Question

show that
$3 + \sqrt{5}$
is an irattional number , assume that
$\sqrt{5}$
is an irrational number​

Answer 1

Answer

if a possible, let us assume that

$3 + \sqrt{5}$

be a rational number so, there exists positive integers a and b such that ,

$3 + \sqrt{5} = \frac{a}{b}$

where a and b are integers having no common factor than other than 1.

=>

$\sqrt{5} = \frac{a}{b} - 3$

=>

$\sqrt{5} = \frac{a - 3b}{b}$

since,

$\frac{a - 3b}{b}$

is a rational number ,

$\sqrt{5}$

is a rational number which is a contradiction to fact that " √5 is irrational "

Hence , 3+√5 is irrational.

Answer 2

Answer:

Answer

if a possible, let us assume that

3 + \sqrt{5}3+

5

be a rational number so, there exists positive integers a and b such that ,

3 + \sqrt{5} = \frac{a}{b}3+

5

=

b

a

where a and b are integers having no common factor than other than 1.

=>

\sqrt{5} = \frac{a}{b} - 3

5

=

b

a

−3

=>

\sqrt{5} = \frac{a - 3b}{b}

5

=

b

a−3b

since,

\frac{a - 3b}{b}

b

a−3b

is a rational number ,

\sqrt{5}

5

is a rational number which is a contradiction to fact that " √5 is irrational "

Hence , 3+√5 is irrational.