Mixing
How do I divide in $Bbb{Z}/pBbb Z $?
$begingroup$
How do I divide in $Bbb{Z}/pBbb Z $ ?
lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?
The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings
abstract-algebra
edited Jan 16 at 4:02
Andrews
5191318
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
$endgroup$
add a comment |
$begingroup$
How do I divide in $Bbb{Z}/pBbb Z $ ?
lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?
The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings
abstract-algebra
edited Jan 16 at 4:02
Andrews
5191318
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
$endgroup$
$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33
add a comment |
$begingroup$
How do I divide in $Bbb{Z}/pBbb Z $ ?
lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?
The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings
abstract-algebra
edited Jan 16 at 4:02
Andrews
5191318
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
$endgroup$
How do I divide in $Bbb{Z}/pBbb Z $ ?
lets assume I'm in $Bbb{Z}/5Bbb Z $ so how do I calculate $17/3$ as an example?
The reason I'm asking is later it is becoming difficult when we calculate with polynomials in quotient rings
abstract-algebra
abstract-algebra
edited Jan 16 at 4:02
Andrews
5191318
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
edited Jan 16 at 4:02
Andrews
5191318
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
edited Jan 16 at 4:02
Andrews
5191318
edited Jan 16 at 4:02
Andrews
5191318
edited Jan 16 at 4:02
Andrews
5191318
5191318
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
asked Jan 15 at 23:41
MathsGuyMathsGuy
295
295
$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33
add a comment |
$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33
$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33
$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$
answered Jan 15 at 23:46
user289143user289143
903313
$endgroup$
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
1
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
add a comment |
$begingroup$
To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.
Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
$;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$
edited Jan 16 at 4:27
Chris Custer
13.7k3827
answered Jan 15 at 23:51
BernardBernard
121k740116
$endgroup$
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
|
show 2 more comments
$begingroup$
Well, in any commutative ring $R$, the division is defined as
$$a:b = frac{a}{b} = acdot b^{-1},$$
where $a,bin R$ and $b$ is a unit (invertible) in $R$.
In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
$endgroup$
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
$begingroup$
Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$
answered Jan 15 at 23:46
user289143user289143
903313
$endgroup$
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
1
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
add a comment |
$begingroup$
Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$
answered Jan 15 at 23:46
user289143user289143
903313
$endgroup$
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
1
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
add a comment |
$begingroup$
Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$
answered Jan 15 at 23:46
user289143user289143
903313
$endgroup$
Since $mathbb{Z}/pmathbb{Z}$ is a field for $p$ prime, every element has a multiplicative inverse. Hence $frac{a}{b}=ab^{-1}$. In our case $3^{-1}=2$ since $3 cdot 2= 6 =1$ and $frac{17}{3}=17 cdot 2= 34= 4$
answered Jan 15 at 23:46
user289143user289143
903313
answered Jan 15 at 23:46
user289143user289143
903313
answered Jan 15 at 23:46
user289143user289143
903313
answered Jan 15 at 23:46
user289143user289143
903313
903313
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
1
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
add a comment |
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
1
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
$begingroup$
what if we had $Z/4Z$ or in general with polynomials
$endgroup$
– MathsGuy
Jan 15 at 23:47
1
1
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
If $n$ is not prime, $mathbb{Z}/nmathbb{Z}$ is not a field, so not every element has a multiplicative inverse and there are also zero divisors (take for example $3$ in $mathbb{Z}/6mathbb{Z}$). What do you mean with polynomials?
$endgroup$
– user289143
Jan 15 at 23:50
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
so i cannot calcute it when my construct is not a field right ? i mean if i have something like $Z/2Z}[X] / (x+1) $ as an example
$endgroup$
– MathsGuy
Jan 15 at 23:55
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
$begingroup$
Every nonzero element in Z/pZ has an inverse. In general $ain Z/nZ$ has a multiplicative inverse if and only if $gcd(a,n)=1$.
$endgroup$
– JavaMan
Jan 16 at 4:32
add a comment |
$begingroup$
To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.
Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
$;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$
edited Jan 16 at 4:27
Chris Custer
13.7k3827
answered Jan 15 at 23:51
BernardBernard
121k740116
$endgroup$
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
|
show 2 more comments
$begingroup$
To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.
Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
$;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$
edited Jan 16 at 4:27
Chris Custer
13.7k3827
answered Jan 15 at 23:51
BernardBernard
121k740116
$endgroup$
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
|
show 2 more comments
$begingroup$
To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.
Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
$;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$
edited Jan 16 at 4:27
Chris Custer
13.7k3827
answered Jan 15 at 23:51
BernardBernard
121k740116
$endgroup$
To divide, you multiply by the inverse, which is known as soon as you have a Bézout's relation with $p$.
Her for instance, in, $mathbf Z/5mathbf Z$, you have $17equiv 2mod 5$ and
$;3cdot 2-5=1$,hence $;3^{-1}equiv 2mod 5$, so$$frac{17}3text{ denotes actually } ;2cdot 2equiv 4equiv -1mod 5.$$
edited Jan 16 at 4:27
Chris Custer
13.7k3827
answered Jan 15 at 23:51
BernardBernard
121k740116
edited Jan 16 at 4:27
Chris Custer
13.7k3827
edited Jan 16 at 4:27
Chris Custer
13.7k3827
edited Jan 16 at 4:27
Chris Custer
13.7k3827
13.7k3827
answered Jan 15 at 23:51
BernardBernard
121k740116
answered Jan 15 at 23:51
BernardBernard
121k740116
answered Jan 15 at 23:51
BernardBernard
121k740116
121k740116
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
|
show 2 more comments
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
what if my construct was not a field such as $mathbf Z/4mathbf Z$
$endgroup$
– MathsGuy
Jan 16 at 0:04
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
The denomi,ator should be coprime to the modulus to be invertible, and you still have a Bézout's relation. Other than that, you cannot always divide when you're not in a field.
$endgroup$
– Bernard
Jan 16 at 0:06
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
so if my denominator is not invertible i cannot do that. $mathbf Z/4mathbf Z$ if my denominator is 3 it is not possible ?
$endgroup$
– MathsGuy
Jan 16 at 0:12
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
$3$ is its own inverse module $3$, so you ‘divide’ by $3$. However, I wouldnt use this vobulary, which is rather confusional with what happens in $mathbf R$.
$endgroup$
– Bernard
Jan 16 at 0:18
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
$begingroup$
sorry i meant $2$
$endgroup$
– MathsGuy
Jan 16 at 0:20
|
show 2 more comments
$begingroup$
Well, in any commutative ring $R$, the division is defined as
$$a:b = frac{a}{b} = acdot b^{-1},$$
where $a,bin R$ and $b$ is a unit (invertible) in $R$.
In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
$endgroup$
add a comment |
$begingroup$
Well, in any commutative ring $R$, the division is defined as
$$a:b = frac{a}{b} = acdot b^{-1},$$
where $a,bin R$ and $b$ is a unit (invertible) in $R$.
In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
$endgroup$
add a comment |
$begingroup$
Well, in any commutative ring $R$, the division is defined as
$$a:b = frac{a}{b} = acdot b^{-1},$$
where $a,bin R$ and $b$ is a unit (invertible) in $R$.
In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
$endgroup$
Well, in any commutative ring $R$, the division is defined as
$$a:b = frac{a}{b} = acdot b^{-1},$$
where $a,bin R$ and $b$ is a unit (invertible) in $R$.
In your case, $17equiv 2mod 5$ and so $17/3 = 2/3$. Thus $2/3 = 2cdot 3^{-1} = 2cdot 2 = 4$.
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
answered Jan 16 at 8:43
WuestenfuxWuestenfux
4,7311513
4,7311513
add a comment |
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$begingroup$
There is no such thing as division in general for these rings. Here the analogous operation is multiplying by the multiplicative inverse.
$endgroup$
– JavaMan
Jan 16 at 4:33