Inverse of a equation with modulo operator The Next CEO of Stack OverflowHow to find the inverse modulo m?Notation for modulo: congruence relation vs operatorSolving equations with modSolving to find the general equation with a “mod” equationRules for modulo of bitwise xorRelation between powers of inverse modulo n.Solve equation with modulo operatorreversing modulo equationAlgebra rules when finding inverse moduloSolving modulo Equation with denominator
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Inverse of a equation with modulo operator
The Next CEO of Stack OverflowHow to find the inverse modulo m?Notation for modulo: congruence relation vs operatorSolving equations with modSolving to find the general equation with a “mod” equationRules for modulo of bitwise xorRelation between powers of inverse modulo n.Solve equation with modulo operatorreversing modulo equationAlgebra rules when finding inverse moduloSolving modulo Equation with denominator
$begingroup$
I have this equation:
$y=ax+b quad pmod26$
where a, b are two parameters. I would like to calculate the inverse of this equation, but I don't know which algebra rules I have to apply. Can you help me?
modular-arithmetic
edited Mar 18 at 12:07
Bernard
123k741117
asked Mar 18 at 11:45
JhdoeJhdoe
537
$endgroup$
add a comment |
$begingroup$
I have this equation:
$y=ax+b quad pmod26$
where a, b are two parameters. I would like to calculate the inverse of this equation, but I don't know which algebra rules I have to apply. Can you help me?
modular-arithmetic
edited Mar 18 at 12:07
Bernard
123k741117
asked Mar 18 at 11:45
JhdoeJhdoe
537
$endgroup$
1
$begingroup$
By "the inverse," do you mean $x=cy+dpmod26$? If so, it may not be doable. It depends on $a$. Is there anything else you can tell us about $a$?
$endgroup$
– Cameron Buie
Mar 18 at 11:51
1
$begingroup$
The usual rules: subtract $b$ from both sides, and multiply both sides by the multiplicative inverse of $a$. Here, the multiplicative inverse of $a$ is the element $c$ such that $ac$ is the multiplicative identity. And the multiplicative identity is the element $1bmod26$.
$endgroup$
– Gerry Myerson
Mar 18 at 11:55
$begingroup$
As @Gerry points out, it isn't much different from what you're used to. The big difference is that $a$ may not have a multiplicative inverse modulo 26.
$endgroup$
– Cameron Buie
Mar 18 at 12:02
add a comment |
$begingroup$
I have this equation:
$y=ax+b quad pmod26$
where a, b are two parameters. I would like to calculate the inverse of this equation, but I don't know which algebra rules I have to apply. Can you help me?
modular-arithmetic
edited Mar 18 at 12:07
Bernard
123k741117
asked Mar 18 at 11:45
JhdoeJhdoe
537
$endgroup$
I have this equation:
$y=ax+b quad pmod26$
where a, b are two parameters. I would like to calculate the inverse of this equation, but I don't know which algebra rules I have to apply. Can you help me?
modular-arithmetic
modular-arithmetic
edited Mar 18 at 12:07
Bernard
123k741117
asked Mar 18 at 11:45
JhdoeJhdoe
537
edited Mar 18 at 12:07
Bernard
123k741117
asked Mar 18 at 11:45
JhdoeJhdoe
537
edited Mar 18 at 12:07
Bernard
123k741117
edited Mar 18 at 12:07
Bernard
123k741117
edited Mar 18 at 12:07
Bernard
123k741117
123k741117
asked Mar 18 at 11:45
JhdoeJhdoe
537
asked Mar 18 at 11:45
JhdoeJhdoe
537
asked Mar 18 at 11:45
JhdoeJhdoe
537
537
1
$begingroup$
By "the inverse," do you mean $x=cy+dpmod26$? If so, it may not be doable. It depends on $a$. Is there anything else you can tell us about $a$?
$endgroup$
– Cameron Buie
Mar 18 at 11:51
1
$begingroup$
The usual rules: subtract $b$ from both sides, and multiply both sides by the multiplicative inverse of $a$. Here, the multiplicative inverse of $a$ is the element $c$ such that $ac$ is the multiplicative identity. And the multiplicative identity is the element $1bmod26$.
$endgroup$
– Gerry Myerson
Mar 18 at 11:55
$begingroup$
As @Gerry points out, it isn't much different from what you're used to. The big difference is that $a$ may not have a multiplicative inverse modulo 26.
$endgroup$
– Cameron Buie
Mar 18 at 12:02
add a comment |
1
$begingroup$
By "the inverse," do you mean $x=cy+dpmod26$? If so, it may not be doable. It depends on $a$. Is there anything else you can tell us about $a$?
$endgroup$
– Cameron Buie
Mar 18 at 11:51
1
$begingroup$
The usual rules: subtract $b$ from both sides, and multiply both sides by the multiplicative inverse of $a$. Here, the multiplicative inverse of $a$ is the element $c$ such that $ac$ is the multiplicative identity. And the multiplicative identity is the element $1bmod26$.
$endgroup$
– Gerry Myerson
Mar 18 at 11:55
$begingroup$
As @Gerry points out, it isn't much different from what you're used to. The big difference is that $a$ may not have a multiplicative inverse modulo 26.
$endgroup$
– Cameron Buie
Mar 18 at 12:02
1
1
$begingroup$
By "the inverse," do you mean $x=cy+dpmod26$? If so, it may not be doable. It depends on $a$. Is there anything else you can tell us about $a$?
$endgroup$
– Cameron Buie
Mar 18 at 11:51
$begingroup$
By "the inverse," do you mean $x=cy+dpmod26$? If so, it may not be doable. It depends on $a$. Is there anything else you can tell us about $a$?
$endgroup$
– Cameron Buie
Mar 18 at 11:51
1
1
$begingroup$
The usual rules: subtract $b$ from both sides, and multiply both sides by the multiplicative inverse of $a$. Here, the multiplicative inverse of $a$ is the element $c$ such that $ac$ is the multiplicative identity. And the multiplicative identity is the element $1bmod26$.
$endgroup$
– Gerry Myerson
Mar 18 at 11:55
$begingroup$
The usual rules: subtract $b$ from both sides, and multiply both sides by the multiplicative inverse of $a$. Here, the multiplicative inverse of $a$ is the element $c$ such that $ac$ is the multiplicative identity. And the multiplicative identity is the element $1bmod26$.
$endgroup$
– Gerry Myerson
Mar 18 at 11:55
$begingroup$
As @Gerry points out, it isn't much different from what you're used to. The big difference is that $a$ may not have a multiplicative inverse modulo 26.
$endgroup$
– Cameron Buie
Mar 18 at 12:02
$begingroup$
As @Gerry points out, it isn't much different from what you're used to. The big difference is that $a$ may not have a multiplicative inverse modulo 26.
$endgroup$
– Cameron Buie
Mar 18 at 12:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can find an inverse equation only if $gcd(a,26)=1$, then $a$ has an inverse $a^-1$ such that$$acdot a^-1equiv 1mod26$$therefore$$yequiv ax+bmod26iff a^-1yequiv x+a^-1bmod 26\iff\ xequiv a^-1y-a^-1bmod 26$$
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
1
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
You can find an inverse equation only if $gcd(a,26)=1$, then $a$ has an inverse $a^-1$ such that$$acdot a^-1equiv 1mod26$$therefore$$yequiv ax+bmod26iff a^-1yequiv x+a^-1bmod 26\iff\ xequiv a^-1y-a^-1bmod 26$$
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
1
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
add a comment |
$begingroup$
You can find an inverse equation only if $gcd(a,26)=1$, then $a$ has an inverse $a^-1$ such that$$acdot a^-1equiv 1mod26$$therefore$$yequiv ax+bmod26iff a^-1yequiv x+a^-1bmod 26\iff\ xequiv a^-1y-a^-1bmod 26$$
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
1
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
add a comment |
$begingroup$
You can find an inverse equation only if $gcd(a,26)=1$, then $a$ has an inverse $a^-1$ such that$$acdot a^-1equiv 1mod26$$therefore$$yequiv ax+bmod26iff a^-1yequiv x+a^-1bmod 26\iff\ xequiv a^-1y-a^-1bmod 26$$
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
You can find an inverse equation only if $gcd(a,26)=1$, then $a$ has an inverse $a^-1$ such that$$acdot a^-1equiv 1mod26$$therefore$$yequiv ax+bmod26iff a^-1yequiv x+a^-1bmod 26\iff\ xequiv a^-1y-a^-1bmod 26$$
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 18 at 12:02
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
1
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
add a comment |
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
1
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
$begingroup$
If i find an $a^-1$ such as $a^-1aequiv 1 quad (mod26)$, what assures me that $a^-1axequiv xquad (mod26)$ ?
$endgroup$
– Jhdoe
Mar 18 at 14:21
1
1
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
$begingroup$
Two things: (1) the definition of inverse $$acdot a^-1equiv 1$$and (2) the neutrality of $1$ $$1cdot xequiv x$$
$endgroup$
– Mostafa Ayaz
Mar 18 at 14:45
add a comment |
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1
$begingroup$
By "the inverse," do you mean $x=cy+dpmod26$? If so, it may not be doable. It depends on $a$. Is there anything else you can tell us about $a$?
$endgroup$
– Cameron Buie
Mar 18 at 11:51
1
$begingroup$
The usual rules: subtract $b$ from both sides, and multiply both sides by the multiplicative inverse of $a$. Here, the multiplicative inverse of $a$ is the element $c$ such that $ac$ is the multiplicative identity. And the multiplicative identity is the element $1bmod26$.
$endgroup$
– Gerry Myerson
Mar 18 at 11:55
$begingroup$
As @Gerry points out, it isn't much different from what you're used to. The big difference is that $a$ may not have a multiplicative inverse modulo 26.
$endgroup$
– Cameron Buie
Mar 18 at 12:02