Solving for n: Natural Number Fraction Equality

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SUMMARY

The discussion focuses on determining the natural number values of \( n \) for which the fraction \( \frac{n^2 - n + 2}{n + 1} \) is also a natural number. Participants suggest using polynomial long division to express the fraction in a more manageable form, ultimately leading to the conclusion that the fraction simplifies to \( n - 2 + \frac{4}{n + 1} \). The values of \( n \) that yield natural numbers for the fraction are confirmed to be \( n = 1 \) and \( n = 3 \).

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  • #31
kbaumen said:
Ah, ok, I'll try not to say that much next time. How 'bout this?

A newbie not so good in Math may get confused. Perhaps the best would be to give him just the first step of your derivation; or something like

\frac{n^{2} - n + 2}{n + 1} = \frac{n(n+1) - 2n + 2}{n + 1}, so that he again makes a factor of (n+1) from the rest of the terms in the numerator.

It all depends on how it's going.
 
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  • #32
Shooting Star said:
A newbie not so good in Math may get confused. Perhaps the best would be to give him just the first step of your derivation; or something like

\frac{n^{2} - n + 2}{n + 1} = \frac{n(n+1) - 2n + 2}{n + 1}, so that he again makes a factor of (n+1) from the rest of the terms in the numerator.

It all depends on how it's going.
I doubt that would help him much
\frac{n^2- n+ 2}{n+1}= \frac{(n-2)(n+1)+ 4}{n+1}
would be much better.
Of course, it would be best if he learned to divide!
 
  • #33
Ok. I understand now, thanks everyone.
n^2- n+ 2= (n-2)(n+1)+4

So i divide the whole equation with \frac{1}{n+1}

\frac{n^2-n+2}{n+1}= n-2 + \frac{4}{n+1}

So only for n=1 and n=3, I get natural numbers for the fraction, right?
 
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  • #34
Theofilius said:
So only for n=1 and n=3, I get natural numbers for the fraction, right?

That's it!

So just remember the moral - turn any nasty fraction into a nice one! (and practise your multiplication …) :smile:
 
  • #35
Shooting Star said:
... or something like

\frac{n^{2} - n + 2}{n + 1} = \frac{n(n+1) - 2n + 2}{n + 1}, so that he again makes a factor of (n+1) from the rest of the terms in the numerator.

It all depends on how it's going.

HallsofIvy said:
I doubt that would help him much
\frac{n^2- n+ 2}{n+1}= \frac{(n-2)(n+1)+ 4}{n+1}
would be much better.

It depends, I repeat, on how you feel the student is responding. I want to give a hint, not the answer. He may miss the point even if you give the hint your way, as happened in this thread.

Shooting Star said:
When you divide (n^2-n+2) by (n+1), what is the quotient and the remainder? .

HallsofIvy said:
Of course, it would be best if he learned to divide!

Ah, somebody who finally agrees with me. But I was considered dated. :rolleyes: Division takes care of all the problems.

HallsofIvy said:
And, as you have already been told, your fraction is the same as 1+ 10/(n-3).

:confused:
I think a perusal of the complete thread is necessary before posting; otherwise OPs may get confused.
 
  • #36
:confused:
™​
 

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