Question

4. Show that
$6 + \sqrt{2} \: is \: irrational \: \\ using \: the \: fact \: that \: \sqrt{2} \: is \\ irrational.$
Standard:- 10

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Answer 1

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By the method of Contradiction,

=> let 6 + √2 be a rational number

Then 6 + √2 = a where a is a rational number

=> 6 + √2 = a

=> a - 6 = √2

We know the fact that √2 is an irrational number but a is a rational number

Thus our assumption is wrong

Thus 6 + √2 is an irrational number

HENCE PROVED

Easy ?

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Answer 2

$Heya$‼

$Let\: 6 + \sqrt{2} \: is \: a \: rational \: number.$

$6 + \sqrt{2}=a/b$ , where a and b are integers and (b ≠ 0).

$\sqrt{2}=a/b-6$

$\sqrt{2}=(a-6b)/b$

Since $(a-6b)/b$ is a rational number so, $\sqrt{2}$ is also a rational number.

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

Thus, our incorrect consumption is wrong.

So, we include that $6 + \sqrt{2}$ is irrational.
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