# Re: you are attempting to look at a specific case... this is general case

[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Joel on October 25, 2002 at 22:37:50:

In Reply to: you are attempting to look at a specific case... this is general case posted by T.Gracken on October 25, 2002 at 21:12:01:

: You still don't get it...

: remember the remainder is still applied to the divisor...

: so when you use 5 as an x value, the problem reduces to the quotient + the remainder over the divisor... that is (in this case), the quotient + 6/4 which is "the quotient plus (1 and 1/2)".

: so it may not be apparent that the remainder is 6 (since the denominator plays a part in this and you can't 'know' the denominator and reduce at first)

: So quit trying to look at specific values. fractions are involved. and there are a million (or more) ways to write a fraction.

You're right. I don't get it.

[(-1/3) * x^3 - x^2 - x + 25/3] / (x-1) = (-1/3) * x^2 - (4/3) * x - 7/3 + (6/(x-1))
so, if x=5 this becomes:
(-190/3) / 4 = -52/3 + 6/4
This actually IS true.
But what the hell does it mean? When I read that problem I expected the answer to consist of some nice, well-behaved integers where one gets divided by another leaving a nice integer remainder.

It's a thoroughly unsatisfying result.

I won't hold it against you, though. :)

Name:
E-Mail:

Subject: