3

Question

Prove that if $a\equiv b \pmod m$ , then $a \bmod m = b \bmod m$

Approach

Given,

$a\equiv b \pmod m$

$\implies m\mid (a-b)$

$\implies (a-b)=m\cdot k$

$\implies a=b+m\cdot k$

Now,

$a \bmod m$ can be written as

$\implies b+m\cdot k \pmod m$

No idea how to move forward to get

$$b \pmod m \impliedby b +m\cdot k \pmod m$$

Please help me out!!!

Thanks!

Community
  • 1
laura
  • 2,530

5 Answers5

4

Hint $\, $ Show $\ c\equiv d\pmod{\!m}\,$ and $\,|c-d|<m\,\Rightarrow\, c= d$

Apply that to $\,\ \underbrace{a\bmod m}_{\Large c}\!\equiv a\equiv b\equiv \underbrace{b\bmod m}_{\Large d}\,\ \pmod{\!m}$

Remark $ $ Any sequence of $m$ consecutive integers forms a complete system $S$ of representatives (normal forms) of all equivalence classes mod $m,\,$ so the above implies that congruence reduces to equality on these normal forms, i.e. $\ a\equiv b\iff \bar a = \bar b,\,$ where $\bar n =$ unique normal-form rep in $S$ congruent to $n\,$ (which is analogous to: $ $ fractions are equivalent iff their least form rep's are equal, i.e. least terms fractions are normal forms for equivalence of fractions).

An alternative convenient choice of normal forms for modular arithmetic arises by choosing the least magnitude reps (signed or balanced residue system), e.g. $\,\{-2,\color{#c00}{-1},0,1,2\}\pmod{\! 5}.\,$ In these signed systems the normal form of $\,m\!-\!1\,$ is $\,-1,\,$ e.g. above $\,4\equiv\color{#c00}{-1}\pmod{\!5},\,$ simplifying common calculations such as $\,(m\!-\!1)^k\equiv (-1)^k,\,$ e.g. when casting out elevens.

Bill Dubuque
  • 272,048
  • I understand the circularity in my solution. Yours is the correct solution of this problem. Thanks for pointing out my mistake. – Dhruv Kohli Jun 14 '17 at 04:47
  • @expiTTp1z0 The proof in your answer may or may not be circular - depending on how one proves the addition Lemma used (which may depend on the order that topics are presented in a number theory course). I thought it worthwhile to point out that possible issue since it might not be obvious to many readers. – Bill Dubuque Jun 14 '17 at 14:55
  • True but can you please provide me link to the proof of $(a+b)\bmod m = (a\bmod m + b\bmod m)\bmod m$ which does not use congruence sum rule. I mean a proof that breaks the circular dependence. I searched for one but couldn't find any. Thanks. – Dhruv Kohli Jun 14 '17 at 15:00
  • @expiTTp1z0 There are various ways to prove $\ (b+km)\bmod m = b\bmod m,$ without using the general addition rule, e.g. by the division algorithm: $, b = qm+r,\Rightarrow, b+km = (q+k)m+r\ $ – Bill Dubuque Jun 14 '17 at 15:15
  • Wow. My trail of thoughts didn't touch the division algorithm. Thanks I'll update my answer with this proof and will state the issue too. – Dhruv Kohli Jun 14 '17 at 15:40
  • I edited my answer. Please check. Thanks. – Dhruv Kohli Jun 14 '17 at 16:08
  • @expiTTp1z0 It's not clear you understood the proof in my prior comment. It shows that both $b$ and $,b+km,$ have the same remainder $r$ when divided by $m$, i.e. $,b+km\bmod m = b\bmod m.,$ The point was to do that without using any properties of mod (other than its definition), to avoid possible circularity. But you're still using mod in your answer. – Bill Dubuque Jun 14 '17 at 16:29
  • Please check again. Thanks. – Dhruv Kohli Jun 14 '17 at 16:45
  • @expiTTp1z0 That's better, but I'd remove the $,(m(k+q)+r)\bmod m = r,$ to avoid any confusion that this may come from applying the mod addition rule when, instead, we are using the definition/uniqueness of remainders/mod. – Bill Dubuque Jun 14 '17 at 16:51
  • Thanks for all those informative suggestions. I'll remember them. – Dhruv Kohli Jun 14 '17 at 16:55
2

$$a\equiv b \pmod m$$ $$\implies (a-b) = m k $$ $$\implies a = m k + b $$ $$\implies a \bmod m = (mk\bmod m + b\bmod m)\bmod m = b\bmod m \bmod m = b \bmod m$$ $$\implies a \bmod m = b \bmod m$$

Here is an important point raised by @(Bill Dubuque) in this proof. In line $4$, I used $(mk+b)\bmod m = (mk\bmod m + b\bmod m)\bmod m$ which must be proved for the completion of the proof of the original result. There are multiple ways to prove this equality, however if one uses the approach used in this answer, then this leads to a circular dependency i.e. usage of this equality to prove the original result and the usage of the original result in proving this equality. So, to complete the proof of original result one must prove this equality with some other approach. Based on the suggestion by Bill, here is a revised proof,

Let $b = mq + r \implies b \bmod m = r$, then,

$$a\equiv b \pmod m$$ $$\implies (a-b) = m k $$ $$\implies a = m k + b = mk + mq + r = m(k+q) + r $$ $$\implies a \bmod m = r = b \bmod m$$ $$\implies a \bmod m = b \bmod m$$

Note: Here is an issues found by @(Bill Dubuque) in my first proof.

$$a\equiv b \pmod m$$ $$\implies (a-b) = m k $$ $$\implies (a-b)\bmod m = m k \bmod m $$ $$\implies a \bmod m - b \bmod m = 0$$ $$\implies a \bmod m = b \bmod m$$

But the $4$th equation above cannot follow from the $3$rd equation in general (counterexample $a=m$ and $b=1$).

Dhruv Kohli
  • 5,216
  • 1
    Incorrect: generally $\ (a-b)\bmod m ,\neq, a\bmod m - b\bmod m,\ $ e.g. $,a=m, ,b = 1$ – Bill Dubuque Jun 12 '17 at 21:47
  • When $a=m$ then $b$ can never be equal to $1$. – Dhruv Kohli Jun 13 '17 at 02:01
  • It seems you misunderstand my point. You appeared to use the equality in my prior comment to derive the 4th equation in your proof. If you are not using that but something else then you need elaborate in order to justify how your derived the 4th equation from the 3rd. – Bill Dubuque Jun 13 '17 at 02:06
  • I now understand what you meant. Thanks. I made a correction. Please check. – Dhruv Kohli Jun 13 '17 at 10:23
  • Beware that the revised approach could be circular, since the identity that you use in the fourth line is often proved using the sought result, e.g. see here.. So you'd need to give a different proof that $,(j+k)\bmod m = (j\bmod m + k\bmod m)\bmod m.,$ But there are simpler ways. – Bill Dubuque Jun 13 '17 at 14:38
  • I am sorry, but I don't understand the fallacy in my proof based on your comment. Can you please elaborate more. Thanks. – Dhruv Kohli Jun 13 '17 at 19:29
1

Write $a=qm+r$ and $b=pm+s$, where $0\leq s,r<m$. Then $a\;\underline{mod}\;m = r$ and $b\;\underline{mod}\;m = s$. On the other hand, $a\equiv b\; mod \;m$ iff $m|a-b$ iff (above) $m|r-s$ iff ($r,s$ remainders) $r=s$.

Bill Dubuque
  • 272,048
Wuestenfux
  • 20,964
  • 1
    In mathematics, a residue class is not equal to an integer. So both the statement and the "proof" do not belong here. –  Jun 12 '17 at 08:29
1

The phrase $a \mod m$ as some specific numerical value is not a universally understood or accepted concept and so the phrase $a \mod m = b\mod m$ may not have any sensible meaning depending on which mathematician you ask.

It can be taken to mean: For any $m \in \mathbb N$ (not including 0) and any integer $a$ there will exist unique integers $q, r$ so that $a = qm + r$ and $0 \le r < m$. We define $r$ as the remainder where $a$ is divided by $m$.

I have seen this called $r = a \mod m$ or more commonly $r = a \% m$. In my opinion, I can accept the latter if it is properly defined but I detest the former as it is an abuse of notation and concept.

The notation $a \equiv b \mod m$ means $m$ divides $a- b$ or equivalently $a$ and $b$ have the same remainder when divided by $m$. But the important thing is that this is not a statement about specific integers-- it is a statement about equivalence classes of sets of integers. $a \equiv b \mod m$ mean that both $a$ and $b$ are elements of the the same set $M = \{ x\in \mathbb Z| m|(x-a)\}$.

Anyway. Assuming that you mean $a \mod m$ to mean the unique integer remainder as I desribed in the second paragraph then the proof is trivial:

Suppose $a \equiv b \mod m$ Let $r_1 = a \mod m$ so $a = jm + r_1$ or $r_1 = a = jm$ for some integer $j$ and let $r_2 = b \mod m$ so $b = lm + r_2$ o $r_2 = b- lm$ and $0 \le r_1 < m; 0 \le r_2 < m$. So $r_1 - r_2= (a-b) + (j- l)m$ So as $m|(a-b)$ and $m|(j-l)n$ then $m|r_1 - r_2$.

Let $r_1 - r_2 = dm$ for some integer $d$.

then $m > r_1 > r_1 - r_2 = dm$ and $dm = r_1 - r_2 > -r_2 > m$ so $m > dm > -m$ and $1 > d > -1$ but $d$ is an integer.

So $d= 0$ and $r_1 - r_2 = 0*m = 0$ and $a \mod m = r_1 = r_2 = b \mod m$.

fleablood
  • 124,253
  • I'm afraid you are quite confused about what notation is correct and/or standard. Note that I (partly) corrected the notation in the question. – Bill Dubuque Jun 13 '17 at 00:23
-2

I think it's obvious:

You have that $a \equiv b+mk \pmod m$.

But, $mk \equiv 0 \pmod m$ , so $a \equiv b+mk \equiv b\pmod m$.

Done.

knm
  • 97
  • No, what you need to prove is that $a$ and $b$ have the same remainder when divided by $m,,$ i.e. $\ a\bmod m = b\bmod m\ \ $ – Bill Dubuque Jun 12 '17 at 23:06