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i mod n Z = r | i mod n Z = r | ||
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| + | ---- | ||
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| + | Let a = i mod nZ, then a - i = nZ. This shows that Z divides a - i by n. | ||
| + | We can also look at the formula as a = nZ + i which tells us that a is a product of Z and the remainder of it. | ||
| + | |||
| + | ---- | ||
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| + | I understand the whole i mod nZ, but I am wondering what is the difference when the operator is addition instead of multiplicaton. | ||
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| + | The Example in class (Z mod 6Z, +). We said a=1 but is not the identity. Is the only difference with the operator that 0 is the identity in addition, and 1 the identity in multiplication? | ||
| + | -Neely | ||
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| + | ---- | ||
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| + | Yes, just like in normal arithemetic, 1 is the multiplicative identity, and 0 is the additive identity. It works the same way in modular arithmetic. | ||
| + | |||
| + | --[[User:Dfreidin|Dfreidin]] 22:08, 14 September 2008 (UTC) | ||
Latest revision as of 18:08, 14 September 2008
It's been a while since I've taken Discrete Math... How do you do i mod n Z?
I believe this is how you do it:
n = iq + r, where q is the quotient and r is the remainder.
i mod n Z = r
Let a = i mod nZ, then a - i = nZ. This shows that Z divides a - i by n.
We can also look at the formula as a = nZ + i which tells us that a is a product of Z and the remainder of it.
I understand the whole i mod nZ, but I am wondering what is the difference when the operator is addition instead of multiplicaton.
The Example in class (Z mod 6Z, +). We said a=1 but is not the identity. Is the only difference with the operator that 0 is the identity in addition, and 1 the identity in multiplication? -Neely
Yes, just like in normal arithemetic, 1 is the multiplicative identity, and 0 is the additive identity. It works the same way in modular arithmetic.
--Dfreidin 22:08, 14 September 2008 (UTC)
