Question

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Answer 1

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Answer 2

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

We have to proof that ${\sf{5+\sqrt{3}}}$ is an irrational number.

$\qquad \qquad{\pmb{\sf{\underline{KnowlEdge \; required...}}}}$

Rational number: Rational number are those numbers which can be written in the form of ${\sf{\dfrac{p}{q}}}$ where q ≠ 0 i.e., q is not equal to zero. Some example of rational number are ${\sf{\dfrac{23}{9} \: , \dfrac{777}{44432}}}$ etc etc etc.

Irrational number: Irrational number are the inverse of rational numbers. These numbers can't be written in the form of ${\sf{\dfrac{p}{q}}}$ The bestest example for irrational numbes are ${\sf{\pi}}$ and ${\sf{\sqrt{}}}$

Now let us prove that ${\sf{5+\sqrt{3}}}$ is an irrational number.

Firstly let us assume that ${\sf{5+\sqrt{3}}}$ is rational number. Meanforth, it can be written in the form of ${\sf{\dfrac{p}{q}}}$ and q is not equal to zero. Therefore,

${\sf{:\implies 5+\sqrt{3} \: = \dfrac{a}{b}}}$

• (Where, a and b are co-primes)

${\sf{:\implies \sqrt{3} \: = \dfrac{a}{b} - 5}}$

Meanforth, a and b are integers and square root 3 is rational number. But remember that square root is irrational?

Therefore, this contradiction ia arises due to wrong assumption that ${\sf{5+\sqrt{3}}}$ is rational number.

Henceforth, proved that ${\sf{5+\sqrt{3}}}$ is an irrational number.