Question

prove that 7-6√3 is irrational​

Answer 1

Answer:

$\mathbb{\pink{answer = 7 - \sqrt[6]{3} }}$ is irrational number

Step-by-step explanation:

LET us assume that 7-6root3 is irrational number,then

$(0 \leqslant b < r)$

So,

$7 - \sqrt[6]{3} = r$

$\sqrt{3} = \frac{ - r - 7}{6}$

since,' r ' is a rational number, so -r-7/6 is also a rational number, but as we know that there is contradiction with LHS which is root 3 .

as we all know that root 3 is a irrational number

Thus root 3 is not a rational number

it means root 3 is irrational number.

Hope it HELPS you

Answer 2

to proof:- 7-6√3

as we know that

√3 which is irrational because√3 is not a perfect square and it is positive also.

so, it is proof that it is irrational