Question

probe that root 2+ root 7 is an irrational number



plzzz answer my question fast ​

Answer 1

Answer:

Suppose (sqrt(2) + sqrt(7)) is not irrational but rational number and equal to r. Then ,

sqrt(2) + sqrt(7) = r ==> sqrt(2) = r - sqrt(7) ==> 2 = r^(2 ) + 7 - 2r sqrt(7) or

sqrt(7) = (r^(2) + 5)/2r . But r.h.s. here is a rational number while l.h.s. is an irrational number, a contradiction .

see the image will be helpful for

Answer 2

Need to prove:

• $\sqrt{2} + \sqrt{7}$is an irrational number

━━━━━━━━━━━━━━━━

Solution:

Suppose $\sqrt{2} + \sqrt{7}$ is not an irrational number. Let it be a rational number.

Then,

$\sqrt{2} + \sqrt{7} = \dfrac{p}{q}$

where p and q has no common factor other than 1 and q isn't equal to 0.

$\implies \sqrt{2} = \dfrac{p}{q} - \sqrt{7}$

$\implies {( \sqrt{2} )}^{2} = {( \dfrac{p}{q} - \sqrt{7} ) }^{2}$ (squaring both sides)

$\implies 2 = {( \dfrac{ {p}^{2} }{ {q}^{2} }) } - 2 \times \dfrac{p}{q} \times \sqrt{7} + 7$

$\implies \dfrac{ {p}^{2} }{ {q}^{2} } - \dfrac{p \times 2\sqrt{7} }{q} = 2 - 7$

$\implies 2 \sqrt{7} \times \dfrac{p}{q} = 5 - \dfrac{ {p}^{2} }{ {q}^{2} }$

$\implies \sqrt{7} \dfrac{p}{q} = \dfrac{5 {q}^{2} - {p}^{2} }{2q}$

$\implies \sqrt{7} = \dfrac{5 {q}^{2} - {p}^{2} }{p}$

━━━━━━━━━━━━━━━━

Here

$\dfrac{5 {q}^{2} - {p}^{2} }{p}$is rational as p, and q has no common factor other than 1

But we know, $\sqrt{7}$ is irrational.

As LHS is not equal to RHS, we can say,

Our assumption is wrong!

━━━━━━━━━━━━━━━━

Hence $\sqrt{2} + \sqrt{7}$ is an irrational number