# real number problem 3

#### saha.subham

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

#### kompik

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

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

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• saha.subham

#### saha.subham

its root n + 1 not whole root n + 1

#### kompik

its root n + 1 not whole root n + 1
So you want to work with $$\displaystyle \sqrt{n-1}+\sqrt{n}+1$$?

#### saha.subham

yeah exactly plzz solve it #### kompik

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

You get:
$$\displaystyle b=\sqrt{n-1}+\sqrt{n}$$
$$\displaystyle b^2=2n-1+2\sqrt{(n-1)n}$$
$$\displaystyle \frac{b^2-2n+1}2=\sqrt{(n-1)n}$$
meaning that $$\displaystyle \sqrt{(n-1)n}$$ 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...)

#### saha.subham

$$\displaystyle a=\sqrt{n}-1+\sqrt{n}+1$$

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

#### kompik

$$\displaystyle a=\sqrt{n}-1+\sqrt{n}+1$$

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:
$$\displaystyle a=2\sqrt{n}$$
$$\displaystyle \sqrt{n}=\frac a2$$
Hence $$\displaystyle \sqrt{n}$$ is rational if and only if n is a perfect square.

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

• saha.subham
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