Question

Prove that the sum of a natural number and it's reciprocal would not be 61/30?

Answer 1

Let there exists a natural number $n$ which when added to its reciprocal gives $\frac{61}{30}.$

Thus we have the equation,

$\longrightarrow n+\dfrac{1}{n}=\dfrac{61}{30}$

$\longrightarrow \dfrac{n^2+1}{n}=\dfrac{61}{30}$

By cross multiplication,

$\longrightarrow 30(n^2+1)=61n$

$\longrightarrow 30n^2+30=61n$

$\longrightarrow 30n^2-61n+30=0$

Now we formed a quadratic equation, where,

• $a=30$

• $b=-61$

• $c=30$

Finding the discriminant,

$\longrightarrow D=(-61)^2-(4\times30\times30)$

$\longrightarrow D=3721-3600$

$\longrightarrow D=121$

Now, solving the quadratic equation,

$\longrightarrow n=\dfrac{-b\pm\sqrt D}{2a}$

$\longrightarrow n=\dfrac{61\pm\sqrt{121}}{2\times30}$

$\longrightarrow n=\dfrac{61\pm11}{60}$

$\longrightarrow n=\dfrac{6}{5}\quad OR\quad n=\dfrac{5}{6}$

Here $\frac{6}{5}$ and $\frac{5}{6}$ are not natural numbers.

This contradicts our assumption that $n$ is a natural number.

Hence Proved!