I read that the modulo operation finds remainder after division.
My misconseption here is about the remainder.
Is the remainder the last digit of the result? I thought it was the first.

Example: $5/7 = 0.7142857142857142857...$

$5 \mod 8 = 7$.

$7/3 = 2.33333..$ but $7 \mod 3 = 1$. Remainder of the first result is point 3333...

$9/4 = 2.25$ $9 \mod 4 = 1$. Ok same as first but different remainder.

$9/7 = 1.2857142857142...$ $9 \mod 7 = 2$. What? Remainder is two, the first digit. This is probably "random" for reals? But not for integers?

$111111111$ : If there are nine ones we should divide by $7$, it gives $2$ remnants. Since $9-2=7$ ones.
So using modulo operation on the reals is not something one should do?
['https://math.stackexchange.com/questions/3002562', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/-1/']
When you want to compute $a \bmod b$ with $a,b$ positive integers, you do division like you learned in school and write $a=qb+r$ where $q$ is the quotient and $r$ is the remainder. We typically choose $r$ to be in the range $0$ to $b-1$. Then we write $r=a \bmod b$, so is you want $1234 \bmod 7$, you write $1234=176\cdot 7 + 2$, so $1234 \bmod 7=2$. You don't need to do anything with the decimal expansion. 

The main time I have seen the modulo operation used with the modulus $b$ other than an integer is reducing angles of a circle when the angles are measured in radians. You compute an angle, then take it $\mod 2\pi$ to put the angle in the range $[0,2\pi)$. You can use other moduli, but this is the common one.
$\frac 73 = 2\frac 13 = 2.333333....$ but neither $\frac 13$ nor $0.33333....$ are the "remainder". $\frac 13$ and $0.3333....$ are the *values* of the result of dividing a value by a divisor that is larger than it. 

The "remainder" is what *remains* after you have divided everything that will fit in in evenly. $2\*3 = 6 < 7$ and $3\*3 = 9 > 7$. So the most we can evenly fit $3$ into $7$ is $2$ times. That leaves $7- 2\*3 = 1$ that we haven't dealt with and can't deal with unless we get a hammer. So that is the *remainder*.

We can then take the remainder and a a big hammer and *pound* it into a fractional bit of $\frac 13$ or $0.33333....$ and take care of it that way. But then ... we have dealt with it. And *nothing* remains. Because we took care of it.

So by definition, the remainder is the *integer* that is left after we divide things evenly *before* we take the hammer and meat grinder to get the fractional parts.

Anyway... the actual digits of the decimal representations have nothing to do with anything. 

...

To your examples:

> 
> 5/7=0.7142857142857142857...
> 
> 
> 

Um... you never do anything with it:

$5 = 0\times 7 + 5; 0\*7 < 5 < 1\*7$ so $5\div 7 = 0$ with $5$ remainder. $\frac 57 = 0\frac 57$. The remainder is $5$. 

$5\equiv 5 \pmod 7$.

> 
> 5mod8=7
> 
> 
> 

That is incorrect.

$5 = 0\times 8 + 5; 0\*8 < 5 < 1\*8$ so $5\div 8 = 0$ with $5$ remainder. $\frac 58 = 0\frac 58$. The remainder is $5$. 

$5\equiv 5 \pmod 8$.

> 
> 7/3=2.33333.. but 7mod3=1. Remainder of the first result is point 3333...
> 
> 
> 

$7 = 2\times 3 + 1; 2\*3 < 7 < 3\*3$ so $7\div 3 = 2$ with $1$ remainder. $\frac 73 = 2\frac 13$. The remainder is $1$. 

$7\equiv 1 \pmod 3$.

> 
> 9/4=2.25 9mod4=1. Ok same as first but different remainder.
> 
> 
> 

$9 = 2\times 4 + 1; 2\*4 < 9 < 3\*4$ so $9\div 4 = 2$ with $1$ remainder. $\frac 94 = 2\frac 14$. The remainder is $1$. 

$9\equiv 1 \pmod 4$.

> 
> 9/7=1.2857142857142... 9mod7=2. What? Remainder is two, the first digit. This is probably "random" for reals? But not for integers?
> 
> 
> 

$9 = 1\times 7 + 2; 1\*7 < 9 < 2\*7$ so $9\div 7 = 1$ with $2$ remainder. $\frac 97 = 1\frac 27$. The remainder is $2$. 

$9\equiv 2 \pmod 7$.

The fact that the first digit of $\frac 27$ is $2$ is an irrelevent coincidence.

Decimals are not important. There are days where I fantasize that it should be illegal to teach decimals outside of engineering and accounting classes. Any way decimal expansion has *NOTHING* to do with this question.

> 
> 111111111 : If there are nine ones we should divide by 7, it gives 2 remnants. Since 9−2=7 ones. So using modulo operation on the reals is not something one should do?
> 
> 
> 

That is .... incomprehensible.

$1111111111 = 15873015\times 7 + 6;15873015\*7 < 111111111 < 15873016\*7$ so $111111111\div 7 = 15873015$ with $6$ remainder. $\frac {111111111}7 = 15873015 \frac 67$. The remainder is $6$. 

$111111111\equiv 6 \pmod 7$.