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while dividing $1\over11$ we get Quotient$ = 0.090909$ and Remainder$ = 1$ but when we apply Dividend$ = $Divisor$\times$Quotient$ + $Remainder, this equation is not satisfied.

What is the reason behind this?

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  • 1
    $\begingroup$ What irrational number are you referring to? $\endgroup$
    – littleO
    Jan 2 '18 at 6:25
  • $\begingroup$ sorry it is rational number. 1/11 $\endgroup$
    – codenut
    Jan 2 '18 at 6:27
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    $\begingroup$ while dividing 1/11 we get quotient = 0.090909 and reminder = 1 How do you get those? Usual integer division would mean quotient $0$ and remainder $1$. $\endgroup$
    – dxiv
    Jan 2 '18 at 6:32
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"while dividing 1/11 we get quotient = 0.090909 and reminder = 1"

No. You do not.

If you divide $11$ into $1$ we find $11 < 1$ so so $11$ goes into $1$, $0$ times so the quotient is $0$ and we have remainder $1$.

And we apply that to $dividend = divisor*quotient + remainder$ we get:

$1 = 11*0 + 1$

which is ... perfect.

You probably went past the decimal point while dividing.

The thing is if you go past the decimal point you are multiplying the dividend by a power of $10$ for every decimal place. That means you will have to divide the remainder by the same powers of $1$.

So if you divided and got $quotient = 0.09090909$ and stopped you went past the decimal place $8$ places.

What that really means is you divided $10^8$ by eleven and got a quotient of $9090909$ and a remainder of $1$. This works. $10^8 = 11*9090909 + 1 = 99999999 + 1 = 100000000 = 10^8$.

But we divide everything by $10^8$ to have a $quotient = 0.09090909$ and a $remainder = 0.00000001$. ANd our equation is $1 = 11*0.09090909 + 0.00000001 = 0.99999999 + 0.00000001 = 1.00000000$.

And that's just fine.

Everything is fine.

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When you're talking about a remainder, all the values are typically integers. In this case, the quotient would be 0, not 0.090909. Then 1 = 11 * 0 + 1.

If you want the quotient to be approximately 0.090909, then there would be no remainder to consider. That is the exact value of 1/11.

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  • $\begingroup$ Actually the remainder would be $0.00000001$. And indeed, $\frac 1{11} = 0.09090909 + \frac {0.00000001}{11}$. Or more familiarly $1 = 11*0.09090909 + 0.00000001$. $\endgroup$
    – fleablood
    Jan 2 '18 at 8:27
  • $\begingroup$ @fleablood That's also technically correct, but there is rarely a case when you're considering a non-integer remainder. I think he meant "0.090909" to be exact, or at least close enough to not care. $\endgroup$
    – nog642
    Jan 2 '18 at 17:24
  • $\begingroup$ You may be right, but then I wouldn't see how s/he got any remainder at all. If s/he did it by hand (then again it's hard to see why anyone would got 8 decimal places by hand and arbitrarily stop there) s/he'd have a "remainder" of "1" written on the paper. Would she'd be overlooking is that the "1" is in the "0.0000000x's position" and is therefore representing "0.00000001" and not "1" $\endgroup$
    – fleablood
    Jan 2 '18 at 18:53
  • $\begingroup$ @fleablood I think they probably did a calculation (maybe by hand, maybe on a calculator) to get 1/11 = 0 R 1. That's where the remainder of 1 came from, but I think they got confused with the term "quotient", and thought that was the exact value of the division, rather than the 0 they had just got, since "quotient" is often used interchangeably to mean both. $\endgroup$
    – nog642
    Jan 3 '18 at 2:08
  • $\begingroup$ Ah... that makes sense too.... $\endgroup$
    – fleablood
    Jan 3 '18 at 2:59
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1 divide 11 equals to 0 remainder 11. Here, the dividend is 1, divisor is 11, quotient is 0, and remainder is 1. 1 equals 0 x 11 + 1; dividend equals quotient x divisor + remainder.

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  • $\begingroup$ remainder is 1. $\endgroup$
    – codenut
    Jan 2 '18 at 6:50
  • $\begingroup$ Oops sorry I edited that. Hope it helps $\endgroup$
    – QuIcKmAtHs
    Jan 2 '18 at 6:52
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If you are doing division with remainder, then the quotient will usually (always?) be an integer. If your notion of division allows for infinitely repeating decimals whenever that happens, then you don't get a remainder at all.

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  • $\begingroup$ I think what the op did was do long division past the decimal 8 places and decided to quit. So s/he had a remainder that was written as $1$ on the paper but didn't realize this was in the $10^{-8}$ position and represented $0.00000001$ instead of an actual $1$. So the remainder was $0.00000001$ instead of $1$. $\endgroup$
    – fleablood
    Jan 2 '18 at 8:31
  • $\begingroup$ @fleablood That's why I put "usually" in there, and just an uncertain "always". You can have remainders for other digits than the $1$'s, but usually you don't call it a remainder in that case, you just ignore it and write the answer as $0.090909\ldots$ $\endgroup$
    – Arthur
    Jan 2 '18 at 8:36
  • $\begingroup$ Yeah.... But I have to wonder why did s/he think s/he had a remainder at all. When you do a division such as $\frac {61}{7}$ and you get $8$ and a remainder of $5$ you may write the result as $8\frac 57$. You go past the decimal point and get to $8.714285$ and realize you are repeating and that you now have an extra remainder of $5$ it is acceptable (albeit poor form) to write $\frac {61}7 = 8.714285+\frac {5}{7000000}$ $\endgroup$
    – fleablood
    Jan 2 '18 at 19:04

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