Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

To show that there is no positive integer n for which\[\sqrt{n-1}+\sqrt{n+1}\] is rational.

seo-qna
Last updated date: 19th Jul 2024
Total views: 451.2k
Views today: 12.51k
Answer
VerifiedVerified
451.2k+ views
Hint: We use the method of contradiction to solve this problem. One can start by assuming that $\sqrt{n-1}+\sqrt{n+1}$ is rational number, such that,
$\dfrac{a}{b}=\sqrt{n-1}+\sqrt{n+1}$
(Where, a and b are integers with b$\ne $0)


Complete step-by-step answer:
Subsequently, we will prove that this assumption is incorrect, which would then prove that$\sqrt{n-1}+\sqrt{n+1}$ is not rational.
First, we take the reciprocal of $\dfrac{a}{b}$ and then start solving,
\[\begin{align}
  & \dfrac{b}{a}=\dfrac{1}{\sqrt{n+1}+\sqrt{n-1}} \\
 & \dfrac{b}{a}=\dfrac{1}{\sqrt{n+1}+\sqrt{n-1}}\times \dfrac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n+1}-\sqrt{n-1}} \\
\end{align}\]
Using the property, (a+b)(a-b) = ${{a}^{2}}-{{b}^{2}}$on the denominator
\[\begin{align}
  & \dfrac{b}{a}=\dfrac{\sqrt{n+1}-\sqrt{n-1}}{(n+1)-(n-1)} \\
 & \dfrac{b}{a}=\dfrac{\sqrt{n+1}-\sqrt{n-1}}{2}\text{ } \\
 & \dfrac{2b}{a}=\sqrt{n+1}-\sqrt{n-1}\text{ -- (1)} \\
\end{align}\]
Now, we know that,
\[\dfrac{a}{b}=\sqrt{n-1}+\sqrt{n+1}\text{ -- (2)}\]
Adding (1) and (2), to solve, we get,
\[\begin{align}
  & \dfrac{2b}{a}+\dfrac{a}{b}=\sqrt{n+1}-\sqrt{n-1}+\sqrt{n-1}+\sqrt{n+1} \\
 & \dfrac{2b}{a}+\dfrac{a}{b}=2\sqrt{n+1} \\
 & \sqrt{n+1}=\dfrac{1}{2}\left( \dfrac{2b}{a}+\dfrac{a}{b} \right)\text{ -- (3)} \\
\end{align}\]
Subtracting (1) and (2), to solve, we get,
\[\begin{align}
  & \dfrac{2b}{a}-\dfrac{a}{b}=\sqrt{n+1}-\sqrt{n-1}-\sqrt{n-1}-\sqrt{n+1} \\
 & \dfrac{2b}{a}-\dfrac{a}{b}=-2\sqrt{n-1} \\
 & \sqrt{n-1}=\dfrac{1}{2}\left( \dfrac{a}{b}-\dfrac{2b}{a} \right)\text{ -- (4)} \\
\end{align}\]
Adding (3) and (4), we get,
\[\sqrt{n-1}+\sqrt{n+1}=\dfrac{1}{2}\left( \dfrac{2b}{a}+\dfrac{a}{b} \right)+\dfrac{1}{2}\left( \dfrac{a}{b}-\dfrac{2b}{a} \right)\text{ }\]
Now, RHS is a rational number since a and b are integers (according to the assumption made by us in the question). Further, since, all the terms on RHS consist of a and b, thus, it is a rational number.
According to equality (LHS=RHS), even LHS should be a rational number. Thus, (n-1) and (n+1) should be perfect squares for LHS to be a rational number. However,
(n+1) - (n-1) = 2
This contradicts our earlier made assumption. This is because the difference of two perfect squares (which are integers) is either 1 or the difference is greater than equal to 3. The difference of two perfect squares (which are integers) can never be 2. For example, we take a few cases. First, we take 0 and 1, the difference is 1. Then, we take 1 and 4, the difference is 3. Similarly, we can take 4 and 9, the difference is 5. Thus, we observe that the difference keeps increasing and there will be no pair of perfect squares for which the difference is 2.

Note: Generally, for questions in which one has to prove a given expression to be irrational, it is always a good idea to prove it by contradiction (as done in the above problem). After that, one should solve until the point where it can be proved that the equation cannot be true with the assumption made in the problem. (Example in this case, (n+1) - (n-1) = 2 for two perfect squares, (n-1) and (n+1), which is never true. Thus, we could rule out our assumption of the expression being rational and conclude that it is irrational).