# 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

$(7 + 10) \mod 4 \equiv 7 \mod 4 + 10 \mod 4 \equiv 3 + 2 \equiv 5$

but $5$ is not $17 \mod 4 \equiv 1$, so I don't think it applies. Please correct me if I'm wrong.

- 1 month ago

You're wrong. $5\equiv 1 \pmod4$.

- 1 month ago

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

- 1 month ago

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.

- 1 month ago

Now I'm confused.

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

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

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

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

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

- 1 month ago

- 1 month ago

≡ sign here means congruence.

So, referring to your first comment:

17 ≡ 5 ≡ 1 mod 4

- 1 month ago

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?

- 1 month ago

You're confusing the two notations $\pmod {x}$ and $\bmod x$.

If $A \bmod B = C$, then $A \pmod B \equiv C$ as well. Similarly, $A\pmod B \equiv C + Bn$, where $n$ is any integer.

For example, $17\bmod4 = 1$, and $17\pmod4 \equiv 1$ are true.

Similarly,
$17\pmod 4 \equiv 5$,
$17\pmod 4 \equiv 9$,
$17\pmod 4 \equiv -3$,
$17\pmod 4 \equiv -7$ are all true as well.

By definition, if $A \pmod B \equiv C$, then $A-C$ is divisible by $B$.

On the other hand, $A \bmod B = C$ implies that when $A$ is divided by $B$, the remainder is $C$.

Do you see the subtle difference here?

- 1 month ago

- 1 month ago

Thanks @Pi Han Goh, I understood now!

- 1 month ago