Number theory: two relatively prime numbers, necessarily one must be odd and the other even?
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Let s and t be two different positive integers which are relatively prime. Does this property alone (of being relatively prime) NECESSARILY IMPLIES that one of s and t MUST be even and the other odd?
If yes, how would you prove so?
number-theory
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
$endgroup$
add a comment |
$begingroup$
Let s and t be two different positive integers which are relatively prime. Does this property alone (of being relatively prime) NECESSARILY IMPLIES that one of s and t MUST be even and the other odd?
If yes, how would you prove so?
number-theory
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
$endgroup$
1
$begingroup$
No. All it means are the two numbers have no (non-trivial) factors in common. There's utterly no reason either of them need to have $2$ as a factor any more than it'd mean one would have to have $3$ or any other prime as a factor.. For example $91 = 7*13$ and $275 = 5^2* 11$ are relatively prime but neither have $2$ as factor. (nor $3$, nor $17$ .....)
$endgroup$
– fleablood
Dec 27 '18 at 1:54
$begingroup$
You only know that either one of them is even and one of them is odd, or both are odd. You definitely know that one has to be odd, as if both are even, they would share common factor $2$. However, it is not necessary that one has to be even.
$endgroup$
– Haran
Dec 27 '18 at 5:05
$begingroup$
The answer to your question seems logically clear: Imagine $a$ and $b$ such that $gcd(a,b) = 1$. Then we have four possibility regarding to parity of $a,b$: (1) both of them are even, (2)&(3) one of them even, the other odd, (4) both of them are odd. We can observe that it is only in (1)st case that we have a strict conclusion via condition, ie. if both $a,b$ are even, $gcd(a,b) ge 2$ as $2$ is a common factor, while we cannot conclude a thing in other situations. So we show that two relatively prime numbers are necessiraliy NOT both even.
$endgroup$
– freehumorist
Dec 27 '18 at 15:45
$begingroup$
See, that by "necessarily" your reasonment wish to know the implication forward. (Tough, here backwards implication seems no trouble once the first established.)
$endgroup$
– freehumorist
Dec 27 '18 at 15:47
add a comment |
$begingroup$
Let s and t be two different positive integers which are relatively prime. Does this property alone (of being relatively prime) NECESSARILY IMPLIES that one of s and t MUST be even and the other odd?
If yes, how would you prove so?
number-theory
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
$endgroup$
Let s and t be two different positive integers which are relatively prime. Does this property alone (of being relatively prime) NECESSARILY IMPLIES that one of s and t MUST be even and the other odd?
If yes, how would you prove so?
number-theory
number-theory
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
asked Dec 27 '18 at 1:39
Daniel RodriguezDaniel Rodriguez
62
62
1
$begingroup$
No. All it means are the two numbers have no (non-trivial) factors in common. There's utterly no reason either of them need to have $2$ as a factor any more than it'd mean one would have to have $3$ or any other prime as a factor.. For example $91 = 7*13$ and $275 = 5^2* 11$ are relatively prime but neither have $2$ as factor. (nor $3$, nor $17$ .....)
$endgroup$
– fleablood
Dec 27 '18 at 1:54
$begingroup$
You only know that either one of them is even and one of them is odd, or both are odd. You definitely know that one has to be odd, as if both are even, they would share common factor $2$. However, it is not necessary that one has to be even.
$endgroup$
– Haran
Dec 27 '18 at 5:05
$begingroup$
The answer to your question seems logically clear: Imagine $a$ and $b$ such that $gcd(a,b) = 1$. Then we have four possibility regarding to parity of $a,b$: (1) both of them are even, (2)&(3) one of them even, the other odd, (4) both of them are odd. We can observe that it is only in (1)st case that we have a strict conclusion via condition, ie. if both $a,b$ are even, $gcd(a,b) ge 2$ as $2$ is a common factor, while we cannot conclude a thing in other situations. So we show that two relatively prime numbers are necessiraliy NOT both even.
$endgroup$
– freehumorist
Dec 27 '18 at 15:45
$begingroup$
See, that by "necessarily" your reasonment wish to know the implication forward. (Tough, here backwards implication seems no trouble once the first established.)
$endgroup$
– freehumorist
Dec 27 '18 at 15:47
add a comment |
1
$begingroup$
No. All it means are the two numbers have no (non-trivial) factors in common. There's utterly no reason either of them need to have $2$ as a factor any more than it'd mean one would have to have $3$ or any other prime as a factor.. For example $91 = 7*13$ and $275 = 5^2* 11$ are relatively prime but neither have $2$ as factor. (nor $3$, nor $17$ .....)
$endgroup$
– fleablood
Dec 27 '18 at 1:54
$begingroup$
You only know that either one of them is even and one of them is odd, or both are odd. You definitely know that one has to be odd, as if both are even, they would share common factor $2$. However, it is not necessary that one has to be even.
$endgroup$
– Haran
Dec 27 '18 at 5:05
$begingroup$
The answer to your question seems logically clear: Imagine $a$ and $b$ such that $gcd(a,b) = 1$. Then we have four possibility regarding to parity of $a,b$: (1) both of them are even, (2)&(3) one of them even, the other odd, (4) both of them are odd. We can observe that it is only in (1)st case that we have a strict conclusion via condition, ie. if both $a,b$ are even, $gcd(a,b) ge 2$ as $2$ is a common factor, while we cannot conclude a thing in other situations. So we show that two relatively prime numbers are necessiraliy NOT both even.
$endgroup$
– freehumorist
Dec 27 '18 at 15:45
$begingroup$
See, that by "necessarily" your reasonment wish to know the implication forward. (Tough, here backwards implication seems no trouble once the first established.)
$endgroup$
– freehumorist
Dec 27 '18 at 15:47
1
1
$begingroup$
No. All it means are the two numbers have no (non-trivial) factors in common. There's utterly no reason either of them need to have $2$ as a factor any more than it'd mean one would have to have $3$ or any other prime as a factor.. For example $91 = 7*13$ and $275 = 5^2* 11$ are relatively prime but neither have $2$ as factor. (nor $3$, nor $17$ .....)
$endgroup$
– fleablood
Dec 27 '18 at 1:54
$begingroup$
No. All it means are the two numbers have no (non-trivial) factors in common. There's utterly no reason either of them need to have $2$ as a factor any more than it'd mean one would have to have $3$ or any other prime as a factor.. For example $91 = 7*13$ and $275 = 5^2* 11$ are relatively prime but neither have $2$ as factor. (nor $3$, nor $17$ .....)
$endgroup$
– fleablood
Dec 27 '18 at 1:54
$begingroup$
You only know that either one of them is even and one of them is odd, or both are odd. You definitely know that one has to be odd, as if both are even, they would share common factor $2$. However, it is not necessary that one has to be even.
$endgroup$
– Haran
Dec 27 '18 at 5:05
$begingroup$
You only know that either one of them is even and one of them is odd, or both are odd. You definitely know that one has to be odd, as if both are even, they would share common factor $2$. However, it is not necessary that one has to be even.
$endgroup$
– Haran
Dec 27 '18 at 5:05
$begingroup$
The answer to your question seems logically clear: Imagine $a$ and $b$ such that $gcd(a,b) = 1$. Then we have four possibility regarding to parity of $a,b$: (1) both of them are even, (2)&(3) one of them even, the other odd, (4) both of them are odd. We can observe that it is only in (1)st case that we have a strict conclusion via condition, ie. if both $a,b$ are even, $gcd(a,b) ge 2$ as $2$ is a common factor, while we cannot conclude a thing in other situations. So we show that two relatively prime numbers are necessiraliy NOT both even.
$endgroup$
– freehumorist
Dec 27 '18 at 15:45
$begingroup$
The answer to your question seems logically clear: Imagine $a$ and $b$ such that $gcd(a,b) = 1$. Then we have four possibility regarding to parity of $a,b$: (1) both of them are even, (2)&(3) one of them even, the other odd, (4) both of them are odd. We can observe that it is only in (1)st case that we have a strict conclusion via condition, ie. if both $a,b$ are even, $gcd(a,b) ge 2$ as $2$ is a common factor, while we cannot conclude a thing in other situations. So we show that two relatively prime numbers are necessiraliy NOT both even.
$endgroup$
– freehumorist
Dec 27 '18 at 15:45
$begingroup$
See, that by "necessarily" your reasonment wish to know the implication forward. (Tough, here backwards implication seems no trouble once the first established.)
$endgroup$
– freehumorist
Dec 27 '18 at 15:47
$begingroup$
See, that by "necessarily" your reasonment wish to know the implication forward. (Tough, here backwards implication seems no trouble once the first established.)
$endgroup$
– freehumorist
Dec 27 '18 at 15:47
add a comment |
4 Answers
4
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No, for example $3$ and $5$ are relatively prime, with both being odd integers. For that matter, all primes, including $2$, are relatively prime to each other.
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
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add a comment |
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If you know that one of the numbers is even, yes. All even numbers share a common factor of $2$ so can't be relatively prime.
However, if not then no. Use the representations of odd numbers as $En+O$ (where $E$ = even numbers, $O$ = odd) to see why. For example, all odd numbers are of the form $4n-1$ or $4n+1$. Can you show how any of the following pairs:
$$4m-1, 4n-1$$
$$4m-1, 4n+1$$
$$4m+1, 4n+1$$
may or may not be relatively prime?
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
$endgroup$
add a comment |
$begingroup$
Suppose $a = prod p_i^{k_i}$ is the unique prime factorization of $a$ and $b= prod q_j^{m_j}$ is the unique prime factorization of $b$.
$a$ and $b$ being relatively prime means that none of $p_i$ equal any of the $q_j$ and vice versa.
$a$ is odd if none of the $p_i$ equal to $2$ and $a$ is even if one of the $p_i$ is equal to $2$. Likewise $b$ is odd if none of the $q_i$ are equal to $2$ and $b$ is odd if one of the $q_i = 2$.
So your statement is a matter of proving: If ${p_i}$ and ${q_i}$ are collections of primes with no primes in common, if one of the collections does not contain $2$, does that mean the other collection must contain $2$.
And the answer to that is: Of course not.
And to come up with a counter example we can have something as simple as ${3}$ and ${5}$ and $a = 3$ and $b = 5$. They are relatively prime and neither are even.
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Counterexample with composite nunbers: pick several odd prime numbers (i.e. ones other than $2$), and multiply them together to get $a$. Pick several more that you didn't use the first time, and multiply those together to get $b$. $a$ and $b$ are both odd, and have no factors in common.
Two numbers being even means they're both divisible by $2$, but being odd only tells us that $2$ isn't a factor of either of them.
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, for example $3$ and $5$ are relatively prime, with both being odd integers. For that matter, all primes, including $2$, are relatively prime to each other.
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
$endgroup$
add a comment |
$begingroup$
No, for example $3$ and $5$ are relatively prime, with both being odd integers. For that matter, all primes, including $2$, are relatively prime to each other.
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
$endgroup$
add a comment |
$begingroup$
No, for example $3$ and $5$ are relatively prime, with both being odd integers. For that matter, all primes, including $2$, are relatively prime to each other.
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
$endgroup$
No, for example $3$ and $5$ are relatively prime, with both being odd integers. For that matter, all primes, including $2$, are relatively prime to each other.
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
answered Dec 27 '18 at 1:41
John OmielanJohn Omielan
3,9251215
3,9251215
add a comment |
add a comment |
$begingroup$
If you know that one of the numbers is even, yes. All even numbers share a common factor of $2$ so can't be relatively prime.
However, if not then no. Use the representations of odd numbers as $En+O$ (where $E$ = even numbers, $O$ = odd) to see why. For example, all odd numbers are of the form $4n-1$ or $4n+1$. Can you show how any of the following pairs:
$$4m-1, 4n-1$$
$$4m-1, 4n+1$$
$$4m+1, 4n+1$$
may or may not be relatively prime?
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
$endgroup$
add a comment |
$begingroup$
If you know that one of the numbers is even, yes. All even numbers share a common factor of $2$ so can't be relatively prime.
However, if not then no. Use the representations of odd numbers as $En+O$ (where $E$ = even numbers, $O$ = odd) to see why. For example, all odd numbers are of the form $4n-1$ or $4n+1$. Can you show how any of the following pairs:
$$4m-1, 4n-1$$
$$4m-1, 4n+1$$
$$4m+1, 4n+1$$
may or may not be relatively prime?
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
$endgroup$
add a comment |
$begingroup$
If you know that one of the numbers is even, yes. All even numbers share a common factor of $2$ so can't be relatively prime.
However, if not then no. Use the representations of odd numbers as $En+O$ (where $E$ = even numbers, $O$ = odd) to see why. For example, all odd numbers are of the form $4n-1$ or $4n+1$. Can you show how any of the following pairs:
$$4m-1, 4n-1$$
$$4m-1, 4n+1$$
$$4m+1, 4n+1$$
may or may not be relatively prime?
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
$endgroup$
If you know that one of the numbers is even, yes. All even numbers share a common factor of $2$ so can't be relatively prime.
However, if not then no. Use the representations of odd numbers as $En+O$ (where $E$ = even numbers, $O$ = odd) to see why. For example, all odd numbers are of the form $4n-1$ or $4n+1$. Can you show how any of the following pairs:
$$4m-1, 4n-1$$
$$4m-1, 4n+1$$
$$4m+1, 4n+1$$
may or may not be relatively prime?
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
answered Dec 27 '18 at 1:57
Rhys HughesRhys Hughes
6,9901630
6,9901630
add a comment |
add a comment |
$begingroup$
Suppose $a = prod p_i^{k_i}$ is the unique prime factorization of $a$ and $b= prod q_j^{m_j}$ is the unique prime factorization of $b$.
$a$ and $b$ being relatively prime means that none of $p_i$ equal any of the $q_j$ and vice versa.
$a$ is odd if none of the $p_i$ equal to $2$ and $a$ is even if one of the $p_i$ is equal to $2$. Likewise $b$ is odd if none of the $q_i$ are equal to $2$ and $b$ is odd if one of the $q_i = 2$.
So your statement is a matter of proving: If ${p_i}$ and ${q_i}$ are collections of primes with no primes in common, if one of the collections does not contain $2$, does that mean the other collection must contain $2$.
And the answer to that is: Of course not.
And to come up with a counter example we can have something as simple as ${3}$ and ${5}$ and $a = 3$ and $b = 5$. They are relatively prime and neither are even.
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Suppose $a = prod p_i^{k_i}$ is the unique prime factorization of $a$ and $b= prod q_j^{m_j}$ is the unique prime factorization of $b$.
$a$ and $b$ being relatively prime means that none of $p_i$ equal any of the $q_j$ and vice versa.
$a$ is odd if none of the $p_i$ equal to $2$ and $a$ is even if one of the $p_i$ is equal to $2$. Likewise $b$ is odd if none of the $q_i$ are equal to $2$ and $b$ is odd if one of the $q_i = 2$.
So your statement is a matter of proving: If ${p_i}$ and ${q_i}$ are collections of primes with no primes in common, if one of the collections does not contain $2$, does that mean the other collection must contain $2$.
And the answer to that is: Of course not.
And to come up with a counter example we can have something as simple as ${3}$ and ${5}$ and $a = 3$ and $b = 5$. They are relatively prime and neither are even.
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
$endgroup$
add a comment |
$begingroup$
Suppose $a = prod p_i^{k_i}$ is the unique prime factorization of $a$ and $b= prod q_j^{m_j}$ is the unique prime factorization of $b$.
$a$ and $b$ being relatively prime means that none of $p_i$ equal any of the $q_j$ and vice versa.
$a$ is odd if none of the $p_i$ equal to $2$ and $a$ is even if one of the $p_i$ is equal to $2$. Likewise $b$ is odd if none of the $q_i$ are equal to $2$ and $b$ is odd if one of the $q_i = 2$.
So your statement is a matter of proving: If ${p_i}$ and ${q_i}$ are collections of primes with no primes in common, if one of the collections does not contain $2$, does that mean the other collection must contain $2$.
And the answer to that is: Of course not.
And to come up with a counter example we can have something as simple as ${3}$ and ${5}$ and $a = 3$ and $b = 5$. They are relatively prime and neither are even.
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
$endgroup$
Suppose $a = prod p_i^{k_i}$ is the unique prime factorization of $a$ and $b= prod q_j^{m_j}$ is the unique prime factorization of $b$.
$a$ and $b$ being relatively prime means that none of $p_i$ equal any of the $q_j$ and vice versa.
$a$ is odd if none of the $p_i$ equal to $2$ and $a$ is even if one of the $p_i$ is equal to $2$. Likewise $b$ is odd if none of the $q_i$ are equal to $2$ and $b$ is odd if one of the $q_i = 2$.
So your statement is a matter of proving: If ${p_i}$ and ${q_i}$ are collections of primes with no primes in common, if one of the collections does not contain $2$, does that mean the other collection must contain $2$.
And the answer to that is: Of course not.
And to come up with a counter example we can have something as simple as ${3}$ and ${5}$ and $a = 3$ and $b = 5$. They are relatively prime and neither are even.
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
answered Dec 27 '18 at 2:47
fleabloodfleablood
72.3k22687
72.3k22687
add a comment |
add a comment |
$begingroup$
Counterexample with composite nunbers: pick several odd prime numbers (i.e. ones other than $2$), and multiply them together to get $a$. Pick several more that you didn't use the first time, and multiply those together to get $b$. $a$ and $b$ are both odd, and have no factors in common.
Two numbers being even means they're both divisible by $2$, but being odd only tells us that $2$ isn't a factor of either of them.
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
Counterexample with composite nunbers: pick several odd prime numbers (i.e. ones other than $2$), and multiply them together to get $a$. Pick several more that you didn't use the first time, and multiply those together to get $b$. $a$ and $b$ are both odd, and have no factors in common.
Two numbers being even means they're both divisible by $2$, but being odd only tells us that $2$ isn't a factor of either of them.
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
Counterexample with composite nunbers: pick several odd prime numbers (i.e. ones other than $2$), and multiply them together to get $a$. Pick several more that you didn't use the first time, and multiply those together to get $b$. $a$ and $b$ are both odd, and have no factors in common.
Two numbers being even means they're both divisible by $2$, but being odd only tells us that $2$ isn't a factor of either of them.
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
$endgroup$
Counterexample with composite nunbers: pick several odd prime numbers (i.e. ones other than $2$), and multiply them together to get $a$. Pick several more that you didn't use the first time, and multiply those together to get $b$. $a$ and $b$ are both odd, and have no factors in common.
Two numbers being even means they're both divisible by $2$, but being odd only tells us that $2$ isn't a factor of either of them.
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
answered Dec 27 '18 at 3:53
timtfjtimtfj
2,468420
2,468420
add a comment |
add a comment |
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No. All it means are the two numbers have no (non-trivial) factors in common. There's utterly no reason either of them need to have $2$ as a factor any more than it'd mean one would have to have $3$ or any other prime as a factor.. For example $91 = 7*13$ and $275 = 5^2* 11$ are relatively prime but neither have $2$ as factor. (nor $3$, nor $17$ .....)
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– fleablood
Dec 27 '18 at 1:54
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You only know that either one of them is even and one of them is odd, or both are odd. You definitely know that one has to be odd, as if both are even, they would share common factor $2$. However, it is not necessary that one has to be even.
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– Haran
Dec 27 '18 at 5:05
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The answer to your question seems logically clear: Imagine $a$ and $b$ such that $gcd(a,b) = 1$. Then we have four possibility regarding to parity of $a,b$: (1) both of them are even, (2)&(3) one of them even, the other odd, (4) both of them are odd. We can observe that it is only in (1)st case that we have a strict conclusion via condition, ie. if both $a,b$ are even, $gcd(a,b) ge 2$ as $2$ is a common factor, while we cannot conclude a thing in other situations. So we show that two relatively prime numbers are necessiraliy NOT both even.
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– freehumorist
Dec 27 '18 at 15:45
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See, that by "necessarily" your reasonment wish to know the implication forward. (Tough, here backwards implication seems no trouble once the first established.)
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– freehumorist
Dec 27 '18 at 15:47