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Definition Modular Arithmetic: Based on the definition of modular arithmetic, $a \equiv b \pmod{m}$ is meant to say that $a$ will leave the same remainder as $b$ when either $a$ or $b$ are divided by $m$. So the remainder of $\frac{a}{m}$ would be the same as $\frac{b}{m}$.

In a question, it was asked to find the integer $a$ such that $a \equiv 99 \pmod{41}$ and $100 \le a \le 140$.

Question 1: A solution to that was, "It's also good to note that for $x \equiv y \mod z$, $x$ is just the remainder of integer division of $y$ by $z$. For example, $35/14$ is equal to $2$ remainder $7$, and $7 \equiv 35 \mod 14$." But is this is not the same as the definition provided above, is not it?

Avv
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  • Are you just asking whether that quote provides a full solution to the problem? If so, then, No. It doesn't. It doesn't even appear to address the question. Or are you asking something else? – lulu Mar 17 '21 at 16:34
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    If, alternatively, you are asking if the quote is correct, then the answer is also No. At least, not generally. We could have, say, $147\equiv 993\pmod 2$ but it is obviously not true that $147$ is the remainder of integer division of $993$ by $2$. – lulu Mar 17 '21 at 16:37
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    I would say that that is exactly the same as the definition given. What seems different to you? $35\div 14$ is a quotient of $2$ and a remainder of $7$. And $7 \div 14$ is a quotient of $0$ and a remainder of $7$. They both have remainder $7$. So $7\equiv 35 \pmod 35\pmod {14}$. That is exactly what the definition said. Likewis if we considered $35$ and $49$. $35$ divided by $14$ is $2$ with remainder $7$ and $49$ divided by $14$ is $3$ with remainder $7$. They both have the same remainder. So $49 \equiv 35\pmod {14}$. ... to be continued.... – fleablood Mar 18 '21 at 02:17
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    Note: we don't care what the quotient is. We don't care AT ALL. And we don't actually care what the remainder of $35$ and of $49$ and of $7$ actually are.... we only care if the remainder is the same for all of them. (but to know they are the same we implicitly probably need to know what it is.... But actually knowing what it is isn't actually part of the definition.) – fleablood Mar 18 '21 at 02:18

2 Answers2

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Yes, you are right. They meant to say, (as I guess):

Note that if $x$ is the remainder when $y$ is divided by $z$, then $x\equiv y\ (\text{mod}\ z)$.

Hope this helps. Ask anything if not clear :)

ultralegend5385
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  • Thank you. I forgot that $c$ is an integer, so it can not be $\frac{-1}{41}$. I deleted that part from the question, so please delete the answer to question 2. – Avv Mar 17 '21 at 16:49
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There are two related notions, and similar notation; which may be confusing.

  1. "Binary mod operator": given an integer $a$ and a positive integer $m$, the binary mod operator $a\bmod m$ (pronounced "$a$ mod $m$" or "$a$ modulo $m$") takes as value the remainder when dividing $a$ by $m$ (integer division with remainder); this remainder is assumed to satisfy $0\leq a\lt m$. (This may be different in some computer languages, where the value is taken to have least possible absolute value; so for example, $a\bmod 3$ could be $-1$, $0$, or $1$.) This is an operation: it takes to arguments ($a$ and $m$), and returns a value (the remainder).

  2. The relation "mod $m$". We say that $a\equiv b\pmod{m}$ (pronounced "$a$ is congruent to $b$ modulo $m$) if and only if $m|a-b$. This is a binary relation on the integers. You take two integers, and either they are in the relation, or they are not. (In fact, it is an equivalence relation).

The connection between the two notions is the following:

Theorem. Let $m\gt 0$. For integers $a$ and $b$, $$a\equiv b\pmod{m} \iff a\bmod m = b\bmod m.$$ That is: $a$ is congruent to $b$ modulo $m$ if and only if $a$ and $b$ leave the same remainder when you divide them by $m$.

Proof. Write $a=qm+r$, $0\leq r\lt m$, and $b=pm+s$, $0\leq s\lt m$. Without loss of generality, we may assume $s\leq r$.

Since $a-b = (q-p)m + (r-s)$, with $0\leq r-s\lt m$, we have that $m|a-b$ if and only if it divides $r-s$, if and only if $r-s=0$, if and only if $a\bmod m = r = s =b\bmod m$. $\Box$

So you could define "$a\equiv b\pmod{m}$ if and only if $a$ and $b$ have the same remainder when divided by $m$".

Note that you use $\equiv$ when you work with the equivalence relation, but you use $=$ when you work with the operator. Your final paragraph should really say that

if $x= y\bmod z$, then $x$ is the remainder of integer division of $y$ by $z$

But note the use of the "binary mod" (MathJax/$\LaTeX$ code \bmod) and the equal sign. If you say "$x\equiv y \bmod{z}$ means that $x$ is the remainder of integer division of $y$ by $z$", then that's not technically correct, since you are mixing notations, and in addition, for instance $5\equiv 7\pmod{2}$, and $5$ is not the remainder when you divide $7$ by $2$; the remainder is $1$.

So the answer to your question 1 is that (i) as written it is false because you are mixing the relational and the operational mods, and they are not the same. But (ii) written correctly it asserts something about the mod operator, which is not the mod relation which is what the definition is about.

Arturo Magidin
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    For site health, it's generally better not to keep duplicating these common facts, e.g. in my answer in the linked dupe I link to one of your prior incarnations. If we iteratively refine these prior answers then they can often serve well as dupe targets for such FAQs – Bill Dubuque Mar 17 '21 at 17:40