Mixing
In the ring $mathbb Z_5$ (class of modulo 5) compute $[3]^{-4}$.
$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?
abstract-algebra ring-theory
edited Nov 20 '18 at 0:18
amWhy
191k28224439
asked Nov 19 '18 at 23:22
Collin Roberts
32
add a comment |
$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?
abstract-algebra ring-theory
edited Nov 20 '18 at 0:18
amWhy
191k28224439
asked Nov 19 '18 at 23:22
Collin Roberts
32
Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 '18 at 23:23
add a comment |
$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?
abstract-algebra ring-theory
edited Nov 20 '18 at 0:18
amWhy
191k28224439
asked Nov 19 '18 at 23:22
Collin Roberts
32
$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Nov 20 '18 at 0:18
amWhy
191k28224439
asked Nov 19 '18 at 23:22
Collin Roberts
32
edited Nov 20 '18 at 0:18
amWhy
191k28224439
asked Nov 19 '18 at 23:22
Collin Roberts
32
edited Nov 20 '18 at 0:18
amWhy
191k28224439
edited Nov 20 '18 at 0:18
amWhy
191k28224439
edited Nov 20 '18 at 0:18
amWhy
191k28224439
191k28224439
asked Nov 19 '18 at 23:22
Collin Roberts
32
asked Nov 19 '18 at 23:22
Collin Roberts
32
asked Nov 19 '18 at 23:22
Collin Roberts
32
32
Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 '18 at 23:23
add a comment |
Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 '18 at 23:23
Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 '18 at 23:23
Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 '18 at 23:23
add a comment |
3 Answers
3
active
oldest
votes
When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,
$$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$
You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
add a comment |
By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$
Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
add a comment |
By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).
We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.
So $3^{-4}equiv 1 pmod 5$
But that'd pretty ineffectual and ignores many things we should have at our finger tips.
If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.
If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.
But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:
$3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or
$3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
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
When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,
$$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$
You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
add a comment |
When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,
$$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$
You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
add a comment |
When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,
$$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$
You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,
$$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$
You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
answered Nov 19 '18 at 23:25
DonAntonio
177k1491225
177k1491225
add a comment |
add a comment |
By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$
Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
add a comment |
By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$
Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
add a comment |
By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$
Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$
Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
edited Nov 20 '18 at 0:21
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
answered Nov 20 '18 at 0:12
Bill Dubuque
208k29190628
208k29190628
add a comment |
add a comment |
By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).
We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.
So $3^{-4}equiv 1 pmod 5$
But that'd pretty ineffectual and ignores many things we should have at our finger tips.
If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.
If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.
But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:
$3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or
$3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
add a comment |
By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).
We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.
So $3^{-4}equiv 1 pmod 5$
But that'd pretty ineffectual and ignores many things we should have at our finger tips.
If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.
If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.
But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:
$3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or
$3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
add a comment |
By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).
We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.
So $3^{-4}equiv 1 pmod 5$
But that'd pretty ineffectual and ignores many things we should have at our finger tips.
If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.
If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.
But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:
$3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or
$3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).
We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.
So $3^{-4}equiv 1 pmod 5$
But that'd pretty ineffectual and ignores many things we should have at our finger tips.
If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.
If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.
But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:
$3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or
$3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
answered Nov 20 '18 at 0:35
fleablood
68.2k22684
68.2k22684
add a comment |
add a comment |
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Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 '18 at 23:23