Question

prove that 5+4 root 2 is irrational number​

Answer 1

$\sf Let's\ assume\ that\ 5\ +\ 4\sqrt{2}\ is\ rational.\\\\\\5\ +\ 4\sqrt{2}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ is\ \ne\ 0.\\\\\\5\ +\ 4\sqrt{2}\ =\ \dfrac{a}{b}\\\\\\4\sqrt{2}\ =\ \dfrac{a}{b}\ -\ 5\\\\\\4\sqrt{2}\ =\ \dfrac{a\ -\ 5b}{b}\\\\\\\sqrt{2}\ =\ \dfrac{a\ -\ 5b}{4b}$

$\sf Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{a\ -\ 5b}{4b}\ is\ rational,\\\\\\\implies\ \sqrt{2}\ is\ rational\ as\ well.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{2}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\5\ +\ 4\sqrt{2}\ is\ irrational.$

Answer 2

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❚ QuEstiOn ❚

# prove that 5+4 root 2 is irrational number.

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❚ ANsWeR ❚

✺ Given :

• $5+4\sqrt{2}$

✺ To ProVe :

• $5+4\sqrt{2}$ is irrational .

✺ Explanation :

Let , us assume to the contrary that $5+4\sqrt{2}$ is rational ,

Again let ,

✏ $5+4\sqrt{2}=\dfrac{a}{b}$

( where, a and b are co-prime numbers )

✏ $4\sqrt{2}=5-\dfrac{a}{b}$

✏ $4\sqrt{2}=\dfrac{5b-a}{b}$

✏ $\sqrt{2}=\dfrac{5b-a}{4b}$

Now , as a and b are integers so , we can say that , $\dfrac{5b-a}{4b}$ is rational , and so $\sqrt{2}$ is also rational .

But , this contradicts the fact that $\sqrt{2}$ is irrational .

Hence , this contradiction has arisen because of our incorrect assumption that $5+4\sqrt{2}$ is rational ,

✺ Therefore :

So we conclude that $5+4\sqrt{2}$ is irrational . (proved)

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