Modular Arith: How is Remainder 2?

[chwala]

Homework Statement

how is -10≡2 mod 3?

Relevant Equations

modular arithmetic

ok this is a bit confusing to me, long since i did this things...-10/3=quotient -3+ remainder -1. How is the remainder 2?

 

[member 587159]

By definition, this means that $3$ divides $-10-2 = -12$ which is trivially true.

What definition are you using?

 

[chwala]

By definition, this means that $3$ divides $-10-2 = -12$ which is trivially true.

What definition are you using?

i am not using any , is it that a=b mod c and a≡ b mod c imply different things? my interest is on the former. The latter i presume applies to congruence.

 

[PeroK]

Problem Statement: how is -10≡2 mod 3?
Relevant Equations: modular arithmetic

ok this is a bit confusing to me, long since i did this things...-10/3=quotient -3+ remainder -1. How is the remainder 2?

$-1$ means the inverse of $1$. In mod 3 arithmetic $1 + 2 = 0$, hence $-1 = 2$.

More simply, $-1 \equiv 2 (mod \ 3)$

 

[Orodruin]

Problem Statement: how is -10≡2 mod 3?
Relevant Equations: modular arithmetic

ok this is a bit confusing to me, long since i did this things...-10/3=quotient -3+ remainder -1. How is the remainder 2?

If you want to talk about remainders in connection to modular arithmetic you should define the remainder of n/m to be the smallest non-negative number k such that the quotient (n-k)/m is an integer. With this definition you have the quotient -4 and the remainder 2.

Also, as has been pointed out already, -1 = 2 mod 3 so saying -10 = -1 mod 3 is also perfectly fine, -1 and 2 are in the same equivalence class.

 

[fresh_42]

Problem Statement: how is -10≡2 mod 3?
Relevant Equations: modular arithmetic

ok this is a bit confusing to me, long since i did this things...-10/3=quotient -3+ remainder -1. How is the remainder 2?

The reason is that multiples of three doesn't count anymore. The procedure is, that we impose a relation $\sim$ on the set of integers which is defined by: $a\sim b$ iff $3$ divides $a-b$ or iff $a$ and $b$ have the same sort of remainder: $3\mathbb{Z}$, $3\mathbb{Z}+1$, or $3\mathbb{Z}+2$. We consider these sets as one element: $[0]=3\mathbb{Z}$, $[1]=3\mathbb{Z}+1$, $[2]=3\mathbb{Z}+2$. The brackets $[]$ which indicates that entire sets are behind the numbers $0,1,2$ are usually not written, as they don't carry any additional information compared to what has already been said by using $\operatorname{mod}$ or $\equiv$. The same is true for $\equiv$: it is often just written as equality $=$, an equality of sets, the so called equivalence classes with respect to $\sim$.

 

[SammyS]

Problem Statement: how is -10≡2 mod 3?
Relevant Equations: modular arithmetic

ok this is a bit confusing to me, long since i did this things...-10/3=quotient -3+ remainder -1. How is the remainder 2?

In addition to what has already been stated (or perhaps I missed that this, too, has already been stated):

The statmment:

$a \equiv b \mod n$​

means that there is some integer, $m$, such that:

$a - b = mn$ .​

Regarding your question:
$-10 \equiv 2 \mod 3$
because
$(-10) - (2) = (-4)( 3)$

 

[chwala]

If you want to talk about remainders in connection to modular arithmetic you should define the remainder of n/m to be the smallest non-negative number k such that the quotient (n-k)/m is an integer. With this definition you have the quotient -4 and the remainder 2.

Also, as has been pointed out already, -1 = 2 mod 3 so saying -10 = -1 mod 3 is also perfectly fine, -1 and 2 are in the same equivalence class.

ok meaning that say -20/3 will be -20≡1mod 3? where quotient is -7 and k=1
-50/3 will be -50≡1 mod 3? where quotient is -17 and k=1

 

[chwala]

If you want to talk about remainders in connection to modular arithmetic you should define the remainder of n/m to be the smallest non-negative number k such that the quotient (n-k)/m is an integer. With this definition you have the quotient -4 and the remainder 2.

Also, as has been pointed out already, -1 = 2 mod 3 so saying -10 = -1 mod 3 is also perfectly fine, -1 and 2 are in the same equivalence class.

so you're implying that k cannot be a negative integer, rather it has to be positive?

 

[chwala]

$-1$ means the inverse of $1$. In mod 3 arithmetic $1 + 2 = 0$, hence $-1 = 2$.

More simply, $-1 \equiv 2 (mod \ 3)$

i don't understand what you're saying here on inverses...this is a relative new area to me as my area of maths is applied maths...

 

[PeroK]

i don't understand what you're saying here on inverses...this is a relative new area to me as my area of maths is applied maths...

Do you understand why:

$-1 \equiv 2 (mod \ 3)$?

 

[Orodruin]

so you're implying that k cannot be a negative integer, rather it has to be positive?

No, both -10 = -1 mod 3 and -10 = 2 mod 3 are valid statements as -1 = 2 mod 3.

 

[chwala]

i figured out a way of doing this things...lets say you have a≡ k mod b then it follows that,
{a-k}/{b}= integer value...positive critism is welcome.

 

[Orodruin]

i figured out a way of doing this things...lets say you have a≡ k mod b then it follows that,
{a-k}/{b}= integer value...positive critism is welcome.

This is just the definition of a ~ k (mod b). See, for example, post #7.

 

[chwala]

This is just the definition of a ~ k (mod b). See, for example, post #7.

thanks seen, i appreciate.

 

[SammyS]

i figured out a way of doing this things...lets say you have a≡ k mod b then it follows that,
{a-k}/{b}= integer value...positive criticism is welcome.

Yes, this is correct.

In post #2, @Math_QED gives this for the specific case in which $a=-1,\ k=2, \text{ and } b=3$ .

Also, as @Orodruin mentioned, I gave a definition equivalent to yours in Post #7.

The more important issue here is that you now have a very solid way of thinking of modular equivalence (congruence), at least as it applies to integers. Congratulations on this!

 

[MidgetDwarf]

The reason why we define the remainder to positive follows from a few key ideas. Do you know what an equivalence class is? Do you know how elements of congruence mod n look like? and how many are there?

The reason that the reduced residue (the remainder) must be positive follows from the statement of the Division Theorem.

 

[fresh_42]

The reason that the reduced residue (the remainder) must be positive follows from the statement of the Division Theorem.

This is wrong. The representatives (remainders) do not have to be positive. This is out of convenience and not a mathematical necessity. Thus it doesn't follow from anything. One can perfectly calculate with $\mathbb{Z}/2\mathbb{Z} =\{\,[17],[-12]\,\}$ but it makes more fun to write it as $\{\,[0],[1]\,\}$.

 

[MidgetDwarf]

This is wrong. The representatives (remainders) do not have to be positive. This is out of convenience and not a mathematical necessity. Thus it doesn't follow from anything. One can perfectly calculate with $\mathbb{Z}/2\mathbb{Z} =\{\,[17],[-12]\,\}$ but it makes more fun to write it as $\{\,[0],[1]\,\}$.

true. but usually we define remainders using the statement of the Division Theorem and showing that the relation congruence modulo n on Z is an equivalence relation. Using the definition of equivalence classes...

Since the elements of Zn are equivalence classes, then it does not matter which representative we use to denote the distinct equivalence classes (this is what you are saying).

Also, it helps establish the fact that a is congruent to r modulo n, where r is the reduced reduced residue of a.writing the remainders as positive helps us build more results of number theory in a "nice way."

 

[fresh_42]

writing the remainders as positive helps us build more results of number theory in a "nice way."

Right. I just wanted to prevent the impression that it has to be those representatives. This contradicted the entire concept of equivalence classes. Any are as good as specific ones, just not as easy to calculate with. And if we only consider the group property and forget about the ring, then $\mathbb{Z}_2=\{\,0,1\,\}$ can even be written multiplicatively as $\mathbb{Z}_2=\{\,-1,1\,\}$.

 

[chwala]

i took my time to look at the modular arithmetic and i could understand it straight away...i found it to be very easy stuff for me...i believe i can attempt any problem in math if i give it the necessary time and effort.

 

[chwala]

Do you understand why:

$-1 \equiv 2 (mod \ 3)$?

yap

 

[chwala]

...therefore in summary, from my study, if we have for instance, a mod b = c then it follows that increasing $a$ by a multiple of $b$ would result to the same remainder.
therefore,
$a mod b= (a+k.b)mod b$ for any integer $k$
for example,
3mod 10=3
13mod 10=3
23mod 10=3
-7mod10=3
bingo

 

[fresh_42]

i took my time to look at the modular arithmetic and i could understand it straight away...i found it to be very easy stuff for me...i believe i can attempt any problem in math if i give it the necessary time and effort.

Be careful! You will never again agree with people who say $1+1=2$ and expect this as a given truth!