1. ## Natural numbers

n is a natural unzero number, find x natural so that:

$\displaystyle \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}$

I saw that x must be bigger than n, but how I find it?

2. Originally Posted by mexxy
n is a natural unzero number, find x natural so that:

$\displaystyle \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}$

I saw that x must be bigger than n, but how I find it?
As x increases the difference in [tex]\sqrt{x - 1}[/math[ and [tex]\sqer{x}[/math[ decreases. So x is not likely to be that large. By trial and error (probably the only way to do this one) I get n = 1 and x = 2.

-Dan

3. "I get n = 1 and x = 2." I don't.
n=1, x=6;
n=2, x=18;
n=3, x=27;
n=4, x=36;
n=5, x=45.

4. Originally Posted by topsquark
By trial and error (probably the only way to do this one) I get n = 1 and x = 2.

-Dan
Are you sure that's correct?
If x=2 and n=1 is inserted in the original inequality, we get:

$\displaystyle \sqrt{2-1}<\sqrt{1-1}+\sqrt{1}+\sqrt{1+1}<\sqrt{2}$

and I think THAT looks a bit suspicious.

5. Originally Posted by mexxy
n is a natural unzero number, find x natural so that:

$\displaystyle \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}$

I saw that x must be bigger than n, but how I find it?
As seen in veileen's post, there's an abundance of solutions to this inequality.
Therefore, I think that all you can do is simplify this expression a bit by quadrating the whole thing, so that you can get a clearer relationship between n and x.

If you quadrate the whole inequality, I think that this will be the end result:

$\displaystyle x-1<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x$

but I might have been a bit sloppy while I did it, so I can be a little bit off.

6. Mexxy said she/ he have to find x... probably something like: x= *stuff, stuff, stuff* >.>
I mean - that relationship between n and x should be clearer than what you wrote.

7. Originally Posted by veileen
Mexxy said she/ he have to find x... probably something like: x= *stuff, stuff, stuff* >.>
I mean - that relationship between n and x should be clearer than what you wrote.
Do you mean like something like this:

$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1$

Where the only natural number for a given n, which makes the inequality true, is x.
I personally don't know how I should go about determining the exact value of x, i.e. x= 'something', when I'm only given this type of inequality, which contains many different solutions.

8. $\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1$
$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})$ is not natural ~ "I personally don't know how I should go about determining the exact value of x," - using integer part. (?)

9. Originally Posted by scounged
Are you sure that's correct?
If x=2 and n=1 is inserted in the original inequality, we get:

$\displaystyle \sqrt{2-1}<\sqrt{1-1}+\sqrt{1}+\sqrt{1+1}<\sqrt{2}$

and I think THAT looks a bit suspicious.
It was a typo. n = 1 and x = 3. And I see that it was wrong anyway.

-Dan

10. Originally Posted by veileen
"I get n = 1 and x = 2." I don't.
n=1, x=6;
n=2, x=18;
n=3, x=27;
n=4, x=36;
n=5, x=45.
I'll be darned.

-Dan

11. Originally Posted by veileen
$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1$
$\displaystyle 3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})$ is not natural
Does it have to be?

Originally Posted by veileen
~ "I personally don't know how I should go about determining the exact value of x," - using integer part. (?)
I'm not familiar with integer part. I know that my inequality might not be the best mathematical way to determine x, but it gives a definite solution for x with a given n with the conditions in the original post set.

12. Does it have to be? - No, that is why I suggested use the integer part.
$\displaystyle 3n+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]}\leq 3n+2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})<x$

$\displaystyle x<3n+1+2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})\leq3n+1+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]+1$

$\displaystyle =3n+2+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]$

$\displaystyle [2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]\in \mathbb{N} \Rightarrow x=3n+1+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]$ (because it's a natural number between $\displaystyle 3n+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]}$ and $\displaystyle 3n+2+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})])$

* [] - means integer part