State whether the following statement is true:
There is a number $x$ such that $x^{2}$ is irrational but $x^{4}$ is rational. Justify your answer by an example.
True,
Let us take $x=\sqrt[4]{2}$
Now, $\quad x^{2}=(\sqrt[4]{2})^{2}=\sqrt{2},$ an irrational number.
$x^{4}=(\sqrt[4]{2})^{4}=2,$ a rational number.
So, we have a number $x$ such that $x^{2}$ is irrational but $x^{4}$ is rational.
If $a=2+\sqrt{3},$ then find the value of $a-\frac{1}{a}$
Prove that, $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}=1$
Insert a rational number and an irrational number between the following:
$6.375289$ and $6.375738$
Show that $0.142857142857 \ldots=\frac{1}{7}$
Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal.
$\frac{\sqrt{2}}{2+\sqrt{2}}$