Mixing
Qustion about the sum of some specific irrational numbers
$begingroup$
I have been organizing my language, if there is anything not clear about the question, I am sorry.
The question should be:
$forall ninmathbb N^*, a_iinmathbb N$, $a_i$ is not a perfect square for $1leq ileq n$. Then $sum_{i=1}^n sqrt{a_i}$ is not a rational number.
(We assume every real number has a square root, and 0 is a perfect square.)
I saw the proof on the website as $n=2$, it is great, and very clever. And it can even prove $n=3$, this statement is true.
However, I think the key in the proof is to change all the square roots into rational numbers, so we get a contradiction. Yet when $n$ is bigger than $3$, square both sides will not be possible to reduce the number of square roots (at least I think). Thus it may not be effective from here.
I think the statement is true, because I think two positive irrational numbers sum up and get rational numbers will only be in form like $2-sqrt 2$ and $sqrt 2$,
and something like $2-sqrt 2$ cannot be $sqrt n$.
However, I cannot think of a way to prove it. This reminds me that when I want to prove something is a irrational number, all the way I have is to assume it is not, and get contradiction. I think this is not good, so if you have some other methods to prove something irrational, and willing to share with me, I will really be appreciated.
number-theory discrete-mathematics
edited Jan 31 at 20:51
Henrik
6,04592030
asked Jan 31 at 20:35
StAKmodStAKmod
481111
$endgroup$
|
show 1 more comment
$begingroup$
I have been organizing my language, if there is anything not clear about the question, I am sorry.
The question should be:
$forall ninmathbb N^*, a_iinmathbb N$, $a_i$ is not a perfect square for $1leq ileq n$. Then $sum_{i=1}^n sqrt{a_i}$ is not a rational number.
(We assume every real number has a square root, and 0 is a perfect square.)
I saw the proof on the website as $n=2$, it is great, and very clever. And it can even prove $n=3$, this statement is true.
However, I think the key in the proof is to change all the square roots into rational numbers, so we get a contradiction. Yet when $n$ is bigger than $3$, square both sides will not be possible to reduce the number of square roots (at least I think). Thus it may not be effective from here.
I think the statement is true, because I think two positive irrational numbers sum up and get rational numbers will only be in form like $2-sqrt 2$ and $sqrt 2$,
and something like $2-sqrt 2$ cannot be $sqrt n$.
However, I cannot think of a way to prove it. This reminds me that when I want to prove something is a irrational number, all the way I have is to assume it is not, and get contradiction. I think this is not good, so if you have some other methods to prove something irrational, and willing to share with me, I will really be appreciated.
number-theory discrete-mathematics
edited Jan 31 at 20:51
Henrik
6,04592030
asked Jan 31 at 20:35
StAKmodStAKmod
481111
$endgroup$
1
$begingroup$
Please learn to use MathJax to enter mathematical formulas. You can get help on this by clicking on the advanced help option in the window for entering and editing questions and following the link in the section entitled $LaTeX$.
$endgroup$
– Rob Arthan
Jan 31 at 20:44
$begingroup$
I've typeset the question for you, but please learn to do so yourself.
$endgroup$
– Henrik
Jan 31 at 20:53
$begingroup$
Thank you Henrik,I have been trying to learn to use LaTeX,but it did not go well.Next time I will try to type myself.
$endgroup$
– StAKmod
Jan 31 at 20:55
1
$begingroup$
This is definitely a true statement; but in this generality the most natural way to approach the proof would be via algebraic number theory, which is probably more advanced than you are hoping for. It would be nice if somebody could organize a more elementary proof.
$endgroup$
– Greg Martin
Jan 31 at 21:26
1
$begingroup$
Perhaps by strengthening the assertion one could proceed by induction—for example, proving that $sum_{i=1}^n sqrt{a_i}$ is not a root of any polynomial with integer coefficients of degree less than $2^n-1$ (being irrational is this statement for degree-$1$ polynomials). One would have to say something about the $a_i$ for this strengthening to be true, though—we couldn't allow $a_1=a_2=2$, or even $a_1=2,a_2=8$. But there is a way of measuring these dependencies and modifying the degree-bound $2^n-1$ accordingly.
$endgroup$
– Greg Martin
Jan 31 at 21:27
|
show 1 more comment
$begingroup$
I have been organizing my language, if there is anything not clear about the question, I am sorry.
The question should be:
$forall ninmathbb N^*, a_iinmathbb N$, $a_i$ is not a perfect square for $1leq ileq n$. Then $sum_{i=1}^n sqrt{a_i}$ is not a rational number.
(We assume every real number has a square root, and 0 is a perfect square.)
I saw the proof on the website as $n=2$, it is great, and very clever. And it can even prove $n=3$, this statement is true.
However, I think the key in the proof is to change all the square roots into rational numbers, so we get a contradiction. Yet when $n$ is bigger than $3$, square both sides will not be possible to reduce the number of square roots (at least I think). Thus it may not be effective from here.
I think the statement is true, because I think two positive irrational numbers sum up and get rational numbers will only be in form like $2-sqrt 2$ and $sqrt 2$,
and something like $2-sqrt 2$ cannot be $sqrt n$.
However, I cannot think of a way to prove it. This reminds me that when I want to prove something is a irrational number, all the way I have is to assume it is not, and get contradiction. I think this is not good, so if you have some other methods to prove something irrational, and willing to share with me, I will really be appreciated.
number-theory discrete-mathematics
edited Jan 31 at 20:51
Henrik
6,04592030
asked Jan 31 at 20:35
StAKmodStAKmod
481111
$endgroup$
I have been organizing my language, if there is anything not clear about the question, I am sorry.
The question should be:
$forall ninmathbb N^*, a_iinmathbb N$, $a_i$ is not a perfect square for $1leq ileq n$. Then $sum_{i=1}^n sqrt{a_i}$ is not a rational number.
(We assume every real number has a square root, and 0 is a perfect square.)
I saw the proof on the website as $n=2$, it is great, and very clever. And it can even prove $n=3$, this statement is true.
However, I think the key in the proof is to change all the square roots into rational numbers, so we get a contradiction. Yet when $n$ is bigger than $3$, square both sides will not be possible to reduce the number of square roots (at least I think). Thus it may not be effective from here.
I think the statement is true, because I think two positive irrational numbers sum up and get rational numbers will only be in form like $2-sqrt 2$ and $sqrt 2$,
and something like $2-sqrt 2$ cannot be $sqrt n$.
However, I cannot think of a way to prove it. This reminds me that when I want to prove something is a irrational number, all the way I have is to assume it is not, and get contradiction. I think this is not good, so if you have some other methods to prove something irrational, and willing to share with me, I will really be appreciated.
number-theory discrete-mathematics
number-theory discrete-mathematics
edited Jan 31 at 20:51
Henrik
6,04592030
asked Jan 31 at 20:35
StAKmodStAKmod
481111
edited Jan 31 at 20:51
Henrik
6,04592030
asked Jan 31 at 20:35
StAKmodStAKmod
481111
edited Jan 31 at 20:51
Henrik
6,04592030
edited Jan 31 at 20:51
Henrik
6,04592030
edited Jan 31 at 20:51
Henrik
6,04592030
6,04592030
asked Jan 31 at 20:35
StAKmodStAKmod
481111
asked Jan 31 at 20:35
StAKmodStAKmod
481111
asked Jan 31 at 20:35
StAKmodStAKmod
481111
481111
1
$begingroup$
Please learn to use MathJax to enter mathematical formulas. You can get help on this by clicking on the advanced help option in the window for entering and editing questions and following the link in the section entitled $LaTeX$.
$endgroup$
– Rob Arthan
Jan 31 at 20:44
$begingroup$
I've typeset the question for you, but please learn to do so yourself.
$endgroup$
– Henrik
Jan 31 at 20:53
$begingroup$
Thank you Henrik,I have been trying to learn to use LaTeX,but it did not go well.Next time I will try to type myself.
$endgroup$
– StAKmod
Jan 31 at 20:55
1
$begingroup$
This is definitely a true statement; but in this generality the most natural way to approach the proof would be via algebraic number theory, which is probably more advanced than you are hoping for. It would be nice if somebody could organize a more elementary proof.
$endgroup$
– Greg Martin
Jan 31 at 21:26
1
$begingroup$
Perhaps by strengthening the assertion one could proceed by induction—for example, proving that $sum_{i=1}^n sqrt{a_i}$ is not a root of any polynomial with integer coefficients of degree less than $2^n-1$ (being irrational is this statement for degree-$1$ polynomials). One would have to say something about the $a_i$ for this strengthening to be true, though—we couldn't allow $a_1=a_2=2$, or even $a_1=2,a_2=8$. But there is a way of measuring these dependencies and modifying the degree-bound $2^n-1$ accordingly.
$endgroup$
– Greg Martin
Jan 31 at 21:27
|
show 1 more comment
1
$begingroup$
Please learn to use MathJax to enter mathematical formulas. You can get help on this by clicking on the advanced help option in the window for entering and editing questions and following the link in the section entitled $LaTeX$.
$endgroup$
– Rob Arthan
Jan 31 at 20:44
$begingroup$
I've typeset the question for you, but please learn to do so yourself.
$endgroup$
– Henrik
Jan 31 at 20:53
$begingroup$
Thank you Henrik,I have been trying to learn to use LaTeX,but it did not go well.Next time I will try to type myself.
$endgroup$
– StAKmod
Jan 31 at 20:55
1
$begingroup$
This is definitely a true statement; but in this generality the most natural way to approach the proof would be via algebraic number theory, which is probably more advanced than you are hoping for. It would be nice if somebody could organize a more elementary proof.
$endgroup$
– Greg Martin
Jan 31 at 21:26
1
$begingroup$
Perhaps by strengthening the assertion one could proceed by induction—for example, proving that $sum_{i=1}^n sqrt{a_i}$ is not a root of any polynomial with integer coefficients of degree less than $2^n-1$ (being irrational is this statement for degree-$1$ polynomials). One would have to say something about the $a_i$ for this strengthening to be true, though—we couldn't allow $a_1=a_2=2$, or even $a_1=2,a_2=8$. But there is a way of measuring these dependencies and modifying the degree-bound $2^n-1$ accordingly.
$endgroup$
– Greg Martin
Jan 31 at 21:27
1
1
$begingroup$
Please learn to use MathJax to enter mathematical formulas. You can get help on this by clicking on the advanced help option in the window for entering and editing questions and following the link in the section entitled $LaTeX$.
$endgroup$
– Rob Arthan
Jan 31 at 20:44
$begingroup$
Please learn to use MathJax to enter mathematical formulas. You can get help on this by clicking on the advanced help option in the window for entering and editing questions and following the link in the section entitled $LaTeX$.
$endgroup$
– Rob Arthan
Jan 31 at 20:44
$begingroup$
I've typeset the question for you, but please learn to do so yourself.
$endgroup$
– Henrik
Jan 31 at 20:53
$begingroup$
I've typeset the question for you, but please learn to do so yourself.
$endgroup$
– Henrik
Jan 31 at 20:53
$begingroup$
Thank you Henrik,I have been trying to learn to use LaTeX,but it did not go well.Next time I will try to type myself.
$endgroup$
– StAKmod
Jan 31 at 20:55
$begingroup$
Thank you Henrik,I have been trying to learn to use LaTeX,but it did not go well.Next time I will try to type myself.
$endgroup$
– StAKmod
Jan 31 at 20:55
1
1
$begingroup$
This is definitely a true statement; but in this generality the most natural way to approach the proof would be via algebraic number theory, which is probably more advanced than you are hoping for. It would be nice if somebody could organize a more elementary proof.
$endgroup$
– Greg Martin
Jan 31 at 21:26
$begingroup$
This is definitely a true statement; but in this generality the most natural way to approach the proof would be via algebraic number theory, which is probably more advanced than you are hoping for. It would be nice if somebody could organize a more elementary proof.
$endgroup$
– Greg Martin
Jan 31 at 21:26
1
1
$begingroup$
Perhaps by strengthening the assertion one could proceed by induction—for example, proving that $sum_{i=1}^n sqrt{a_i}$ is not a root of any polynomial with integer coefficients of degree less than $2^n-1$ (being irrational is this statement for degree-$1$ polynomials). One would have to say something about the $a_i$ for this strengthening to be true, though—we couldn't allow $a_1=a_2=2$, or even $a_1=2,a_2=8$. But there is a way of measuring these dependencies and modifying the degree-bound $2^n-1$ accordingly.
$endgroup$
– Greg Martin
Jan 31 at 21:27
$begingroup$
Perhaps by strengthening the assertion one could proceed by induction—for example, proving that $sum_{i=1}^n sqrt{a_i}$ is not a root of any polynomial with integer coefficients of degree less than $2^n-1$ (being irrational is this statement for degree-$1$ polynomials). One would have to say something about the $a_i$ for this strengthening to be true, though—we couldn't allow $a_1=a_2=2$, or even $a_1=2,a_2=8$. But there is a way of measuring these dependencies and modifying the degree-bound $2^n-1$ accordingly.
$endgroup$
– Greg Martin
Jan 31 at 21:27
|
show 1 more comment
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1
$begingroup$
Please learn to use MathJax to enter mathematical formulas. You can get help on this by clicking on the advanced help option in the window for entering and editing questions and following the link in the section entitled $LaTeX$.
$endgroup$
– Rob Arthan
Jan 31 at 20:44
$begingroup$
I've typeset the question for you, but please learn to do so yourself.
$endgroup$
– Henrik
Jan 31 at 20:53
$begingroup$
Thank you Henrik,I have been trying to learn to use LaTeX,but it did not go well.Next time I will try to type myself.
$endgroup$
– StAKmod
Jan 31 at 20:55
1
$begingroup$
This is definitely a true statement; but in this generality the most natural way to approach the proof would be via algebraic number theory, which is probably more advanced than you are hoping for. It would be nice if somebody could organize a more elementary proof.
$endgroup$
– Greg Martin
Jan 31 at 21:26
1
$begingroup$
Perhaps by strengthening the assertion one could proceed by induction—for example, proving that $sum_{i=1}^n sqrt{a_i}$ is not a root of any polynomial with integer coefficients of degree less than $2^n-1$ (being irrational is this statement for degree-$1$ polynomials). One would have to say something about the $a_i$ for this strengthening to be true, though—we couldn't allow $a_1=a_2=2$, or even $a_1=2,a_2=8$. But there is a way of measuring these dependencies and modifying the degree-bound $2^n-1$ accordingly.
$endgroup$
– Greg Martin
Jan 31 at 21:27