As an example, why does 1 modulo 2 equal 1?
According to Google's built-in calculator:
1 % 2 = 1
5 % 40 = 5
12 % 2000 = 12
Why is the remainder not "0", "error", or something?
In other words, I don't follow the mathematical reasoning:
Why is the remainder of 5 % 40 set as 5 itself, when, in fact, there is no positive integer (whole-number) remainder, e.g. 5/40 = 0.125?
On
$$1\equiv b \mod 2 \to b-1=2k :k\in\mathbb Z , 0\le b\le2 $$ only $b=1$ satisfy in $b-1=2k :k\in\mathbb Z , 0\le b\le2 $
On
Because; ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
See the space where I gave it a green $0$. For a while consider there is nothing written. What number $\color{green}{n}$ could we put there such that after multiplying by $\color{red}{2}$, say $\color{red}{2}\color{green}{n}$, a substraction $1-\color{red}{2}\color{green}{n}$ makes sense? Search it.... The common number for $\color{green}{n}$ could be $\color{green}{0}$. Now I think, you can make this way generalized for other case.
On
I think the confusion happens because faulty logic that if something can be divided in even parts there should be no remainder (at least that was my first faulty thought which caused me to look at this a littler closer) For example, I reasoned if $10$ can be divided by $2$ and result in $0$ as remainder, that should apply to $\frac12$ where by dividing $1$ into two equal pieces we have nothing left.
Though it is true that $1$ can be divided by $2$ in perfectly equal parts of $\frac12$ but that is not the definition of remainder. One way to think of a remainders is that it is whatever is left when the divisor becomes larger than leftover of a dividend. E.g. $\frac{23}2 = 11$ quotient and $1$ remainder. (since $23-22 = 1$ and $1$ as the current dividend is smaller than $2$ the divisor it becomes the remainder) In cases of any smaller number divided by a larger number, the smaller number (dividend) by being smaller automatically equals the remainder. Thus $12.5 \% 15000 = 12.5$ or $1 \% 2 = 1$. In other words we can't divide into fractions less than the divisor. One more way to think about this is that division is basically a repeated subtraction of one number from another. At some point the number can't be further subtracted as a whole while still resulting in a positive number. whatever that leftover is , is the remainder.
The remainder when $1$ is divided by $2$ is $1$, since $1=(0)(2)+1$ and $0\le 1\lt 2$.
In general, if $0\le a\lt m$ then $a\operatorname{\%}m=a$.
In general, when you divide an integer $a$ by a positive integer $m$, there is a quotient $q$ and a remainder $r$. So $$a=qm+r,$$ where $0\le r\lt m$.
For instance, if $a=30$ and $m=12$, then $q=2$ and $r=6$. If $a=5$ and $m=12$, then $q=0$ and $r=5$.
In the case where $a=1$ and $m=2$, the quotient is $0$ and the remainder is $1$.
Remark: It is useful to have concrete images to go along with more abstract descriptions. Suppose that we have a box that contains $a$ cookies, and we have $m$ kids in the room. We give a cookie to everyone (if we can). Then we do it again, and again, doing a full round each time. The number of cookies left in the box is the remainder when $a$ is divided by $m$, it is what's left over.
For example, if $a=40$ and $m=12$, we do $3$ full rounds, each kid gets $3$ cookies. This $3$ is called the quotient. We will have $4$ cookies left over, the remainder is $4$, in symbols $40\operatorname{\%} 12=4$. If we start with $72$ cookies, the remainder is $0$.
But if we start with $5$ cookies, then we can't even get started, we cannot distribute cookies without causing a riot. So the quotient is $0$, nobody gets a cookie. And all the cookies are left over, the remainder is $5$, that is, $5\operatorname{\%}12=5$.