before i will explain of my question , first of all let us introduce classical definition of division algorithm ,namely for given integer $a$ and integer $d$, here exist unique integers $q$ and $r$ such that

$0 \leq r <d$

so that $a=d*q+r$

from where


and $r$ is remainder when we are dividing $a$ by $d$

now question about one theorem : namely let $a$ and $b$ be integers and let $m$ be positive integers, then

$${a}\equiv b \pmod {m}. $$

if they have same remainder when divided by $m$

this $${a}\equiv b \pmod {m}. $$ means that


or $a=k*m+b$

compare to this equation


$b$ is remainder when $a$ is divided by $m$, on the other hand


from here $b=-k*m+a$

from this equation we know that when b is divided by $m$, quotient part is $-k$ and remainder is $a$, but that means that $a=b$ , as i know it is not necessary that $a$ must be equal to $b$ , if they are equal of course

$${a}\equiv b \pmod {m}. $$

as $a-b=0$ and $0$ mod $m$ is equal to zero, so where i am making mistake in my conclusion ?

  • $\begingroup$ yes yes thanks, i will change $\endgroup$ – dato datuashvili Apr 12 '17 at 14:02
  • $\begingroup$ And I wouldn't write $q=\frac{a}{d}$. That asserts something different from the sentence "$q$ is the quotient when $a$ is divided by $d$, with remainder $r$." $\endgroup$ – Thomas Andrews Apr 12 '17 at 14:03
  • $\begingroup$ i could not express in latex a div d and that why $\endgroup$ – dato datuashvili Apr 12 '17 at 14:04
  • $\begingroup$ Better to use language when notation fails, rather than the wrong symbols. $\endgroup$ – Thomas Andrews Apr 12 '17 at 14:05
  • $\begingroup$ thanks @ThomasAndrews now what about theorem itself? $\endgroup$ – dato datuashvili Apr 12 '17 at 14:05

Your confusion is due to the statement

'$b$ is remainder when $a$ is divided by $m$, ....'

which is not necessarily true. It is only true if it happens to be the case that $b$ < $m$, which is not required in general. What is true is that $a$ and $b$ will have the same remainder $r$ under the division algorithm (given $m$).


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