Proof relatively prime numbers
$begingroup$
Let $p,q,r$ be three distinct prime numbers and $m = ptimes qtimes r$.
How many of the numbers {$1,2,...,m$} are relatively prime to $m$?
My attempt:
$m=2 times 3 times 5=30$,
$m=2times 3 times 7=42$,
$m=3times 5times 7=105$,
and I see the pattern that relatively prime numbers to $m$ are prime numbers(and their powers) except {$p,q,r$}. Prime factorization of $m$ is $ptimes qtimes r$ and I can maybe argue that $m$ cannot be divided by any other prime number.
But how do I prove this more formally ? I can't see any formal way to do it.
number-theory prime-numbers
$endgroup$
add a comment |
$begingroup$
Let $p,q,r$ be three distinct prime numbers and $m = ptimes qtimes r$.
How many of the numbers {$1,2,...,m$} are relatively prime to $m$?
My attempt:
$m=2 times 3 times 5=30$,
$m=2times 3 times 7=42$,
$m=3times 5times 7=105$,
and I see the pattern that relatively prime numbers to $m$ are prime numbers(and their powers) except {$p,q,r$}. Prime factorization of $m$ is $ptimes qtimes r$ and I can maybe argue that $m$ cannot be divided by any other prime number.
But how do I prove this more formally ? I can't see any formal way to do it.
number-theory prime-numbers
$endgroup$
1
$begingroup$
Are you familiar with Euler’s totient function(en.m.wikipedia.org/wiki/Euler%27s_totient_function)?
$endgroup$
– Anurag A
Dec 9 '18 at 19:17
$begingroup$
I didn't realize it was Euler's totien function, thank you!
$endgroup$
– Glion
Dec 9 '18 at 19:31
add a comment |
$begingroup$
Let $p,q,r$ be three distinct prime numbers and $m = ptimes qtimes r$.
How many of the numbers {$1,2,...,m$} are relatively prime to $m$?
My attempt:
$m=2 times 3 times 5=30$,
$m=2times 3 times 7=42$,
$m=3times 5times 7=105$,
and I see the pattern that relatively prime numbers to $m$ are prime numbers(and their powers) except {$p,q,r$}. Prime factorization of $m$ is $ptimes qtimes r$ and I can maybe argue that $m$ cannot be divided by any other prime number.
But how do I prove this more formally ? I can't see any formal way to do it.
number-theory prime-numbers
$endgroup$
Let $p,q,r$ be three distinct prime numbers and $m = ptimes qtimes r$.
How many of the numbers {$1,2,...,m$} are relatively prime to $m$?
My attempt:
$m=2 times 3 times 5=30$,
$m=2times 3 times 7=42$,
$m=3times 5times 7=105$,
and I see the pattern that relatively prime numbers to $m$ are prime numbers(and their powers) except {$p,q,r$}. Prime factorization of $m$ is $ptimes qtimes r$ and I can maybe argue that $m$ cannot be divided by any other prime number.
But how do I prove this more formally ? I can't see any formal way to do it.
number-theory prime-numbers
number-theory prime-numbers
edited Dec 17 '18 at 16:57
Klangen
1,72311334
1,72311334
asked Dec 9 '18 at 19:15
GlionGlion
334
334
1
$begingroup$
Are you familiar with Euler’s totient function(en.m.wikipedia.org/wiki/Euler%27s_totient_function)?
$endgroup$
– Anurag A
Dec 9 '18 at 19:17
$begingroup$
I didn't realize it was Euler's totien function, thank you!
$endgroup$
– Glion
Dec 9 '18 at 19:31
add a comment |
1
$begingroup$
Are you familiar with Euler’s totient function(en.m.wikipedia.org/wiki/Euler%27s_totient_function)?
$endgroup$
– Anurag A
Dec 9 '18 at 19:17
$begingroup$
I didn't realize it was Euler's totien function, thank you!
$endgroup$
– Glion
Dec 9 '18 at 19:31
1
1
$begingroup$
Are you familiar with Euler’s totient function(en.m.wikipedia.org/wiki/Euler%27s_totient_function)?
$endgroup$
– Anurag A
Dec 9 '18 at 19:17
$begingroup$
Are you familiar with Euler’s totient function(en.m.wikipedia.org/wiki/Euler%27s_totient_function)?
$endgroup$
– Anurag A
Dec 9 '18 at 19:17
$begingroup$
I didn't realize it was Euler's totien function, thank you!
$endgroup$
– Glion
Dec 9 '18 at 19:31
$begingroup$
I didn't realize it was Euler's totien function, thank you!
$endgroup$
– Glion
Dec 9 '18 at 19:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can clearly see it from Euler's totient function.
$endgroup$
add a comment |
$begingroup$
The solution to this is given by Euler's totient function, which can be computed explicitly using Euler's product formula (see: here)
If this machinery seems too complex, since the number you're dealing with only has three primes in its prime factorization, you could consider a "Brute Force" approach:
Let $A^{ell}_{m} :={{x : x=ccdot ell, cin mathbb{N}, xleq m}}$
Note that the sets $A^{p}_{m}, A^{q}_{m}, A^{r}_{m}, A^{pq}_{m}, A^{pr}_{m}, A^{qr}_{m}$ contain all natural numbers that are NOT relatively prime to $m$.
So how many of the numbers ${{1,2,...,m}}$ are relatively prime to $m$?
$$m-vert A^{p}_{m} cup A^{q}_{m}cup A^{r}_{m}cup A^{pq}_{m}cup A^{pr}_{m}cup A^{qr}_{m} vert $$
Note that there is some overlap between these sets, but the size of the union can easily be determined using inclusion-exclusion.
$endgroup$
add a comment |
$begingroup$
This is a simple counting problem disguised as number theory. Among the numbers $1,2,ldots,m$, it is easy to count the number of multiples of $p$, those which are multiples of $q$, and those which are multiples of $r$. Now we need to exclude numbers which we counted twice, and add back those we excluded too many times. This idea is known as the Principle of Inclusion and Exclusion and is a well-known method of counting. This can be applied to your case to find the answer, and can also be generalised to an arbitrary number of primes.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can clearly see it from Euler's totient function.
$endgroup$
add a comment |
$begingroup$
You can clearly see it from Euler's totient function.
$endgroup$
add a comment |
$begingroup$
You can clearly see it from Euler's totient function.
$endgroup$
You can clearly see it from Euler's totient function.
answered Jan 13 at 16:13
scarfacescarface
707
707
add a comment |
add a comment |
$begingroup$
The solution to this is given by Euler's totient function, which can be computed explicitly using Euler's product formula (see: here)
If this machinery seems too complex, since the number you're dealing with only has three primes in its prime factorization, you could consider a "Brute Force" approach:
Let $A^{ell}_{m} :={{x : x=ccdot ell, cin mathbb{N}, xleq m}}$
Note that the sets $A^{p}_{m}, A^{q}_{m}, A^{r}_{m}, A^{pq}_{m}, A^{pr}_{m}, A^{qr}_{m}$ contain all natural numbers that are NOT relatively prime to $m$.
So how many of the numbers ${{1,2,...,m}}$ are relatively prime to $m$?
$$m-vert A^{p}_{m} cup A^{q}_{m}cup A^{r}_{m}cup A^{pq}_{m}cup A^{pr}_{m}cup A^{qr}_{m} vert $$
Note that there is some overlap between these sets, but the size of the union can easily be determined using inclusion-exclusion.
$endgroup$
add a comment |
$begingroup$
The solution to this is given by Euler's totient function, which can be computed explicitly using Euler's product formula (see: here)
If this machinery seems too complex, since the number you're dealing with only has three primes in its prime factorization, you could consider a "Brute Force" approach:
Let $A^{ell}_{m} :={{x : x=ccdot ell, cin mathbb{N}, xleq m}}$
Note that the sets $A^{p}_{m}, A^{q}_{m}, A^{r}_{m}, A^{pq}_{m}, A^{pr}_{m}, A^{qr}_{m}$ contain all natural numbers that are NOT relatively prime to $m$.
So how many of the numbers ${{1,2,...,m}}$ are relatively prime to $m$?
$$m-vert A^{p}_{m} cup A^{q}_{m}cup A^{r}_{m}cup A^{pq}_{m}cup A^{pr}_{m}cup A^{qr}_{m} vert $$
Note that there is some overlap between these sets, but the size of the union can easily be determined using inclusion-exclusion.
$endgroup$
add a comment |
$begingroup$
The solution to this is given by Euler's totient function, which can be computed explicitly using Euler's product formula (see: here)
If this machinery seems too complex, since the number you're dealing with only has three primes in its prime factorization, you could consider a "Brute Force" approach:
Let $A^{ell}_{m} :={{x : x=ccdot ell, cin mathbb{N}, xleq m}}$
Note that the sets $A^{p}_{m}, A^{q}_{m}, A^{r}_{m}, A^{pq}_{m}, A^{pr}_{m}, A^{qr}_{m}$ contain all natural numbers that are NOT relatively prime to $m$.
So how many of the numbers ${{1,2,...,m}}$ are relatively prime to $m$?
$$m-vert A^{p}_{m} cup A^{q}_{m}cup A^{r}_{m}cup A^{pq}_{m}cup A^{pr}_{m}cup A^{qr}_{m} vert $$
Note that there is some overlap between these sets, but the size of the union can easily be determined using inclusion-exclusion.
$endgroup$
The solution to this is given by Euler's totient function, which can be computed explicitly using Euler's product formula (see: here)
If this machinery seems too complex, since the number you're dealing with only has three primes in its prime factorization, you could consider a "Brute Force" approach:
Let $A^{ell}_{m} :={{x : x=ccdot ell, cin mathbb{N}, xleq m}}$
Note that the sets $A^{p}_{m}, A^{q}_{m}, A^{r}_{m}, A^{pq}_{m}, A^{pr}_{m}, A^{qr}_{m}$ contain all natural numbers that are NOT relatively prime to $m$.
So how many of the numbers ${{1,2,...,m}}$ are relatively prime to $m$?
$$m-vert A^{p}_{m} cup A^{q}_{m}cup A^{r}_{m}cup A^{pq}_{m}cup A^{pr}_{m}cup A^{qr}_{m} vert $$
Note that there is some overlap between these sets, but the size of the union can easily be determined using inclusion-exclusion.
answered Dec 9 '18 at 19:37
mm8511mm8511
54328
54328
add a comment |
add a comment |
$begingroup$
This is a simple counting problem disguised as number theory. Among the numbers $1,2,ldots,m$, it is easy to count the number of multiples of $p$, those which are multiples of $q$, and those which are multiples of $r$. Now we need to exclude numbers which we counted twice, and add back those we excluded too many times. This idea is known as the Principle of Inclusion and Exclusion and is a well-known method of counting. This can be applied to your case to find the answer, and can also be generalised to an arbitrary number of primes.
$endgroup$
add a comment |
$begingroup$
This is a simple counting problem disguised as number theory. Among the numbers $1,2,ldots,m$, it is easy to count the number of multiples of $p$, those which are multiples of $q$, and those which are multiples of $r$. Now we need to exclude numbers which we counted twice, and add back those we excluded too many times. This idea is known as the Principle of Inclusion and Exclusion and is a well-known method of counting. This can be applied to your case to find the answer, and can also be generalised to an arbitrary number of primes.
$endgroup$
add a comment |
$begingroup$
This is a simple counting problem disguised as number theory. Among the numbers $1,2,ldots,m$, it is easy to count the number of multiples of $p$, those which are multiples of $q$, and those which are multiples of $r$. Now we need to exclude numbers which we counted twice, and add back those we excluded too many times. This idea is known as the Principle of Inclusion and Exclusion and is a well-known method of counting. This can be applied to your case to find the answer, and can also be generalised to an arbitrary number of primes.
$endgroup$
This is a simple counting problem disguised as number theory. Among the numbers $1,2,ldots,m$, it is easy to count the number of multiples of $p$, those which are multiples of $q$, and those which are multiples of $r$. Now we need to exclude numbers which we counted twice, and add back those we excluded too many times. This idea is known as the Principle of Inclusion and Exclusion and is a well-known method of counting. This can be applied to your case to find the answer, and can also be generalised to an arbitrary number of primes.
answered Dec 9 '18 at 23:59
YiFanYiFan
3,0161424
3,0161424
add a comment |
add a comment |
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1
$begingroup$
Are you familiar with Euler’s totient function(en.m.wikipedia.org/wiki/Euler%27s_totient_function)?
$endgroup$
– Anurag A
Dec 9 '18 at 19:17
$begingroup$
I didn't realize it was Euler's totien function, thank you!
$endgroup$
– Glion
Dec 9 '18 at 19:31