real number problem 3

Jul 2009
68
0
prove that there is no positive integer n for which root n - 1 + root n + 1 is rational.
 
Mar 2010
116
41
Bratislava
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.
 
Last edited:
  • Like
Reactions: saha.subham
Mar 2010
116
41
Bratislava
\(\displaystyle a=\sqrt{n-1}+\sqrt{n}+1\) 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...)
 
Jul 2009
68
0
\(\displaystyle
a=\sqrt{n}-1+\sqrt{n}+1
\)


sorry i meant to said this one. extremely sorry ti disturb u again
 
Mar 2010
116
41
Bratislava
\(\displaystyle
a=\sqrt{n}-1+\sqrt{n}+1
\)


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.
 
  • Like
Reactions: saha.subham
Similar Math Discussions Math Forum Date
Algebra
Algebra
Algebra
Number Theory