Finding $a$ in modular arithmetic while $a<0$

Question

I'm having a problem understanding this equation in modular arithmetic I have tried searching the internet but I haven't found a solution, I hope you can help.

$a = k(26) + b\; \text{ for }a > 0\:$ (26 is just what he uses in the book as he is explaining the Caesar cipher)

The author then goes on to say that even if $a$ were negative, we could easily find a positive number $b$ in the set $\{1,2,\ldots,26\}$ such that $a$ is congruent to $b$ by dividing the positive number $-a$ by $26$, obtaining:

$-a = q(26) + r = (q+1)26 - (26-r),\quad q\ge 0, \enspace 0\le r<26$.

My question is: How did he get to that equation, I seem to have tried anything, it might be that I am really tired, but I have to know the answer before I sleep.

Thank you

Answer 1

We usually write

$$a = k(26) + b\; \text{ for }a > 0\:$$ as $$a = 26 \cdot k + b\; \text{ for }a > 0\:$$ where $k$ is the quotient when divided by 26 and $b$ is the remainder.

Now look at

$$-a = q(26) + r = (q+1)26 - (26-r),\quad q\ge 0, \enspace 0\le r<26.$$

it is used to find the representative of a negative $-a$ number

$$-a = 26\cdot q + r = 26 (q+1) - (26-r),\quad q\ge 0, \enspace 0\le r<26.$$

The trick is, add and subtract 26. Take the $\mod 26$ in the last equation. And, the shortest way is $26-a$ if $a < 26$.

Answer 2

All he is saying is that if $a < 0$ is it possible to say

$a = 26k + b$ where $b = \{1,....., 26\}$.

He does:

$a < 0$ so $-a > 0$. Then if you divide $-a$ by $26$ you we get a quotient $q$ and a remainder $r$ so that

$\frac {-a}{26} = q + \frac r{26}$ and

$-a = 26q + r$ and $ 0 \le r < 26$.

Then $-a = 26(q+1) -26 + r = 26(q+1) - (26-r)$. So $0 \le r < 26$ we have $0 < 26-r \le 26$ so if we let $b = 26-r \in \{1,..., 26\}$ and let $k = -q-1$ we get

$a = 26(-q-1) +(26-r) = 26k + b$.

It works.

If $a > 0$ then there is a $b \in \{1....26\}$ where $a \equiv b \pmod {26}$.

And if $a < 0$ then there is a $b \in \{1... 26\}$ where $a \equiv b\pmod {26}$.

And if $a = 0$ then $a \equiv 26\pmod{26}$.

....

It's not deep.

I'd have done it simply by saying. "Just add or subtract $26$ until you get some between $1$ and $26$ inclusive".

====== old answer===

If $a > 0$ there is a $k$ and a $b > 0$ so that $a = 26k + b$. [The book seems to have established that.]

But what if $a < 0$. Can we find a $k$ and $b > 0$ so that $a = 26k +b$.

Yes.

If $a < 0$ then $-a > 0$ and we can find $q$ and $r: 0 \le r < 26$ so that

$-a = 26q + r$

$= 26(q+1) - 26 + r$

$= 26(q+1) - (26-r)$ and note that $r < 26$ so $ 26 -r > 0$.

But then

$a = 26(-q-1) + (26-r)$.

Let $k = -q-1$ and $b = 26-r$ and this works fine

$a = 26k + b$ where $b > 0$ even though $a < 0$.

[ I suppose for $a = 0$ the book uses $0 = 26(-1) + 26$? ]

[It's not entirely clear what the book is trying to show.]