Solving for n: Natural Number Fraction Equality

[Theofilius]

Homework Statement

Hello! I am new user on this forum.

I have one problem with Algebra.

Here it is:

For which n \in \mathbb{N}, this fraction is also natural number?

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

Homework Equations

The Attempt at a Solution

I don't have any idea, where to start from. Thanks for the help.

 

[tiny-tim]

Hi Theofilius! Welcome to PF!

Whenever you have a nasty fraction …

turn it into a nice one! ​

So … do the long division … try expressing \frac{n^2-n+2}{n+1} as a sum of an ordinary polynomial (that isn't a fraction), and a number over (n + 1).

And then …

 

[Theofilius]

Like this?

(n^2-n+2)(n+1)^-^1

And then what?

 

[tiny-tim]

No.

Like this: (n+7)/(n - 3) = 1 + 10/(n - 3).

Try again!

 

[Theofilius]

How did you find this? And what did we prove with this equation?

 

[tiny-tim]

How did you find this? And what did we prove with this equation?

Ah!

Obviously, 1 = (n - 3)/(n - 3).

So 1 + 10/(n - 3)
= (n - 3)/(n - 3) + 10/(n - 3)
= (n - 3 + 10)/(n - 3)
= (n - 7)/(n - 3).

Similarly, you can use n = n(n - 3)/(n - 3), n^2 = n^2(n - 3)/(n - 3), and so on …

Have a go!

 

[Theofilius]

I think you have some error. We can't write (n+7)/(n - 3) = 1 + 10/(n - 3), because if I write n=1,
(1+7)/(1-3)=1+10/(1-3)
8/-2=1+10/-2
-4=-4

And if we get n=1 in the fraction, we will have \frac{1^2-1+2}{1+1}=1

 

[tiny-tim]

I think you have some error. We can't write (n+7)/(n - 3) = 1 + 10/(n - 3), because if I write n=1,
(1+7)/(1-3)=1+10/(1-3)
8/-2=1+10/-2
-4=-4

But -4=-4 is correct!

Theofilius, you are somehow going to have to convince yourself that (n+7)/(n - 3) = 1 + 10/(n - 3) , or you'll never be able to deal with fractions of polynomials.

First, try some numbers other than 1.

Then go though my proof again:

Obviously, 1 = (n - 3)/(n - 3)\,.

So 1\,+\.10/(n - 3)<br /> = (n\,-\,3)/(n\,-\,3)\,+\,10/(n\,-\,3)<br /> = (n\,-\,3\,+\,10)/(n\.-\,3)<br /> = (n\,-\,7)/(n\,-\,3)\,.

and tell me which bit of the proof you don't follow.

 

[Theofilius]

Look, we need to find out for which number n which belongs to the natural numbers, we get also natural number for the fraction. Not an equation -4=-4 or whatever.

 

[tiny-tim]

Theofilius, I repeat:

you are somehow going to have to convince yourself that (n+7)/(n - 3) = 1 + 10/(n - 3) , or you'll never be able to deal with fractions of polynomials.

(a) That's very true!

(b) Take my word for it - once you've rewritten \frac{n^2-n+2}{n+1} as a sum of an ordinary polynomial (that isn't a fraction), and a number over (n + 1), the answer will be obvious.

 

[Theofilius]

And what have I done with it? With every n, number I get natural number?

 

[tiny-tim]

And what have I done with it? With every n, number I get natural number?

No, you'll find that with most n, you don't. But with some you do …

Try, and you'll see …

 

[Theofilius]

And my task says, for which n we get natural number for the fraction?

 

[HallsofIvy]

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

1 is, of course, a natural number. For what values of n is 10/(n-3) a natural number?

If you can't see it immediately try n= 1, 2, 3, 4, 5, etc.

 

[Theofilius]

\frac{n^2-n+2}{n+1} \neq 1+ \frac{10}{n-3}

let's get n=2

\frac{2^2-2+2}{2+1} \neq 1+ \frac{10}{2-3}

\frac{4}{3} \neq 1+ \frac{10}{-1}

\frac{4}{3} \neq -9

 

[Shooting Star]

I think you have some error. We can't write (n+7)/(n - 3) = 1 + 10/(n - 3),

Why? Have you tried to simplify the RHS and see whether you get back (n+7)/(n - 3)?

(b) Take my word for it - once you've rewritten \frac{n^2-n+2}{n+1} as a sum of an ordinary polynomial (that isn't a fraction), and a number over (n + 1), the answer will be obvious.

What tiny-tim is trying to do is teach you algebraic division again.

When you divide (n^2-n+2) by (n+1), what is the quotient and the remainder? In the example he has given, when you divide n+7 by n-3, you get a quotient of 1 and a remainder of 10. So, (n+7)/(n - 3) = 1 + 10/(n - 3). He wants you to express your original fraction in this way.

 

[Theofilius]

Oh my god! It was its example? I thought it was connected with my example. Lol!
When I'll divide n^2-n+2:(n+1)=n-2
So I will get natural number for every n-2 number, right?

 

[Shooting Star]

No. Write it as quotient and remainder form.

 

[Theofilius]

So, I've done it.

(n+1)(n-2)+0=n^2-n+2

(n+1)(n-2)+0=n^2-n+2 /: \frac{1}{n+1}

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

 

[Shooting Star]

So, I've done it.

(n+1)(n-2)+0=n^2-n+2

(n+1)(n-2)+0=n^2-n+2/:\frac{1}{n+1}

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

Not correct...

What you have to do is write (n^2-n+2)/(n+1) as Q + R/n, where Q would be a polynomial in n whose degree here is 1, and R is a polynomial whose degree is less than that of n-1.

Go back to the example again.

 

[tiny-tim]

Nearly there …

(n+1)(n-2)+0=n^2-n+2

Nearly there!

But you have to be more careful with your multiplication:

(n+1)(n-2)\,=\,n^2\,-\,n\,-\,2, not n^2\,-\,n\,+\,2.

So that gives you … ?

 

[Shooting Star]

Hello Physicsissuef,

I think the OP would have eventually got there by himself! That was the purpose of nudging him gently toward the right direction, without spelling it out for him...

(EDIT: This was in response to a post by Physicsissuef, which he later deleted.)

 

[Physicsissuef]

Hello Physicsissuef,

I think the OP would have eventually got there by himself! That was the purpose of nudging him gently toward the right direction, without spelling it out for him...

Now, he will understand that, his error was in his multiplication, and nothing else.

 

[tiny-tim]

It should be …

Hi Physicsissuef!

Could you possibly edit your post so as not to give the actual answer - in the hope that Theofilius hasn't seen it yet?

Thankx.

 

[Physicsissuef]

Ok, I deleted it.

 

[kbaumen]

I'll give a BIG hint. \frac{n^{2} - n + 2}{n + 1} = \frac{n^{2} - n - 2 + 4}{n + 1}= \frac{(n + 1)(n - 2) + 4}{n + 1} = \frac{(n + 1)(n - 2)}{n + 1} + \frac{4}{n + 1} = n - 2 + \frac{4}{n + 1}.
You should be able to solve it from here.

 

[Shooting Star]

Hi Physicsissuef!

Could you possibly edit your post so as not to give the actual answer - in the hope that Theofilius hasn't seen it yet?

Thankx.

The reply would have automatically gone to the OP's mailbox anyway. But thanks to Physicsissuef for trying. And now, after presumably having read the whole thread, here is a BIG "hinter".

Give it up, tiny-tim. Be satisfied thinking that the OP must have got the answer by himself, and learned something. That's what PF's for.

 

[kbaumen]

Well, I didn't show the complete solution. I just showed a little bit different approach. I saw that some people here recommended doing the division, looking at the reminder etc, when the equation could simply be rearranged to a form in which it is quite easy to understand.

 

[Shooting Star]

Well, I didn't show the complete solution. I just showed a little bit different approach. I saw that some people here recommended doing the division, looking at the reminder etc, when the equation could simply be rearranged to a form in which it is quite easy to understand.

Some people seem to be very old fashioned indeed!

The "people" here meant to ultimately guide the OP to do exactly what you have done, but since the OP perhaps seemed a bit confused by member tiny-tim's approach through an arbitrary example, the "people" here decided to give it in a way that maybe a young OP could relate to, saying quotients and remainders.

To be honest, your hint was a bit too big for a hint. But good job.

 

[kbaumen]

Ah, ok, I'll try not to say that much next time. How 'bout this?

To solve this exercise, you need to rearrange the equation. n^{2} - n + 2 can't be factorized, but what about n^{2} - n - 2?

 

[Shooting Star]

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.

 

[HallsofIvy]

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!

 

[Theofilius]

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?

 

[tiny-tim]

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 …)

 

[Shooting Star]

... 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.

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.

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

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

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

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

I think a perusal of the complete thread is necessary before posting; otherwise OPs may get confused.

 

[tiny-tim]

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