Does the distributed property of modular arithematic have a proof?

There is a property in modular arithematic and that is:

(a + b) mod N ≡ a mod N + b mod N

I wonder if there is a proof why this mod distributes inside brackets.

Any help would be really appreciated.

Pardon if this is a silly question.

Note by Shiran Abbasi
1 month ago

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I don't think it will because

(7+10)mod47mod4+10mod43+25(7 + 10) \mod 4 \equiv 7 \mod 4 + 10 \mod 4 \equiv 3 + 2 \equiv 5

but 55 is not 17mod4117 \mod 4 \equiv 1, so I don't think it applies. Please correct me if I'm wrong.

@Shiran Abbasi

Percy Jackson - 1 month ago

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You're wrong. 51(mod4)5\equiv 1 \pmod4.

Pi Han Goh - 1 month ago

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I was disproving the property that Shiran provided. By separating 17 into 7+10 and modding both of them.

Percy Jackson - 1 month ago

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@Percy Jackson The equation he provided is correct.

17 mod 4 == (7 mod 4) + (10 mod4)

Notice the equivalent symbol in the middle, it's not an equal sign.

Pi Han Goh - 1 month ago

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@Pi Han Goh Now I'm confused.

17mod4117 \mod 4 \equiv 1 because 17 = 4x4 + 1

So to find - 17mod4(7mod4)+(10mod4)17 \mod 4 \equiv (7 \mod 4) + (10 \mod 4)

we know (7mod4)3 and (10mod4)2(7 \mod 4) \equiv 3 \text{ and } (10 \mod 4) \equiv 2

Shouldn't (7mod4)+(10mod4)3+25(7 \mod 4) + (10 \mod 4) \equiv 3 + 2 \equiv 5???

If we calculate (7 mod 4) and (10 mod 4) separately, and then add them, we get 5 don't we???

Percy Jackson - 1 month ago

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@Percy Jackson the source is this: https://brilliant.org/wiki/modular-arithmetic/#modular-arithmetic-addition

Shiran Abbasi - 1 month ago

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@Percy Jackson ≡ sign here means congruence.

So, referring to your first comment:

17 ≡ 5 ≡ 1 mod 4

Shiran Abbasi - 1 month ago

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@Percy Jackson i also asked this question on math.stackexchange but they closed my question. Unfortunately i can't find an answer to my question.

Can anyone help?

Shiran Abbasi - 1 month ago

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@Shiran Abbasi You're confusing the two notations (modx)\pmod {x} and modx\bmod x.

If AmodB=CA \bmod B = C, then A(modB)CA \pmod B \equiv C as well. Similarly, A(modB)C+BnA\pmod B \equiv C + Bn , where nn is any integer.

For example, 17mod4=117\bmod4 = 1, and 17(mod4)117\pmod4 \equiv 1 are true.

Similarly,
17(mod4)517\pmod 4 \equiv 5 ,
17(mod4)917\pmod 4 \equiv 9 ,
17(mod4)317\pmod 4 \equiv -3 ,
17(mod4)717\pmod 4 \equiv -7 are all true as well.


By definition, if A(modB)CA \pmod B \equiv C, then ACA-C is divisible by BB.

On the other hand, AmodB=CA \bmod B = C implies that when AA is divided by BB, the remainder is CC.

Do you see the subtle difference here?

Pi Han Goh - 1 month ago

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@Pi Han Goh thanks for your answer. Now i got it.

Shiran Abbasi - 1 month ago

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@Pi Han Goh Thanks @Pi Han Goh, I understood now!

Percy Jackson - 1 month ago

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