Question

please help in this of rd sharma ​

Answer 1

(1)

Assume to reach the contradiction that $\displaystyle\sf{\dfrac{2+\sqrt3}{5}}$ is a rational number.

Let,

$\displaystyle\longrightarrow\sf{x=\dfrac{2+\sqrt3}{5}}$

where $\displaystyle\sf{x}$ is assumed to be a rational number as earlier.

Then,

$\displaystyle\longrightarrow\sf{x^2=\dfrac{(2+\sqrt{3})^2}{5^2}}$

$\displaystyle\longrightarrow\sf{x^2=\dfrac{7+4\sqrt3}{25}}$

$\displaystyle\longrightarrow\sf{\sqrt3=\dfrac{25x^2-7}{4}}$

In this equation, since $\displaystyle\sf{x}$ is rational, so is $\displaystyle\sf{x^2,}$ and hence the RHS. But the LHS, i.e., $\displaystyle\sf{\sqrt3,}$ is irrational and cannot be expressed as a fraction.

This contradicts our earlier assumption and hence proved that $\displaystyle\sf{\dfrac{2+\sqrt3}{5}}$ is irrational.

(2) We have,

$\displaystyle\longrightarrow\sf{a^n-b^n=(a-b)\sum_{k=1}^na^{n-k}b^{k-1}}$

Therefore,

$\displaystyle\longrightarrow\sf{9^{2n}-4^{2n}=(9-4)\sum_{k=1}^{2n}9^{2n-k}4^{k-1}}$

$\displaystyle\longrightarrow\sf{9^{2n}-4^{2n}=5\sum_{k=1}^{2n}9^{2n-k}4^{k-1}}$

$\displaystyle\longrightarrow\sf{9^{2n}-4^{2n}=5(9^{2n-1}+9^{2n-2}\cdot4+9^{2n-3}\cdot4^2+\dots\!\ +9\cdot4^{2n-2}+4^{2n-1})}$

Hence $\displaystyle\sf{9^{2n}-4^{2n}}$ is always divisible by 5.