Question

prove that √5 + 7√2 is irrational. do not copy

Answer 1

$\bf\underbrace\red{answer:}$

Let us assume (5+7√2) is a rational number. Since , a,b are integers, (a-5b)/7b is a rational . So , √2 is rational .

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Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\sf \: Let \: assume \: that \: \sqrt{5} + 7 \sqrt{2} \: is \: not \: irrational$

$\rm :\implies\: \sqrt{5} + 7 \sqrt{2} \: is \: rational.$

$\rm :\longmapsto\:Let \: \sqrt{5} + 7 \sqrt{2} = \dfrac{x}{y}$

where x and y are integers and HCF (x, y) = 1.

$\rm :\longmapsto\:\: 7 \sqrt{2} = \dfrac{x}{y} - \sqrt{5}$

On squaring both sides, we get

$\rm :\longmapsto\: {(7 \sqrt{2})}^{2} = {\bigg(\dfrac{x}{y} - \sqrt{5} \bigg) }^{2}$

$\rm :\longmapsto\:98 = \dfrac{ {x}^{2} }{ {y}^{2} } + 5 - 2 \times \sqrt{5} \times \dfrac{x}{y}$

$\rm :\longmapsto\:\dfrac{2 \sqrt{5}x}{y} = \dfrac{ {x}^{2} }{ {y}^{2} } + 5 - 98$

$\rm :\longmapsto\:\dfrac{2 \sqrt{5}x}{y} = \dfrac{ {x}^{2} }{ {y}^{2} } - 93$

$\rm :\longmapsto\:\dfrac{2 \sqrt{5}x}{y} = \dfrac{ {x}^{2} - 93 {y}^{2} }{ {y}^{2} }$

$\rm :\longmapsto\: \sqrt{5} = \dfrac{ {x}^{2} - 93 {y}^{2} }{ 2x{y}}$

Since, x and y are integers,

So,

$\rm :\longmapsto\: \dfrac{ {x}^{2} - 93 {y}^{2} }{ 2x{y}} \: is \: rational \:number$

$\bf\implies \: \sqrt{5} \: is \: rational$

$\rm :\longmapsto\:which \: is \: contradiction \: to \: the \: fact \: that \: \sqrt{5} \: is \: irrational.$

Hence, our assumption is wrong.

$\bf\implies \:\sqrt{5} + 7 \sqrt{2} \: \: is \: \: irrational$

Hence, proved.

Additional Information :-

Irrational numbers :-

Those numbers whose decimal representation is non terminating and non repeating. Basically, the square root of prime numbers are always Irrational.

Rational number :-

Those numbers whose decimal representation is either terminating or non terminating but repeating.