Â
up vote
2
down vote
Reduce all the numbers $bmod 3$. The property that any three neighboring numbers can be arranged to be in AP is retained. If there are two different values next to each other the value clockwise must be the third residue to make the AP work. We must therefore have the circle be $a,b,c,a,b,c,ldots c$ assigning one residue to each letter so that when it closes we still meet the AP requirement, so the number of terms is a multiple of $3$.
If there are not two different values next to each other all the numbers are equivalent $bmod 3$. Subtract off the residue and all the numbers will be multiples of $3$. Divide by $3$ and repeat the argument. If the numbers are not all the same, there is a multiple of $3$ of them. If the numbers are all the same $bmod 3$, subtract the residue and divide by $3$. After enough subtractions and divisions, if you don't find a difference all the numbers will be $0$ and all the numbers started out the same.
answered Jul 25 at 14:53
Ross Millikan
275k21186351
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
4
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The problem can be solved in an elementary way.
Let $bf x=(x_1,x_2,ldots,x_n)$ be the given cyclic sequence of numbers and $bf d$ the cyclic sequence of their first differences $d_k:=x_k+1-x_k$. The basic condition on the sequence $bf x$ then implies
$$d_k=hquadRightarrow quad d_k+1inlefth,-hover2,-2hrightqquadforall> kin[n] .$$
It follows that there is an $hinmathbb R$ such that $$d_k=pm 2^j_kh,quad j_kinmathbb Z,qquadforall,kin[n] .$$ If $h=0$ all $x_k$ are equal. If $hne0$ then after multiplying $bf x$, hence $bf d$, with a suitable constant we may assume that the $d_k$ of smallest absolute value is $=1$. We then know that
$$d_kinbigl1,-2,4,-8,16,-32,ldotsbigr .$$
Now $sum_k=1^n d_k=0$, hence this sum is divisible by $3$. But all $d_k$ have remainder $1$ mod $3$, hence $n$ has to be divisible by $3$.
answered Jul 26 at 6:47
Christian Blatter
163k7107306
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
add a comment |Â
up vote
5
down vote
The problem can be solved in an elementary way.
Let $bf x=(x_1,x_2,ldots,x_n)$ be the given cyclic sequence of numbers and $bf d$ the cyclic sequence of their first differences $d_k:=x_k+1-x_k$. The basic condition on the sequence $bf x$ then implies
$$d_k=hquadRightarrow quad d_k+1inlefth,-hover2,-2hrightqquadforall> kin[n] .$$
It follows that there is an $hinmathbb R$ such that $$d_k=pm 2^j_kh,quad j_kinmathbb Z,qquadforall,kin[n] .$$ If $h=0$ all $x_k$ are equal. If $hne0$ then after multiplying $bf x$, hence $bf d$, with a suitable constant we may assume that the $d_k$ of smallest absolute value is $=1$. We then know that
$$d_kinbigl1,-2,4,-8,16,-32,ldotsbigr .$$
Now $sum_k=1^n d_k=0$, hence this sum is divisible by $3$. But all $d_k$ have remainder $1$ mod $3$, hence $n$ has to be divisible by $3$.
answered Jul 26 at 6:47
Christian Blatter
163k7107306
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The problem can be solved in an elementary way.
Let $bf x=(x_1,x_2,ldots,x_n)$ be the given cyclic sequence of numbers and $bf d$ the cyclic sequence of their first differences $d_k:=x_k+1-x_k$. The basic condition on the sequence $bf x$ then implies
$$d_k=hquadRightarrow quad d_k+1inlefth,-hover2,-2hrightqquadforall> kin[n] .$$
It follows that there is an $hinmathbb R$ such that $$d_k=pm 2^j_kh,quad j_kinmathbb Z,qquadforall,kin[n] .$$ If $h=0$ all $x_k$ are equal. If $hne0$ then after multiplying $bf x$, hence $bf d$, with a suitable constant we may assume that the $d_k$ of smallest absolute value is $=1$. We then know that
$$d_kinbigl1,-2,4,-8,16,-32,ldotsbigr .$$
Now $sum_k=1^n d_k=0$, hence this sum is divisible by $3$. But all $d_k$ have remainder $1$ mod $3$, hence $n$ has to be divisible by $3$.
answered Jul 26 at 6:47
Christian Blatter
163k7107306
The problem can be solved in an elementary way.
Let $bf x=(x_1,x_2,ldots,x_n)$ be the given cyclic sequence of numbers and $bf d$ the cyclic sequence of their first differences $d_k:=x_k+1-x_k$. The basic condition on the sequence $bf x$ then implies
$$d_k=hquadRightarrow quad d_k+1inlefth,-hover2,-2hrightqquadforall> kin[n] .$$
It follows that there is an $hinmathbb R$ such that $$d_k=pm 2^j_kh,quad j_kinmathbb Z,qquadforall,kin[n] .$$ If $h=0$ all $x_k$ are equal. If $hne0$ then after multiplying $bf x$, hence $bf d$, with a suitable constant we may assume that the $d_k$ of smallest absolute value is $=1$. We then know that
$$d_kinbigl1,-2,4,-8,16,-32,ldotsbigr .$$
Now $sum_k=1^n d_k=0$, hence this sum is divisible by $3$. But all $d_k$ have remainder $1$ mod $3$, hence $n$ has to be divisible by $3$.
answered Jul 26 at 6:47
Christian Blatter
163k7107306
edited Jul 26 at 9:35
answered Jul 26 at 6:47
Christian Blatter
163k7107306
answered Jul 26 at 6:47
Christian Blatter
163k7107306
answered Jul 26 at 6:47
Christian Blatter
163k7107306
163k7107306
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
add a comment |Â
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
What doe thi sentence mean: If $hne0$ we may WLG assume that the $d_k$ of smallest absolute value is $=1$.
– greedoid
Jul 26 at 9:01
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
@Angle - the AP property is unchanged if we multiply all $x_i$ by a real constant $k$. And if all $x_i$ are multipled by $k$ then all $d_i$ are also multiplied by $k$. So we find the $d_i$ with the smallest absolute value - say this is $d_j$ - and we mutiply all $x_i$ (and hence all $d_i$ too) by $d_j^-1$. Each $d_i$ is now an integer and must be a power of $-2$, hence each $d_i$ is now $1 mod 3$.
– gandalf61
Jul 26 at 11:28
add a comment |Â
up vote
4
down vote
This question has a lot in common with Prove that the elements of X all have the same weight:
- It was originally meant to be solved using linear algebra.
- Someone mistakenly assumed the numbers were integers.
- Christian Blatter pointed out the mistake.
- The answer can be salvaged by reducing to the integer case.
You can essentially apply the same reasoning as in my answer to that question:
Assume the elements are not all the same. Scale them by a factor large enough that their nearest integers are not all the same. Now apply the simultaneous version of Dirichlet's approximation theorem to find an integer $q$ to multiply them by, such that the resulting numbers all differ by less than $frac14$ from the nearest integer. These rescalings preserve the premise. Since $4$ differences from the nearest integers add up to less than $1$, the arithmetic-mean condition $a+b=2c$ in the premise must also hold for the nearest integers, which are by construction not all the same. Thus it suffices to prove the claim for integers, as Ross has done (in a deleted answer which I hope he'll undelete upon seeing this).
answered Jul 25 at 22:01
joriki
164k10180328
add a comment |Â
up vote
4
down vote
This question has a lot in common with Prove that the elements of X all have the same weight:
- It was originally meant to be solved using linear algebra.
- Someone mistakenly assumed the numbers were integers.
- Christian Blatter pointed out the mistake.
- The answer can be salvaged by reducing to the integer case.
You can essentially apply the same reasoning as in my answer to that question:
Assume the elements are not all the same. Scale them by a factor large enough that their nearest integers are not all the same. Now apply the simultaneous version of Dirichlet's approximation theorem to find an integer $q$ to multiply them by, such that the resulting numbers all differ by less than $frac14$ from the nearest integer. These rescalings preserve the premise. Since $4$ differences from the nearest integers add up to less than $1$, the arithmetic-mean condition $a+b=2c$ in the premise must also hold for the nearest integers, which are by construction not all the same. Thus it suffices to prove the claim for integers, as Ross has done (in a deleted answer which I hope he'll undelete upon seeing this).
answered Jul 25 at 22:01
joriki
164k10180328
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This question has a lot in common with Prove that the elements of X all have the same weight:
- It was originally meant to be solved using linear algebra.
- Someone mistakenly assumed the numbers were integers.
- Christian Blatter pointed out the mistake.
- The answer can be salvaged by reducing to the integer case.
You can essentially apply the same reasoning as in my answer to that question:
Assume the elements are not all the same. Scale them by a factor large enough that their nearest integers are not all the same. Now apply the simultaneous version of Dirichlet's approximation theorem to find an integer $q$ to multiply them by, such that the resulting numbers all differ by less than $frac14$ from the nearest integer. These rescalings preserve the premise. Since $4$ differences from the nearest integers add up to less than $1$, the arithmetic-mean condition $a+b=2c$ in the premise must also hold for the nearest integers, which are by construction not all the same. Thus it suffices to prove the claim for integers, as Ross has done (in a deleted answer which I hope he'll undelete upon seeing this).
answered Jul 25 at 22:01
joriki
164k10180328
This question has a lot in common with Prove that the elements of X all have the same weight:
- It was originally meant to be solved using linear algebra.
- Someone mistakenly assumed the numbers were integers.
- Christian Blatter pointed out the mistake.
- The answer can be salvaged by reducing to the integer case.
You can essentially apply the same reasoning as in my answer to that question:
Assume the elements are not all the same. Scale them by a factor large enough that their nearest integers are not all the same. Now apply the simultaneous version of Dirichlet's approximation theorem to find an integer $q$ to multiply them by, such that the resulting numbers all differ by less than $frac14$ from the nearest integer. These rescalings preserve the premise. Since $4$ differences from the nearest integers add up to less than $1$, the arithmetic-mean condition $a+b=2c$ in the premise must also hold for the nearest integers, which are by construction not all the same. Thus it suffices to prove the claim for integers, as Ross has done (in a deleted answer which I hope he'll undelete upon seeing this).
answered Jul 25 at 22:01
joriki
164k10180328
answered Jul 25 at 22:01
joriki
164k10180328
answered Jul 25 at 22:01
joriki
164k10180328
answered Jul 25 at 22:01
joriki
164k10180328
164k10180328
add a comment |Â
add a comment |Â
up vote
2
down vote
Reduce all the numbers $bmod 3$. The property that any three neighboring numbers can be arranged to be in AP is retained. If there are two different values next to each other the value clockwise must be the third residue to make the AP work. We must therefore have the circle be $a,b,c,a,b,c,ldots c$ assigning one residue to each letter so that when it closes we still meet the AP requirement, so the number of terms is a multiple of $3$.
If there are not two different values next to each other all the numbers are equivalent $bmod 3$. Subtract off the residue and all the numbers will be multiples of $3$. Divide by $3$ and repeat the argument. If the numbers are not all the same, there is a multiple of $3$ of them. If the numbers are all the same $bmod 3$, subtract the residue and divide by $3$. After enough subtractions and divisions, if you don't find a difference all the numbers will be $0$ and all the numbers started out the same.
answered Jul 25 at 14:53
Ross Millikan
275k21186351
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
4
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
add a comment |Â
up vote
2
down vote
Reduce all the numbers $bmod 3$. The property that any three neighboring numbers can be arranged to be in AP is retained. If there are two different values next to each other the value clockwise must be the third residue to make the AP work. We must therefore have the circle be $a,b,c,a,b,c,ldots c$ assigning one residue to each letter so that when it closes we still meet the AP requirement, so the number of terms is a multiple of $3$.
If there are not two different values next to each other all the numbers are equivalent $bmod 3$. Subtract off the residue and all the numbers will be multiples of $3$. Divide by $3$ and repeat the argument. If the numbers are not all the same, there is a multiple of $3$ of them. If the numbers are all the same $bmod 3$, subtract the residue and divide by $3$. After enough subtractions and divisions, if you don't find a difference all the numbers will be $0$ and all the numbers started out the same.
answered Jul 25 at 14:53
Ross Millikan
275k21186351
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
4
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Reduce all the numbers $bmod 3$. The property that any three neighboring numbers can be arranged to be in AP is retained. If there are two different values next to each other the value clockwise must be the third residue to make the AP work. We must therefore have the circle be $a,b,c,a,b,c,ldots c$ assigning one residue to each letter so that when it closes we still meet the AP requirement, so the number of terms is a multiple of $3$.
If there are not two different values next to each other all the numbers are equivalent $bmod 3$. Subtract off the residue and all the numbers will be multiples of $3$. Divide by $3$ and repeat the argument. If the numbers are not all the same, there is a multiple of $3$ of them. If the numbers are all the same $bmod 3$, subtract the residue and divide by $3$. After enough subtractions and divisions, if you don't find a difference all the numbers will be $0$ and all the numbers started out the same.
answered Jul 25 at 14:53
Ross Millikan
275k21186351
Reduce all the numbers $bmod 3$. The property that any three neighboring numbers can be arranged to be in AP is retained. If there are two different values next to each other the value clockwise must be the third residue to make the AP work. We must therefore have the circle be $a,b,c,a,b,c,ldots c$ assigning one residue to each letter so that when it closes we still meet the AP requirement, so the number of terms is a multiple of $3$.
If there are not two different values next to each other all the numbers are equivalent $bmod 3$. Subtract off the residue and all the numbers will be multiples of $3$. Divide by $3$ and repeat the argument. If the numbers are not all the same, there is a multiple of $3$ of them. If the numbers are all the same $bmod 3$, subtract the residue and divide by $3$. After enough subtractions and divisions, if you don't find a difference all the numbers will be $0$ and all the numbers started out the same.
answered Jul 25 at 14:53
Ross Millikan
275k21186351
answered Jul 25 at 14:53
Ross Millikan
275k21186351
answered Jul 25 at 14:53
Ross Millikan
275k21186351
answered Jul 25 at 14:53
Ross Millikan
275k21186351
275k21186351
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
4
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
add a comment |Â
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
4
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
The numbers are supposed to be real. I don't know whether reducing mod $3$ keeps the AP property.
– Christian Blatter
Jul 25 at 15:37
4
4
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
Undeleted per joriki's request. He shows how to extend the argument from integers to reals.
– Ross Millikan
Jul 25 at 22:08
add a comment |Â
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
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Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password