Congruence multiplication property
0
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I was looking over some properties of modular arithmetic and noticed the following one:
If $a equiv a'$ (mod $n$) and $b equiv b'$ (mod $n$), then $ab equiv a'b'$ (mod n)
Now this is ok and I understand the proofs that prove that property, but I have also noticed that reducing only one variable modulo $n$ the congruence will also be satisfied. Namely:
$ab equiv ab'$ (mod $n$)
Can anyone tell me why this also works? I am not very good with proofs, but I think it is just a corollary of the first property.
elementary-number-theory proof-writing modular-arithmetic
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
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add a comment |
0
$begingroup$
I was looking over some properties of modular arithmetic and noticed the following one:
If $a equiv a'$ (mod $n$) and $b equiv b'$ (mod $n$), then $ab equiv a'b'$ (mod n)
Now this is ok and I understand the proofs that prove that property, but I have also noticed that reducing only one variable modulo $n$ the congruence will also be satisfied. Namely:
$ab equiv ab'$ (mod $n$)
Can anyone tell me why this also works? I am not very good with proofs, but I think it is just a corollary of the first property.
elementary-number-theory proof-writing modular-arithmetic
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
$endgroup$
1
$begingroup$
Yes, it's just the special case of the first where $,a' = a,$ (so $,a'equiv a).,$ Both versions of the Congruence Product Rule are equivalent.
$endgroup$
– Bill Dubuque
Jan 11 at 16:33
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$a equiv a pmod n$, so of course it's fine.
$endgroup$
– Randall
Jan 11 at 16:41
$begingroup$
@BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence?
$endgroup$
– Michael Munta
Jan 15 at 8:51
$begingroup$
@MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11equiv 1pmod {10}$ becomes $[11]_{10} = [1]_{10},,$ i.e. $, 11+10Bbb Z = 1+10Bbb Z) $
$endgroup$
– Bill Dubuque
Jan 15 at 15:27
add a comment |
0
0
0
$begingroup$
I was looking over some properties of modular arithmetic and noticed the following one:
If $a equiv a'$ (mod $n$) and $b equiv b'$ (mod $n$), then $ab equiv a'b'$ (mod n)
Now this is ok and I understand the proofs that prove that property, but I have also noticed that reducing only one variable modulo $n$ the congruence will also be satisfied. Namely:
$ab equiv ab'$ (mod $n$)
Can anyone tell me why this also works? I am not very good with proofs, but I think it is just a corollary of the first property.
elementary-number-theory proof-writing modular-arithmetic
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
$endgroup$
I was looking over some properties of modular arithmetic and noticed the following one:
If $a equiv a'$ (mod $n$) and $b equiv b'$ (mod $n$), then $ab equiv a'b'$ (mod n)
Now this is ok and I understand the proofs that prove that property, but I have also noticed that reducing only one variable modulo $n$ the congruence will also be satisfied. Namely:
$ab equiv ab'$ (mod $n$)
Can anyone tell me why this also works? I am not very good with proofs, but I think it is just a corollary of the first property.
elementary-number-theory proof-writing modular-arithmetic
elementary-number-theory proof-writing modular-arithmetic
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
asked Jan 11 at 16:30
Michael MuntaMichael Munta
8011
8011
1
$begingroup$
Yes, it's just the special case of the first where $,a' = a,$ (so $,a'equiv a).,$ Both versions of the Congruence Product Rule are equivalent.
$endgroup$
– Bill Dubuque
Jan 11 at 16:33
$begingroup$
$a equiv a pmod n$, so of course it's fine.
$endgroup$
– Randall
Jan 11 at 16:41
$begingroup$
@BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence?
$endgroup$
– Michael Munta
Jan 15 at 8:51
$begingroup$
@MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11equiv 1pmod {10}$ becomes $[11]_{10} = [1]_{10},,$ i.e. $, 11+10Bbb Z = 1+10Bbb Z) $
$endgroup$
– Bill Dubuque
Jan 15 at 15:27
add a comment |
1
$begingroup$
Yes, it's just the special case of the first where $,a' = a,$ (so $,a'equiv a).,$ Both versions of the Congruence Product Rule are equivalent.
$endgroup$
– Bill Dubuque
Jan 11 at 16:33
$begingroup$
$a equiv a pmod n$, so of course it's fine.
$endgroup$
– Randall
Jan 11 at 16:41
$begingroup$
@BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence?
$endgroup$
– Michael Munta
Jan 15 at 8:51
$begingroup$
@MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11equiv 1pmod {10}$ becomes $[11]_{10} = [1]_{10},,$ i.e. $, 11+10Bbb Z = 1+10Bbb Z) $
$endgroup$
– Bill Dubuque
Jan 15 at 15:27
1
1
$begingroup$
Yes, it's just the special case of the first where $,a' = a,$ (so $,a'equiv a).,$ Both versions of the Congruence Product Rule are equivalent.
$endgroup$
– Bill Dubuque
Jan 11 at 16:33
$begingroup$
Yes, it's just the special case of the first where $,a' = a,$ (so $,a'equiv a).,$ Both versions of the Congruence Product Rule are equivalent.
$endgroup$
– Bill Dubuque
Jan 11 at 16:33
$begingroup$
$a equiv a pmod n$, so of course it's fine.
$endgroup$
– Randall
Jan 11 at 16:41
$begingroup$
$a equiv a pmod n$, so of course it's fine.
$endgroup$
– Randall
Jan 11 at 16:41
$begingroup$
@BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence?
$endgroup$
– Michael Munta
Jan 15 at 8:51
$begingroup$
@BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence?
$endgroup$
– Michael Munta
Jan 15 at 8:51
$begingroup$
@MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11equiv 1pmod {10}$ becomes $[11]_{10} = [1]_{10},,$ i.e. $, 11+10Bbb Z = 1+10Bbb Z) $
$endgroup$
– Bill Dubuque
Jan 15 at 15:27
$begingroup$
@MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11equiv 1pmod {10}$ becomes $[11]_{10} = [1]_{10},,$ i.e. $, 11+10Bbb Z = 1+10Bbb Z) $
$endgroup$
– Bill Dubuque
Jan 15 at 15:27
add a comment |
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1
$begingroup$
Yes, it's just the special case of the first where $,a' = a,$ (so $,a'equiv a).,$ Both versions of the Congruence Product Rule are equivalent.
$endgroup$
– Bill Dubuque
Jan 11 at 16:33
$begingroup$
$a equiv a pmod n$, so of course it's fine.
$endgroup$
– Randall
Jan 11 at 16:41
$begingroup$
@BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence?
$endgroup$
– Michael Munta
Jan 15 at 8:51
$begingroup$
@MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11equiv 1pmod {10}$ becomes $[11]_{10} = [1]_{10},,$ i.e. $, 11+10Bbb Z = 1+10Bbb Z) $
$endgroup$
– Bill Dubuque
Jan 15 at 15:27