Mixing
Meaning of dividing every side of congruence
0
$begingroup$
When I am faced with the following:
$15 equiv x$ (mod 10)
Ignore the simplicity of this. What I am interested is is the following in any way connected to the solution for $x$:
$frac{15}{5} equiv frac{x}{5}$ (mod 2)
The congruence was divided by $5$. Does solving this for $x$ in any way connect to the first congruence?
elementary-number-theory modular-arithmetic
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
$endgroup$
add a comment |
0
$begingroup$
When I am faced with the following:
$15 equiv x$ (mod 10)
Ignore the simplicity of this. What I am interested is is the following in any way connected to the solution for $x$:
$frac{15}{5} equiv frac{x}{5}$ (mod 2)
The congruence was divided by $5$. Does solving this for $x$ in any way connect to the first congruence?
elementary-number-theory modular-arithmetic
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
$endgroup$
$begingroup$
May be useful - $$axequiv aypmod miff xequiv yleft(bmod frac m{text{gcd}(a,m)}right) $$
$endgroup$
– Thomas Shelby
Jan 27 at 16:32
$begingroup$
So $x$ would be $1$ but it is not $1$ in the first congruence.
$endgroup$
– Michael Munta
Jan 27 at 16:44
$begingroup$
I know about this, but that $x$ is then a soultion in different modulo.
$endgroup$
– Michael Munta
Jan 27 at 17:10
add a comment |
0
0
0
$begingroup$
When I am faced with the following:
$15 equiv x$ (mod 10)
Ignore the simplicity of this. What I am interested is is the following in any way connected to the solution for $x$:
$frac{15}{5} equiv frac{x}{5}$ (mod 2)
The congruence was divided by $5$. Does solving this for $x$ in any way connect to the first congruence?
elementary-number-theory modular-arithmetic
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
$endgroup$
When I am faced with the following:
$15 equiv x$ (mod 10)
Ignore the simplicity of this. What I am interested is is the following in any way connected to the solution for $x$:
$frac{15}{5} equiv frac{x}{5}$ (mod 2)
The congruence was divided by $5$. Does solving this for $x$ in any way connect to the first congruence?
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
asked Jan 27 at 15:59
Michael MuntaMichael Munta
99111
99111
$begingroup$
May be useful - $$axequiv aypmod miff xequiv yleft(bmod frac m{text{gcd}(a,m)}right) $$
$endgroup$
– Thomas Shelby
Jan 27 at 16:32
$begingroup$
So $x$ would be $1$ but it is not $1$ in the first congruence.
$endgroup$
– Michael Munta
Jan 27 at 16:44
$begingroup$
I know about this, but that $x$ is then a soultion in different modulo.
$endgroup$
– Michael Munta
Jan 27 at 17:10
add a comment |
$begingroup$
May be useful - $$axequiv aypmod miff xequiv yleft(bmod frac m{text{gcd}(a,m)}right) $$
$endgroup$
– Thomas Shelby
Jan 27 at 16:32
$begingroup$
So $x$ would be $1$ but it is not $1$ in the first congruence.
$endgroup$
– Michael Munta
Jan 27 at 16:44
$begingroup$
I know about this, but that $x$ is then a soultion in different modulo.
$endgroup$
– Michael Munta
Jan 27 at 17:10
$begingroup$
May be useful - $$axequiv aypmod miff xequiv yleft(bmod frac m{text{gcd}(a,m)}right) $$
$endgroup$
– Thomas Shelby
Jan 27 at 16:32
$begingroup$
May be useful - $$axequiv aypmod miff xequiv yleft(bmod frac m{text{gcd}(a,m)}right) $$
$endgroup$
– Thomas Shelby
Jan 27 at 16:32
$begingroup$
So $x$ would be $1$ but it is not $1$ in the first congruence.
$endgroup$
– Michael Munta
Jan 27 at 16:44
$begingroup$
So $x$ would be $1$ but it is not $1$ in the first congruence.
$endgroup$
– Michael Munta
Jan 27 at 16:44
$begingroup$
I know about this, but that $x$ is then a soultion in different modulo.
$endgroup$
– Michael Munta
Jan 27 at 17:10
$begingroup$
I know about this, but that $x$ is then a soultion in different modulo.
$endgroup$
– Michael Munta
Jan 27 at 17:10
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
$x equiv ca pmod{!cn},Rightarrow, cmid x, $ so $, c(x/c)equiv ca pmod{!cn} iff x/c equiv a pmod{! n}$
because $,cnmid cX-caiff nmid X-a,,$ since $ dfrac{cX-ca}{cn} = dfrac{X-a}n,$
Remark $ $ For the general case (and its fractional viewpoint) see this answer.
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
|
show 4 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
$x equiv ca pmod{!cn},Rightarrow, cmid x, $ so $, c(x/c)equiv ca pmod{!cn} iff x/c equiv a pmod{! n}$
because $,cnmid cX-caiff nmid X-a,,$ since $ dfrac{cX-ca}{cn} = dfrac{X-a}n,$
Remark $ $ For the general case (and its fractional viewpoint) see this answer.
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
|
show 4 more comments
1
$begingroup$
$x equiv ca pmod{!cn},Rightarrow, cmid x, $ so $, c(x/c)equiv ca pmod{!cn} iff x/c equiv a pmod{! n}$
because $,cnmid cX-caiff nmid X-a,,$ since $ dfrac{cX-ca}{cn} = dfrac{X-a}n,$
Remark $ $ For the general case (and its fractional viewpoint) see this answer.
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
|
show 4 more comments
1
1
1
$begingroup$
$x equiv ca pmod{!cn},Rightarrow, cmid x, $ so $, c(x/c)equiv ca pmod{!cn} iff x/c equiv a pmod{! n}$
because $,cnmid cX-caiff nmid X-a,,$ since $ dfrac{cX-ca}{cn} = dfrac{X-a}n,$
Remark $ $ For the general case (and its fractional viewpoint) see this answer.
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
$endgroup$
$x equiv ca pmod{!cn},Rightarrow, cmid x, $ so $, c(x/c)equiv ca pmod{!cn} iff x/c equiv a pmod{! n}$
because $,cnmid cX-caiff nmid X-a,,$ since $ dfrac{cX-ca}{cn} = dfrac{X-a}n,$
Remark $ $ For the general case (and its fractional viewpoint) see this answer.
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
edited Jan 27 at 17:57
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
answered Jan 27 at 17:22
Bill DubuqueBill Dubuque
213k29195654
213k29195654
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
|
show 4 more comments
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
Here $,cneq 0 $ (usually by convention for moduli)
$endgroup$
– Bill Dubuque
Jan 27 at 17:23
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
So having this $frac{15}{5} equiv frac{x}{5} equiv$ -> $3 equiv frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right?
$endgroup$
– Michael Munta
Jan 28 at 10:39
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true.
$endgroup$
– Michael Munta
Jan 28 at 11:05
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
@Michael $xequiv 15equiv 5pmod{!10}iff x = 5+10n = 5(1+2n)iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!bmod 10 $
$endgroup$
– Bill Dubuque
Jan 28 at 15:36
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
$begingroup$
But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence?
$endgroup$
– Michael Munta
Jan 28 at 15:46
|
show 4 more comments
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$begingroup$
May be useful - $$axequiv aypmod miff xequiv yleft(bmod frac m{text{gcd}(a,m)}right) $$
$endgroup$
– Thomas Shelby
Jan 27 at 16:32
$begingroup$
So $x$ would be $1$ but it is not $1$ in the first congruence.
$endgroup$
– Michael Munta
Jan 27 at 16:44
$begingroup$
I know about this, but that $x$ is then a soultion in different modulo.
$endgroup$
– Michael Munta
Jan 27 at 17:10