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It's been a while since I've taken Discrete Math... How do you do i mod n Z?
 
It's been a while since I've taken Discrete Math... How do you do i mod n Z?
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I believe this is how you do it:
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n = iq + r, where q is the quotient and r is the remainder.
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i mod n Z = r
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Let a = i mod nZ, then a - i = nZ. This shows that Z divides a - i by n.
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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?
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-Neely
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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.
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--[[User:Dfreidin|Dfreidin]] 22:08, 14 September 2008 (UTC)

Latest revision as of 17:08, 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)

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