Question

Show that 5✓3 irrational number​

Answer 1

$\huge\mathfrak{\underline{\underline{\red {Question}}}}$

Show that 5✓3 irrational number

$\huge\mathfrak{\underline{\underline{\red {Answer}}}}</p><p>$

$\sf \: Let \: us \: assume \: that \: 5 \sqrt{3} \: is \: rational \: number$

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

here a and b are co-prime nimbers

$\sf\sqrt{5} =[( \frac{a}{b} )−3]$

$\sf\sqrt{5} =[( \frac{a - 3b}{b} )]$

$\sf \: [ \frac{a - 3b}{b} ] is \: a \: rational \: number.</p><p></p><p></p><p>$

But we know that square root of 5 is an irrational number.

$\sf \: [( \frac{a - 3b}{b} )] is \: also \: a \: irrational \: number.</p><p></p><p></p><p>$

$\sf \: 3+ \sqrt{5}  is \: an \: irrational \: number.</p><p></p><p></p><p>$

Hence, proved.

Answer 2

Letusassumethat5

3

isrationalnumber

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

3

=

b

a

here a and b are co-prime nimbers

\sf\sqrt{5} =[( \frac{a}{b} )−3]

5

=[(

b

a

)−3]

\sf\sqrt{5} =[( \frac{a - 3b}{b} )]

5

=[(

b

a−3b

)]

\sf \: [ \frac{a - 3b}{b} ] is \: a \: rational \: number.[

b

a−3b

] isarationalnumber.

But we know that square root of 5 is an irrational number.

\sf \: [( \frac{a - 3b}{b} )] is \: also \: a \: irrational \: number.[(

b

a−3b

)] isalsoairrationalnumber.

\sf \: 3+ \sqrt{5} is \: an \: irrational \: number.3+

5

isanirrationalnumber.