Homework Help: Natural number

1. Mar 7, 2008

Theofilius

1. The problem statement, all variables and given/known data

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?

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

2. Relevant equations

3. The attempt at a solution

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

2. Mar 7, 2008

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 …

3. Mar 8, 2008

Theofilius

Like this?

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

And then what?

4. Mar 8, 2008

tiny-tim

No.

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

Try again!

5. Mar 8, 2008

Theofilius

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

6. Mar 8, 2008

tiny-tim

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!

7. Mar 8, 2008

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

8. Mar 8, 2008

tiny-tim

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

9. Mar 8, 2008

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.

10. Mar 8, 2008

tiny-tim

Theofilius, I repeat:
(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.

11. Mar 8, 2008

Theofilius

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

12. Mar 8, 2008

tiny-tim

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

Try, and you'll see …

13. Mar 8, 2008

Theofilius

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

14. Mar 8, 2008

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.

15. Mar 8, 2008

Theofilius

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

let's get n=2

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

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

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

Last edited: Mar 8, 2008
16. Mar 8, 2008

Shooting Star

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

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.

17. Mar 8, 2008

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?

18. Mar 8, 2008

Shooting Star

No. Write it as quotient and remainder form.

19. Mar 8, 2008

Theofilius

So, I've done it.

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

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

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

Last edited: Mar 8, 2008
20. Mar 8, 2008

Shooting Star

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.