Is $\frac{\left(4k-4n\right)\bmod 4N}{4}=\left(k-n\right)\bmod N$?

Question

I'm trying to understand the modulo operation.

Let $$k,n \in \mathbb{Z}$$ At firt I thought:

$$\left(4k-4n\right)\bmod 4N=\left(k-n\right)\bmod N$$

But after a few simple examples I realized that this is not true. On the other hand, for each example I find that the following statement is true:

$$\frac{\left(4k-4n\right)\bmod 4N}{4}=\left(k-n\right)\bmod N$$

Is this statement really true? If so, why?

The $k,n$ aren't serving any purpose here...your question gets simpler if you take $m=k-n$ and just speak of, say, $m\pmod N$. — lulu, Dec 31 '22 at 11:48
To your question: If $m=qN+r$ then $4m=q(4N)+4r$ so you just need to check the, straight forward, inequalities you need. — lulu, Dec 31 '22 at 11:49
By the mod distributive law: $,(4a \bmod 4n)/4 = 4(a\bmod n)/4 = a\bmod n,,$ so what remains is straightforward. — Bill Dubuque, Dec 31 '22 at 14:47

score -1 · Answer 1 · edited Dec 31 '22 at 13:17

-1

Let $\frac{(4k−4n)\bmod 4N}{4}=a$. Then $(4k-4n)\bmod4N=4a$ and so $4a=4k-4n+4Nm$ for some integer $m$. So dividing both sides by $4$ we get $$a=\frac{4k-4n+4Nm}{4}=k-n+Nm$$ Taking both sides $\bmod N$ we get $a=(k-n)\bmod N$ and therefore $$\frac{(4k−4n)\bmod 4N}{4}=a=(k-n)\bmod N$$

edited Dec 31 '22 at 13:17

Gary

31,845

answered Dec 31 '22 at 11:55

SFA

117

Is $\frac{\left(4k-4n\right)\bmod 4N}{4}=\left(k-n\right)\bmod N$?

1 Answers1