# Thread: real number problem 3

1. ## real number problem 3

prove that there is no positive integer n for which root n - 1 + root n + 1 is rational.

2. Originally Posted by saha.subham
prove that there is no positive integer n for which root n - 1 + root n + 1 is rational.
Suppose $a=\sqrt{n-1}+\sqrt{n+1}$ is rational.
Then
$a^2=2n+2\sqrt{(n+1)(n-1)}$
$\frac{a^2-2n}2=\sqrt{(n+1)(n-1)}$
Hence $\sqrt{(n+1)(n-1)}$ is rational, which implies that $(n+1)(n-1)=n^2-1$ is a perfect square.
We get $n^2-1=m^2$ for some integer m.
But the only solutions in integere are $n=\pm1$ and $m=0$.
But for n=1 we have $a=\sqrt2$, which is irrational.

EDIT: Corrected typo n(n-1) to (n+1)(n-1) on one place.

3. its root n + 1 not whole root n + 1

4. Originally Posted by saha.subham
its root n + 1 not whole root n + 1
So you want to work with $\sqrt{n-1}+\sqrt{n}+1$?

5. yeah exactly plzz solve it

6. $a=\sqrt{n-1}+\sqrt{n}+1$ is rational if and only $b=a-1=\sqrt{n-1}+\sqrt{n}$ is rational.

You get:
$b=\sqrt{n-1}+\sqrt{n}$
$b^2=2n-1+2\sqrt{(n-1)n}$
$\frac{b^2-2n+1}2=\sqrt{(n-1)n}$
meaning that $\sqrt{(n-1)n}$ is rational and therefore (n-1)n is a perfect square.

Now since n-1 and n are coprime, both n-1 and n are perfect squares. So again you have two squares which differ by 1 and you can continue int he same way as I did in my first post. You get that the only solution is n=1.

BTW do you know the result, that a square root of an integer is rational if and only if this integer is a perfect square? If you did not learn this, you were probably supposed to look for a completely different solution. (Perhaps there is a simpler one and I have overlooked it...)

7. $
a=\sqrt{n}-1+\sqrt{n}+1
$

sorry i meant to said this one. extremely sorry ti disturb u again

8. Originally Posted by saha.subham
$
a=\sqrt{n}-1+\sqrt{n}+1
$

sorry i meant to said this one. extremely sorry ti disturb u again
This one is extremely easy. But this expression is rational for many integers. You have:
$a=2\sqrt{n}$
$\sqrt{n}=\frac a2$
Hence $\sqrt{n}$ is rational if and only if n is a perfect square.

You can try it yourself for $1=1^2$, $4=2^2$, $9=3^2$, $16=4^2$, etc.