Question

If $\sqrt{2}=1.414$ , then find the value of $\dfrac{1}{1-\sqrt{2}}$

Answer 1

Hint: First we need to rationalize the number and then substitute the value of the given variable and then we need to find the final value of expression.

Complete step-by-step answer:
First, we will prove that the given variable $\sqrt{2}$ is irrational.
Assume $\sqrt{2}$ is rational.
By basic algebraic concepts, we get: A number which can be written the form $\dfrac{p}{q}$ with p and q being in lowest terms is called rational number
By applying above concept, we get:
$\sqrt{2}=\dfrac{p}{q}$
Here we have p, q in lowest terms, that means there are no common factors between them.
By squaring on both sides, we get:
$2=\dfrac{{{p}^{2}}}{{{q}^{2}}}$
By multiplying ${{q}^{2}}$ on both sides we get: $2{{q}^{2}}={{p}^{2}}$
Since, the left-hand side and right-hand side are equal if 2 divides left hand side it must also divide right hand side.
So, we can write p = 2k, where k is any integer except zero.
By substituting this, we get:
$\begin{align}
  & 2{{q}^{2}}={{\left( 2k \right)}^{2}} \\
 & {{q}^{2}}=2{{k}^{2}} \\
\end{align}$
Using the same logic, we can prove 2 also divides q.
By above, we get: 2 divides both p, q.
Thus, by this, we can say 2 is a common factor of p, q.
But by definition of rational number pq must not have a common factor.
So, our assumption is wrong.
So, $\sqrt{2}$ is irrational
The expression we have
$\dfrac{1}{1-\sqrt{2}}$
By multiplying and dividing with $1+\sqrt{2}$ , we get
$\dfrac{1}{1-\sqrt{2}}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}=\dfrac{1+\sqrt{2}}{\left( 1-\sqrt{2} \right)\left( 1+\sqrt{2} \right)}$
By applying distributive law:
a.(b + c) = a.b + a.c
Our expression turns into
$\dfrac{1+\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-2}=\dfrac{1+\sqrt{2}}{-1}$
By simplifying we get:
$-\left( \sqrt{2}+1 \right)$
By substituting $\sqrt{2}$ value, we get:
$-\left( 1.414+1 \right)=-2.414$
Therefore, value of given expression is -2.414

Note: Alternate method is to use the algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ after rationalization (instead of distributive law).