Question

prove that 3—√2 is irrational number​

Answer 1

Answer:

Let a=root3-root2 be a rational number.

Now squaring both sides-

a^2=3+2-2×root6

(a^2-5)/2=-root6

Since a is a rational number then a^2 is also a rational number

But -root6 is an irrational number.

So root3-root2 is not a rational number

Hence,PROVED

Answer 2

$\large\underline\mathfrak\red{Answer \: :- }$

Let 3 - √2 be a rational number.

$\sf\implies{3 - \sqrt{2} = x }$

Squaring on both sides

We get,

$\sf\implies{(3 - \sqrt{2}) {}^{2} = x {}^{2} } \: \: \: \: \: \: \: \: \: \: \: \:\\ \sf\implies{9 + 2 - 2x3x \sqrt{2} = x {}^{2} } \: \\ \sf\implies{11 - x {}^{2} = 6 \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\implies{ \sqrt{2} = \frac{11 - x {}^{2} }{6} } \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Here, x is a rational number.

Also,

• x² is a rational number.
• 11 - x² is a rational number.
• $\sf{ \frac{11 - x {}^{2} }{6} \: is \: also \: a \: rational \: number }$
• $\sf{ \sqrt{2} = \frac{11 - x {}^{2} }{6} \: is \: a \: rational \: number}$

But √2 is a irrational number.

$\sf\implies{ \frac{11 - x {}^{2} }{6} = \sqrt{2 } \: is \: an \: irrational \: number }$

$\sf\implies{11 - x {}^{2} \: is \: an \: irrational \: number}$

$\sf\implies{x {}^{2} \: is \: an \: irrational \: number}$

$\sf\implies{x \: is \: an \: irrational \: number}$

But we have assume that x is a rational number.

$\sf\therefore{we \: arrive \: at \: a \: contradiction.}$

So, our assumption that 3 - √2 is a rational number is wrong.

$\sf\therefore{3 - \sqrt{2} \: is \: an \: irrational \: number }$

${\large{\underline{\underline{\bf{\red{Hence \: Proved \: ! }}}}}}$