Mixing
The rigorousness of this proof about greatest common divisors.
$begingroup$
My task is to prove that gcd(n, n+1)=1 for all n>0.
It is obvious that 1 is a common divisor of both n and n+1 since
$$
1|n → 1x=n
$$
if x=n, and
$$
1|n+1 → 1y=n+1
$$
if y=n+1.
To prove that 1 is the greatest common divisor, I did as follows:
From the integers n and n+1 the other must be an even integer, and the other an odd integer. Since one of them must be odd, the gcd of the two can't be an even number, because the odd one doesn't have any even divisors. If the gcd of the two were an odd integer greater than 1, for example 3, then
$$
3|n → 3x=n → n=3x
$$
and
$$
3|n+1 → 3y=n+1 → n=3y-1,
$$
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1.
Is the part about the odd integers greater than 1 rigorous enough for a proper proof? I don't think it is; how would you prove this more rigorously? If anyone has any other improvement ideas of any kind, they're more than welcome!
elementary-number-theory proof-writing
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
$endgroup$
add a comment |
$begingroup$
My task is to prove that gcd(n, n+1)=1 for all n>0.
It is obvious that 1 is a common divisor of both n and n+1 since
$$
1|n → 1x=n
$$
if x=n, and
$$
1|n+1 → 1y=n+1
$$
if y=n+1.
To prove that 1 is the greatest common divisor, I did as follows:
From the integers n and n+1 the other must be an even integer, and the other an odd integer. Since one of them must be odd, the gcd of the two can't be an even number, because the odd one doesn't have any even divisors. If the gcd of the two were an odd integer greater than 1, for example 3, then
$$
3|n → 3x=n → n=3x
$$
and
$$
3|n+1 → 3y=n+1 → n=3y-1,
$$
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1.
Is the part about the odd integers greater than 1 rigorous enough for a proper proof? I don't think it is; how would you prove this more rigorously? If anyone has any other improvement ideas of any kind, they're more than welcome!
elementary-number-theory proof-writing
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
$endgroup$
$begingroup$
Seems unnecessarily complex. If $d|a$ and $d|b$ then $d|(a-b)$. Hence, in your case, $d|(n+1-n)implies d|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1.
$endgroup$
– lulu
Nov 8 '15 at 22:34
$begingroup$
$gcd(a,b)le|a-b|$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:36
$begingroup$
It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given
$endgroup$
– ASKASK
Nov 8 '15 at 22:37
$begingroup$
(obv) for $ane b$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:46
add a comment |
$begingroup$
My task is to prove that gcd(n, n+1)=1 for all n>0.
It is obvious that 1 is a common divisor of both n and n+1 since
$$
1|n → 1x=n
$$
if x=n, and
$$
1|n+1 → 1y=n+1
$$
if y=n+1.
To prove that 1 is the greatest common divisor, I did as follows:
From the integers n and n+1 the other must be an even integer, and the other an odd integer. Since one of them must be odd, the gcd of the two can't be an even number, because the odd one doesn't have any even divisors. If the gcd of the two were an odd integer greater than 1, for example 3, then
$$
3|n → 3x=n → n=3x
$$
and
$$
3|n+1 → 3y=n+1 → n=3y-1,
$$
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1.
Is the part about the odd integers greater than 1 rigorous enough for a proper proof? I don't think it is; how would you prove this more rigorously? If anyone has any other improvement ideas of any kind, they're more than welcome!
elementary-number-theory proof-writing
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
$endgroup$
My task is to prove that gcd(n, n+1)=1 for all n>0.
It is obvious that 1 is a common divisor of both n and n+1 since
$$
1|n → 1x=n
$$
if x=n, and
$$
1|n+1 → 1y=n+1
$$
if y=n+1.
To prove that 1 is the greatest common divisor, I did as follows:
From the integers n and n+1 the other must be an even integer, and the other an odd integer. Since one of them must be odd, the gcd of the two can't be an even number, because the odd one doesn't have any even divisors. If the gcd of the two were an odd integer greater than 1, for example 3, then
$$
3|n → 3x=n → n=3x
$$
and
$$
3|n+1 → 3y=n+1 → n=3y-1,
$$
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1.
Is the part about the odd integers greater than 1 rigorous enough for a proper proof? I don't think it is; how would you prove this more rigorously? If anyone has any other improvement ideas of any kind, they're more than welcome!
elementary-number-theory proof-writing
elementary-number-theory proof-writing
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
asked Nov 8 '15 at 22:31
user265554user265554
1,3374815
1,3374815
$begingroup$
Seems unnecessarily complex. If $d|a$ and $d|b$ then $d|(a-b)$. Hence, in your case, $d|(n+1-n)implies d|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1.
$endgroup$
– lulu
Nov 8 '15 at 22:34
$begingroup$
$gcd(a,b)le|a-b|$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:36
$begingroup$
It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given
$endgroup$
– ASKASK
Nov 8 '15 at 22:37
$begingroup$
(obv) for $ane b$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:46
add a comment |
$begingroup$
Seems unnecessarily complex. If $d|a$ and $d|b$ then $d|(a-b)$. Hence, in your case, $d|(n+1-n)implies d|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1.
$endgroup$
– lulu
Nov 8 '15 at 22:34
$begingroup$
$gcd(a,b)le|a-b|$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:36
$begingroup$
It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given
$endgroup$
– ASKASK
Nov 8 '15 at 22:37
$begingroup$
(obv) for $ane b$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:46
$begingroup$
Seems unnecessarily complex. If $d|a$ and $d|b$ then $d|(a-b)$. Hence, in your case, $d|(n+1-n)implies d|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1.
$endgroup$
– lulu
Nov 8 '15 at 22:34
$begingroup$
Seems unnecessarily complex. If $d|a$ and $d|b$ then $d|(a-b)$. Hence, in your case, $d|(n+1-n)implies d|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1.
$endgroup$
– lulu
Nov 8 '15 at 22:34
$begingroup$
$gcd(a,b)le|a-b|$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:36
$begingroup$
$gcd(a,b)le|a-b|$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:36
$begingroup$
It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given
$endgroup$
– ASKASK
Nov 8 '15 at 22:37
$begingroup$
It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given
$endgroup$
– ASKASK
Nov 8 '15 at 22:37
$begingroup$
(obv) for $ane b$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:46
$begingroup$
(obv) for $ane b$
$endgroup$
– JonMark Perry
Nov 8 '15 at 22:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you know that $a mid b$ and $a mid c$ implies $a mid b pm c$, then this theorem is easier than you have laid it out to be. Let $a$ be any common divisor of $n$ and $n+1$. Then $a mid (n+1)-n$, so $a mid 1$. Do you think you can take it from here?
In regards to the first part, just note that if $ap=b$ and $aq=c$, then $b pm c = a(p pm q)$.
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
$endgroup$
add a comment |
$begingroup$
"3|n→3x=n→n=3x
and
3|n+1→3y=n+1→n=3y−1,
,
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1."
I really hate to say this, but if you can state that there are no integers that solve these, you could just have easily have stated "There are no integers other than 1 that divide both n and n+1" and saved yourself a lot of trouble.
In fact, that's the gyst of the matter, the only number that divides both n and n+1 is 1.
So how do you prove that? If you accept that if a|b and a|c then a|(b - c), it is easy, as a|n+1 and a|n implies a|(n+1 -n) = 1. And if a|1 = 1. So gcd(n, n+1) = 1.
If you don't know that if a|b and a|c then a|(b - c). Well you know that if a|n then n = a*m for some m. So $frac {n+ 1}{a} = m + frac{1}{a}$. If a > 1 then this is not an integer and $a$ does not divide $n + 1$. So no integer other than 1 divides both n and n + 1. So gcd(n, n+1) = 1.
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
$endgroup$
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
add a comment |
$begingroup$
If $n$ is even then $n+1$ is odd, therefore $gcd(n,n+1)=1$
Analogously, if $n$ is odd then $n+1$ is even, therefore $gcd(n,n+1)=1$.
Consecutive Numbers are coprime
and
https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you know that $a mid b$ and $a mid c$ implies $a mid b pm c$, then this theorem is easier than you have laid it out to be. Let $a$ be any common divisor of $n$ and $n+1$. Then $a mid (n+1)-n$, so $a mid 1$. Do you think you can take it from here?
In regards to the first part, just note that if $ap=b$ and $aq=c$, then $b pm c = a(p pm q)$.
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
$endgroup$
add a comment |
$begingroup$
If you know that $a mid b$ and $a mid c$ implies $a mid b pm c$, then this theorem is easier than you have laid it out to be. Let $a$ be any common divisor of $n$ and $n+1$. Then $a mid (n+1)-n$, so $a mid 1$. Do you think you can take it from here?
In regards to the first part, just note that if $ap=b$ and $aq=c$, then $b pm c = a(p pm q)$.
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
$endgroup$
add a comment |
$begingroup$
If you know that $a mid b$ and $a mid c$ implies $a mid b pm c$, then this theorem is easier than you have laid it out to be. Let $a$ be any common divisor of $n$ and $n+1$. Then $a mid (n+1)-n$, so $a mid 1$. Do you think you can take it from here?
In regards to the first part, just note that if $ap=b$ and $aq=c$, then $b pm c = a(p pm q)$.
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
$endgroup$
If you know that $a mid b$ and $a mid c$ implies $a mid b pm c$, then this theorem is easier than you have laid it out to be. Let $a$ be any common divisor of $n$ and $n+1$. Then $a mid (n+1)-n$, so $a mid 1$. Do you think you can take it from here?
In regards to the first part, just note that if $ap=b$ and $aq=c$, then $b pm c = a(p pm q)$.
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
answered Nov 8 '15 at 22:36
ASKASKASKASK
5,87931836
5,87931836
add a comment |
add a comment |
$begingroup$
"3|n→3x=n→n=3x
and
3|n+1→3y=n+1→n=3y−1,
,
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1."
I really hate to say this, but if you can state that there are no integers that solve these, you could just have easily have stated "There are no integers other than 1 that divide both n and n+1" and saved yourself a lot of trouble.
In fact, that's the gyst of the matter, the only number that divides both n and n+1 is 1.
So how do you prove that? If you accept that if a|b and a|c then a|(b - c), it is easy, as a|n+1 and a|n implies a|(n+1 -n) = 1. And if a|1 = 1. So gcd(n, n+1) = 1.
If you don't know that if a|b and a|c then a|(b - c). Well you know that if a|n then n = a*m for some m. So $frac {n+ 1}{a} = m + frac{1}{a}$. If a > 1 then this is not an integer and $a$ does not divide $n + 1$. So no integer other than 1 divides both n and n + 1. So gcd(n, n+1) = 1.
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
$endgroup$
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
add a comment |
$begingroup$
"3|n→3x=n→n=3x
and
3|n+1→3y=n+1→n=3y−1,
,
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1."
I really hate to say this, but if you can state that there are no integers that solve these, you could just have easily have stated "There are no integers other than 1 that divide both n and n+1" and saved yourself a lot of trouble.
In fact, that's the gyst of the matter, the only number that divides both n and n+1 is 1.
So how do you prove that? If you accept that if a|b and a|c then a|(b - c), it is easy, as a|n+1 and a|n implies a|(n+1 -n) = 1. And if a|1 = 1. So gcd(n, n+1) = 1.
If you don't know that if a|b and a|c then a|(b - c). Well you know that if a|n then n = a*m for some m. So $frac {n+ 1}{a} = m + frac{1}{a}$. If a > 1 then this is not an integer and $a$ does not divide $n + 1$. So no integer other than 1 divides both n and n + 1. So gcd(n, n+1) = 1.
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
$endgroup$
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
add a comment |
$begingroup$
"3|n→3x=n→n=3x
and
3|n+1→3y=n+1→n=3y−1,
,
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1."
I really hate to say this, but if you can state that there are no integers that solve these, you could just have easily have stated "There are no integers other than 1 that divide both n and n+1" and saved yourself a lot of trouble.
In fact, that's the gyst of the matter, the only number that divides both n and n+1 is 1.
So how do you prove that? If you accept that if a|b and a|c then a|(b - c), it is easy, as a|n+1 and a|n implies a|(n+1 -n) = 1. And if a|1 = 1. So gcd(n, n+1) = 1.
If you don't know that if a|b and a|c then a|(b - c). Well you know that if a|n then n = a*m for some m. So $frac {n+ 1}{a} = m + frac{1}{a}$. If a > 1 then this is not an integer and $a$ does not divide $n + 1$. So no integer other than 1 divides both n and n + 1. So gcd(n, n+1) = 1.
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
$endgroup$
"3|n→3x=n→n=3x
and
3|n+1→3y=n+1→n=3y−1,
,
and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1."
I really hate to say this, but if you can state that there are no integers that solve these, you could just have easily have stated "There are no integers other than 1 that divide both n and n+1" and saved yourself a lot of trouble.
In fact, that's the gyst of the matter, the only number that divides both n and n+1 is 1.
So how do you prove that? If you accept that if a|b and a|c then a|(b - c), it is easy, as a|n+1 and a|n implies a|(n+1 -n) = 1. And if a|1 = 1. So gcd(n, n+1) = 1.
If you don't know that if a|b and a|c then a|(b - c). Well you know that if a|n then n = a*m for some m. So $frac {n+ 1}{a} = m + frac{1}{a}$. If a > 1 then this is not an integer and $a$ does not divide $n + 1$. So no integer other than 1 divides both n and n + 1. So gcd(n, n+1) = 1.
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
answered Nov 8 '15 at 22:57
fleabloodfleablood
68.8k22685
68.8k22685
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
add a comment |
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
$begingroup$
Er... no positve integer other than 1 divides both n and n+1.....
$endgroup$
– fleablood
Nov 8 '15 at 22:58
add a comment |
$begingroup$
If $n$ is even then $n+1$ is odd, therefore $gcd(n,n+1)=1$
Analogously, if $n$ is odd then $n+1$ is even, therefore $gcd(n,n+1)=1$.
Consecutive Numbers are coprime
and
https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
$endgroup$
add a comment |
$begingroup$
If $n$ is even then $n+1$ is odd, therefore $gcd(n,n+1)=1$
Analogously, if $n$ is odd then $n+1$ is even, therefore $gcd(n,n+1)=1$.
Consecutive Numbers are coprime
and
https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
$endgroup$
add a comment |
$begingroup$
If $n$ is even then $n+1$ is odd, therefore $gcd(n,n+1)=1$
Analogously, if $n$ is odd then $n+1$ is even, therefore $gcd(n,n+1)=1$.
Consecutive Numbers are coprime
and
https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
$endgroup$
If $n$ is even then $n+1$ is odd, therefore $gcd(n,n+1)=1$
Analogously, if $n$ is odd then $n+1$ is even, therefore $gcd(n,n+1)=1$.
Consecutive Numbers are coprime
and
https://proofwiki.org/wiki/Consecutive_Integers_are_Coprime
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
answered Jan 3 at 1:35
Julio Trujillo GonzalezJulio Trujillo Gonzalez
856
856
add a comment |
add a comment |
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Seems unnecessarily complex. If $d|a$ and $d|b$ then $d|(a-b)$. Hence, in your case, $d|(n+1-n)implies d|1$. Thus the only positive common divisor of $n$ and $n+1$ is 1.
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– lulu
Nov 8 '15 at 22:34
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$gcd(a,b)le|a-b|$
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– JonMark Perry
Nov 8 '15 at 22:36
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It should be noted that the solution OP has given really is just a long-winded version of the version @lulu and I have given
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– ASKASK
Nov 8 '15 at 22:37
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(obv) for $ane b$
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– JonMark Perry
Nov 8 '15 at 22:46