breaking down triple congurence relationshipsare ternary relationships associative?Solving systems of basic congruencesDoes this system of congruences have a solution even if they are not relatively prime at first?Given that $kequiv 2 mod 4$, determine the remainder when $5k + 13$ is divided by $4$first-order predicate calculusProve that $24^31$ is congruent to $23^32$ mod 19.What is the difference between equality and congruence outside of geometry?Prove $ ∀a, b, c ∈ Z,$ if $ab + ac ≡ 3$ (mod 6) then b $notequiv$ c (mod 6).Struggling to understand proofs involving modules.Proof of $a equiv b$ mod m => $a equiv b$ mod $m'$ with $d cdot m' = m$
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breaking down triple congurence relationships
are ternary relationships associative?Solving systems of basic congruencesDoes this system of congruences have a solution even if they are not relatively prime at first?Given that $kequiv 2 mod 4$, determine the remainder when $5k + 13$ is divided by $4$first-order predicate calculusProve that $24^31$ is congruent to $23^32$ mod 19.What is the difference between equality and congruence outside of geometry?Prove $ ∀a, b, c ∈ Z,$ if $ab + ac ≡ 3$ (mod 6) then b $notequiv$ c (mod 6).Struggling to understand proofs involving modules.Proof of $a equiv b$ mod m => $a equiv b$ mod $m'$ with $d cdot m' = m$
$begingroup$
How do you read the statement,i.e. what does this mean? example is a congruent to b? there's no mod for b, does this mean they are equal?:
$$a equiv b equiv H mod M$$
And, if we were to break this statement to two simpler statements of congruence, how do we do that?
discrete-mathematics
asked yesterday
mathmaniagemathmaniage
16311
$endgroup$
add a comment |
$begingroup$
How do you read the statement,i.e. what does this mean? example is a congruent to b? there's no mod for b, does this mean they are equal?:
$$a equiv b equiv H mod M$$
And, if we were to break this statement to two simpler statements of congruence, how do we do that?
discrete-mathematics
asked yesterday
mathmaniagemathmaniage
16311
$endgroup$
2
$begingroup$
Just imagine equalities $a=b=H$ in the quotient $Bbb Z/MBbb Z$. So we read this as usual: $a=H$ and $b=H$ and $a=b$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , but in congruence, there no $aequiv b$ without defining a mod world, and congruence and equals are different
$endgroup$
– mathmaniage
yesterday
$begingroup$
Not true! $aequiv b bmod m$ is by definition equivalent to the equation $a=b$ in the ring $Bbb Z/mBbb Z$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , Haven't studied rings, but what does $aequiv b$ mean then? but $a equiv b mod n$ means $n|a-b$
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , in my statement above, is the mod M relavant for all, can we break it down like $a mod m = b mod m = H mod m$ ?
$endgroup$
– mathmaniage
yesterday
add a comment |
$begingroup$
How do you read the statement,i.e. what does this mean? example is a congruent to b? there's no mod for b, does this mean they are equal?:
$$a equiv b equiv H mod M$$
And, if we were to break this statement to two simpler statements of congruence, how do we do that?
discrete-mathematics
asked yesterday
mathmaniagemathmaniage
16311
$endgroup$
How do you read the statement,i.e. what does this mean? example is a congruent to b? there's no mod for b, does this mean they are equal?:
$$a equiv b equiv H mod M$$
And, if we were to break this statement to two simpler statements of congruence, how do we do that?
discrete-mathematics
discrete-mathematics
asked yesterday
mathmaniagemathmaniage
16311
asked yesterday
mathmaniagemathmaniage
16311
asked yesterday
mathmaniagemathmaniage
16311
asked yesterday
mathmaniagemathmaniage
16311
asked yesterday
mathmaniagemathmaniage
16311
16311
2
$begingroup$
Just imagine equalities $a=b=H$ in the quotient $Bbb Z/MBbb Z$. So we read this as usual: $a=H$ and $b=H$ and $a=b$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , but in congruence, there no $aequiv b$ without defining a mod world, and congruence and equals are different
$endgroup$
– mathmaniage
yesterday
$begingroup$
Not true! $aequiv b bmod m$ is by definition equivalent to the equation $a=b$ in the ring $Bbb Z/mBbb Z$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , Haven't studied rings, but what does $aequiv b$ mean then? but $a equiv b mod n$ means $n|a-b$
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , in my statement above, is the mod M relavant for all, can we break it down like $a mod m = b mod m = H mod m$ ?
$endgroup$
– mathmaniage
yesterday
add a comment |
2
$begingroup$
Just imagine equalities $a=b=H$ in the quotient $Bbb Z/MBbb Z$. So we read this as usual: $a=H$ and $b=H$ and $a=b$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , but in congruence, there no $aequiv b$ without defining a mod world, and congruence and equals are different
$endgroup$
– mathmaniage
yesterday
$begingroup$
Not true! $aequiv b bmod m$ is by definition equivalent to the equation $a=b$ in the ring $Bbb Z/mBbb Z$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , Haven't studied rings, but what does $aequiv b$ mean then? but $a equiv b mod n$ means $n|a-b$
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , in my statement above, is the mod M relavant for all, can we break it down like $a mod m = b mod m = H mod m$ ?
$endgroup$
– mathmaniage
yesterday
2
2
$begingroup$
Just imagine equalities $a=b=H$ in the quotient $Bbb Z/MBbb Z$. So we read this as usual: $a=H$ and $b=H$ and $a=b$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Just imagine equalities $a=b=H$ in the quotient $Bbb Z/MBbb Z$. So we read this as usual: $a=H$ and $b=H$ and $a=b$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , but in congruence, there no $aequiv b$ without defining a mod world, and congruence and equals are different
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , but in congruence, there no $aequiv b$ without defining a mod world, and congruence and equals are different
$endgroup$
– mathmaniage
yesterday
$begingroup$
Not true! $aequiv b bmod m$ is by definition equivalent to the equation $a=b$ in the ring $Bbb Z/mBbb Z$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
Not true! $aequiv b bmod m$ is by definition equivalent to the equation $a=b$ in the ring $Bbb Z/mBbb Z$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , Haven't studied rings, but what does $aequiv b$ mean then? but $a equiv b mod n$ means $n|a-b$
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , Haven't studied rings, but what does $aequiv b$ mean then? but $a equiv b mod n$ means $n|a-b$
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , in my statement above, is the mod M relavant for all, can we break it down like $a mod m = b mod m = H mod m$ ?
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , in my statement above, is the mod M relavant for all, can we break it down like $a mod m = b mod m = H mod m$ ?
$endgroup$
– mathmaniage
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$a equiv b equiv H mod M$ can basically be written as:
$$a equiv b mod M $$
$$a equiv H mod M $$
$$b equiv H mod M $$
answered yesterday
mathmaniagemathmaniage
16311
$endgroup$
add a comment |
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$begingroup$
$a equiv b equiv H mod M$ can basically be written as:
$$a equiv b mod M $$
$$a equiv H mod M $$
$$b equiv H mod M $$
answered yesterday
mathmaniagemathmaniage
16311
$endgroup$
add a comment |
$begingroup$
$a equiv b equiv H mod M$ can basically be written as:
$$a equiv b mod M $$
$$a equiv H mod M $$
$$b equiv H mod M $$
answered yesterday
mathmaniagemathmaniage
16311
$endgroup$
add a comment |
$begingroup$
$a equiv b equiv H mod M$ can basically be written as:
$$a equiv b mod M $$
$$a equiv H mod M $$
$$b equiv H mod M $$
answered yesterday
mathmaniagemathmaniage
16311
$endgroup$
$a equiv b equiv H mod M$ can basically be written as:
$$a equiv b mod M $$
$$a equiv H mod M $$
$$b equiv H mod M $$
answered yesterday
mathmaniagemathmaniage
16311
answered yesterday
mathmaniagemathmaniage
16311
answered yesterday
mathmaniagemathmaniage
16311
answered yesterday
mathmaniagemathmaniage
16311
16311
add a comment |
add a comment |
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$begingroup$
Just imagine equalities $a=b=H$ in the quotient $Bbb Z/MBbb Z$. So we read this as usual: $a=H$ and $b=H$ and $a=b$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , but in congruence, there no $aequiv b$ without defining a mod world, and congruence and equals are different
$endgroup$
– mathmaniage
yesterday
$begingroup$
Not true! $aequiv b bmod m$ is by definition equivalent to the equation $a=b$ in the ring $Bbb Z/mBbb Z$.
$endgroup$
– Dietrich Burde
yesterday
$begingroup$
@DietrichBurde , Haven't studied rings, but what does $aequiv b$ mean then? but $a equiv b mod n$ means $n|a-b$
$endgroup$
– mathmaniage
yesterday
$begingroup$
@DietrichBurde , in my statement above, is the mod M relavant for all, can we break it down like $a mod m = b mod m = H mod m$ ?
$endgroup$
– mathmaniage
yesterday