Do the following system of equation have a solution in $mathbb{Z}_{p}$
$begingroup$
Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.
I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?
linear-algebra systems-of-equations
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
asked Dec 7 '18 at 10:40
user46697user46697
215211
$endgroup$
add a comment |
$begingroup$
Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.
I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?
linear-algebra systems-of-equations
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
asked Dec 7 '18 at 10:40
user46697user46697
215211
$endgroup$
add a comment |
$begingroup$
Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.
I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?
linear-algebra systems-of-equations
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
asked Dec 7 '18 at 10:40
user46697user46697
215211
$endgroup$
Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.
I solved these equations in the usual way and found that $x = 1$ and $y = -frac{1}{3}$. I know that $mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $mathbb{Z}_{3}$ or $mathbb{Z}_{7}$.?
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
asked Dec 7 '18 at 10:40
user46697user46697
215211
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
asked Dec 7 '18 at 10:40
user46697user46697
215211
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
edited Dec 7 '18 at 10:43
Sujit Bhattacharyya
1,409519
1,409519
asked Dec 7 '18 at 10:40
user46697user46697
215211
asked Dec 7 '18 at 10:40
user46697user46697
215211
asked Dec 7 '18 at 10:40
user46697user46697
215211
215211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both
but in $pmod3$ or $Z_3,$ there is no inverse of $-3$
In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$
Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both
but in $pmod3$ or $Z_3,$ there is no inverse of $-3$
In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$
Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
add a comment |
$begingroup$
As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both
but in $pmod3$ or $Z_3,$ there is no inverse of $-3$
In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$
Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
add a comment |
$begingroup$
As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both
but in $pmod3$ or $Z_3,$ there is no inverse of $-3$
In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$
Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
As $x=1,y=-dfrac13=(-3)^{-1}$ satisfy both
but in $pmod3$ or $Z_3,$ there is no inverse of $-3$
In $Z_7,$ $$(-3)^{-1}equiv2pmod7$$
Alternatively for obvious reason, $3(x+2y)equiv1pmod3$ is untenable right?
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
edited Dec 7 '18 at 11:03
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 7 '18 at 10:43
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
add a comment |
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
Thank you. Can you please explain how you got 2(mod 7).
$endgroup$
– user46697
Dec 7 '18 at 11:02
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
$begingroup$
@user46697, $$-3cdot2=-6equiv1pmod7$$ right? Typo rectified in the post
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 11:04
add a comment |
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