Question

Prove that 2+3√5 is irrational​

Answer 1

Step-by-step explanation:

let , 2+3√5 be a rational number i.e. p/q

$2 + 3 \sqrt{5} = \frac{x}{y} \\ = > \frac{x}{y} - 3 \sqrt{5} = 2 \\ squaring \: both \: side \\ ({\frac{x}{y} - 3 \sqrt{5} })^{2} = {2}^{2} \\ ( { \frac{x}{y} })^{2} + 45 - 6 \sqrt{5} \frac{x}{y} = 4 \\ = > { (\frac{x}{y} )}^{2} + 45 - 4 = 6 \sqrt{5} \frac{x}{y} \\ = > ({ (\frac{x}{y} )}^{2} + 41) \frac{y}{6x} = \sqrt{5} \\ w e\: know \: \sqrt{5} is \: an \: irrational \: no.$

=> 2+3√5 is an irrational no.

Answer 2

let , 2+3√5 be a rational number i.e. p/q

\begin{gathered}2 + 3 \sqrt{5} = \frac{x}{y} \\ = > \frac{x}{y} - 3 \sqrt{5} = 2 \\ squaring \: both \: side \\ ({\frac{x}{y} - 3 \sqrt{5} })^{2} = {2}^{2} \\ ( { \frac{x}{y} })^{2} + 45 - 6 \sqrt{5} \frac{x}{y} = 4 \\ = > { (\frac{x}{y} )}^{2} + 45 - 4 = 6 \sqrt{5} \frac{x}{y} \\ = > ({ (\frac{x}{y} )}^{2} + 41) \frac{y}{6x} = \sqrt{5} \\ w e\: know \: \sqrt{5} is \: an \: irrational \: no.\end{gathered}

2+3

5

=

y

x

=>

y

x

−3

5

=2

squaring both side

(

y

x

−3

5

)

2

=2

2

(

y

x

)

2

+45−6

5

y

x

=4

=>(

y

x

)

2

+45−4=6

5

y

x

=>((

y

x

)

2

+41)

6x

y

=

5

we know

5

is an irrational no.

=> 2+3√5 is an irrational no.

Hope it helps uu..