Solving for n: Natural Number Fraction Equality

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Homework Help Overview

The original poster presents a problem in algebra, specifically focusing on the conditions under which the fraction \(\frac{n^2-n+2}{n+1}\) yields a natural number for \(n \in \mathbb{N}\).

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss rewriting the fraction through polynomial long division and question the validity of certain algebraic manipulations. There are attempts to clarify the relationship between the original fraction and its polynomial representation.

Discussion Status

The discussion is ongoing, with various participants providing hints and nudges towards understanding the algebra involved. Some participants express confusion about the algebraic steps and seek clarification, while others encourage exploration of specific values of \(n\) to identify when the fraction results in a natural number.

Contextual Notes

There are indications of misunderstanding regarding algebraic manipulation and the need for careful consideration of polynomial division. The original poster is encouraged to explore specific values of \(n\) to see the outcomes of the fraction.

  • #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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