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Is the summation of the same irrational number always irrational?
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I'm asking this question because I was unable to find an answer elsewhere as most questions are about the summation of different irrational numbers, which is not what this question is about. Here, I'm interested in demonstrating that the result of the summation of the same irrational number is always irrational: $sum_i=1^n a$, where $n$ is rational and $a$ is an irrational constant.
proof-verification irrational-numbers
asked 52 mins ago
m_power
18110
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up vote
3
down vote
favorite
I'm asking this question because I was unable to find an answer elsewhere as most questions are about the summation of different irrational numbers, which is not what this question is about. Here, I'm interested in demonstrating that the result of the summation of the same irrational number is always irrational: $sum_i=1^n a$, where $n$ is rational and $a$ is an irrational constant.
proof-verification irrational-numbers
asked 52 mins ago
m_power
18110
Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $natimes sum_i=0^n1$. But I expect you meant something else.
– lulu
51 mins ago
Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well.
– m_power
11 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm asking this question because I was unable to find an answer elsewhere as most questions are about the summation of different irrational numbers, which is not what this question is about. Here, I'm interested in demonstrating that the result of the summation of the same irrational number is always irrational: $sum_i=1^n a$, where $n$ is rational and $a$ is an irrational constant.
proof-verification irrational-numbers
asked 52 mins ago
m_power
18110
I'm asking this question because I was unable to find an answer elsewhere as most questions are about the summation of different irrational numbers, which is not what this question is about. Here, I'm interested in demonstrating that the result of the summation of the same irrational number is always irrational: $sum_i=1^n a$, where $n$ is rational and $a$ is an irrational constant.
proof-verification irrational-numbers
proof-verification irrational-numbers
asked 52 mins ago
m_power
18110
asked 52 mins ago
m_power
18110
edited 2 mins ago
asked 52 mins ago
m_power
18110
asked 52 mins ago
m_power
18110
asked 52 mins ago
m_power
18110
18110
Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $natimes sum_i=0^n1$. But I expect you meant something else.
– lulu
51 mins ago
Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well.
– m_power
11 mins ago
add a comment |Â
Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $natimes sum_i=0^n1$. But I expect you meant something else.
– lulu
51 mins ago
Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well.
– m_power
11 mins ago
Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $natimes sum_i=0^n1$. But I expect you meant something else.
– lulu
51 mins ago
Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $natimes sum_i=0^n1$. But I expect you meant something else.
– lulu
51 mins ago
Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well.
– m_power
11 mins ago
Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well.
– m_power
11 mins ago
add a comment |Â
2 Answers
2
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up vote
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It is trivially so. If you sum the irrational number $x$ $n$ times, you end up with $nx$.
If $nx=frac ab$, with $(a,b)$ integers, then $x=fracanb$, thus is rational, which is not true...
answered 49 mins ago
Martigan
4,434714
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up vote
3
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$sum k_ia = a(sum k_i)$ and $sum k_i$ is rational.
And if $sum k_i ne 0$ then a (non-zero) rational times an irrational is irrational.
====
I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.
But that is the only exception.
answered 50 mins ago
fleablood
61.9k22678
1
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
3
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It is trivially so. If you sum the irrational number $x$ $n$ times, you end up with $nx$.
If $nx=frac ab$, with $(a,b)$ integers, then $x=fracanb$, thus is rational, which is not true...
answered 49 mins ago
Martigan
4,434714
add a comment |Â
up vote
4
down vote
It is trivially so. If you sum the irrational number $x$ $n$ times, you end up with $nx$.
If $nx=frac ab$, with $(a,b)$ integers, then $x=fracanb$, thus is rational, which is not true...
answered 49 mins ago
Martigan
4,434714
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It is trivially so. If you sum the irrational number $x$ $n$ times, you end up with $nx$.
If $nx=frac ab$, with $(a,b)$ integers, then $x=fracanb$, thus is rational, which is not true...
answered 49 mins ago
Martigan
4,434714
It is trivially so. If you sum the irrational number $x$ $n$ times, you end up with $nx$.
If $nx=frac ab$, with $(a,b)$ integers, then $x=fracanb$, thus is rational, which is not true...
answered 49 mins ago
Martigan
4,434714
answered 49 mins ago
Martigan
4,434714
answered 49 mins ago
Martigan
4,434714
answered 49 mins ago
Martigan
4,434714
4,434714
add a comment |Â
add a comment |Â
up vote
3
down vote
$sum k_ia = a(sum k_i)$ and $sum k_i$ is rational.
And if $sum k_i ne 0$ then a (non-zero) rational times an irrational is irrational.
====
I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.
But that is the only exception.
answered 50 mins ago
fleablood
61.9k22678
1
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
3
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
add a comment |Â
up vote
3
down vote
$sum k_ia = a(sum k_i)$ and $sum k_i$ is rational.
And if $sum k_i ne 0$ then a (non-zero) rational times an irrational is irrational.
====
I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.
But that is the only exception.
answered 50 mins ago
fleablood
61.9k22678
1
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
3
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$sum k_ia = a(sum k_i)$ and $sum k_i$ is rational.
And if $sum k_i ne 0$ then a (non-zero) rational times an irrational is irrational.
====
I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.
But that is the only exception.
answered 50 mins ago
fleablood
61.9k22678
$sum k_ia = a(sum k_i)$ and $sum k_i$ is rational.
And if $sum k_i ne 0$ then a (non-zero) rational times an irrational is irrational.
====
I guess it's somewhat important to point out that if the the sum is $0$ such as $a + 2a - 3a$ or $k_1a + k_2 a + ..... + k_n a$ where $k_1 + k_2 + .... + k_n = 0$, then the statement is trivially false.
But that is the only exception.
answered 50 mins ago
fleablood
61.9k22678
edited 43 mins ago
answered 50 mins ago
fleablood
61.9k22678
answered 50 mins ago
fleablood
61.9k22678
answered 50 mins ago
fleablood
61.9k22678
61.9k22678
1
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
3
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
add a comment |Â
1
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
3
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
1
1
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
.. unless that rational factor were $=0$
– Hagen von Eitzen
48 mins ago
3
3
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
Yeah, I was drinking my coffee and eating a bagel and think "oh ###, unless $0$! Better get back and fix it before anyone notices!" .... oh, well.
– fleablood
44 mins ago
add a comment |Â
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Note: $a$ factors out of that expression so, sure. But...is your question clear? There is no $i$ anywhere in your summand so your expression is just $natimes sum_i=0^n1$. But I expect you meant something else.
– lulu
51 mins ago
Good eye. I'll correct this mistake and replace $n$ by $i$ in the equation. Also, $i$ should start at 1 and not 0. I'll correct that as well.
– m_power
11 mins ago