Question

prove that root 6 + root 5 is irrational number​

Answer 1

Answer:

This question can be solved in two ways.

1)Let us assume, to the contrary, that √5+√6 is irrational.

So we can find two integers numbers a and b(≠0), in the following way,

√5+√6 = a/b

Rearranging,

√5 = a/b - √6

= integer/integer - rational ...( as √6 is assume as rational)

So, this means that √5 is rational.

But this contradict the fact that √5 is irrational.

Our assumption is wrong.

Hence, √5 + √6 is irrational.

2) Let us assume, to the contrary, that √5+√6 is irrational.

So we can find two integers numbers a and b(≠0), in the following way,

√5+√6 = a/b

Rearranging,

√6 = a/b - √5

= integer/integer - rational ...( as √5 is assume as rational)

So, this means that √6 is rational.

But this contradict the fact that √6 is irrational.

Our assumption is wrong.

Hence, √5 + √6 is irrational.

Hope it may help you!!!!

Answer 2

First prove that root 6 is irrational.You will have it in your textbook.

let root 6+root5=a rational number,r

$\sqrt{5} ^{2} =(r-\sqrt(6})^{2} \\5^{2} =r^{2} +6 -2\sqrt6} \\25=r^{2} +6-2\sqrt{6} \\19-r^{2} =-2\sqrt{6}$

now since 19 is rational

⇒19-$r^{2}$ is rational

⇒19-r2/-2 is also rational

⇒which implies that √6 is rational

Butit has already been proven at the start that √6 is irrational

Hence our assumption that √6+√5 is a rational is wrong

therefore √6+√5 is irrational