Question

prove that 7+ 3√2 is not a rational number.

Answer 1

Answer:

Step-by-step explanation :

let 7 + 3√2 be an rational number where

7+3√2 = a/b [ a and b are coprime and b is not equal to zero]

3√2= a/b-7

3√2 =( a-7b) /b

√2 = (a-7b) /3b .....(i)

Now ,from equation (i) ,we get that √2 is rational but we know that √2 is irrational. so actually 7 + 3√2 is irrational not rational. thus our assumption is wrong. The number is irrational.

hope it helped u....

Answer 2

SOLUTION :

Let us assume that ( 7 + 3√2 ) is a rational number

so,

$\bf 7+ 3\sqrt{2} = \frac{p}{q} \:\:\:\:\:[in\:\: which\:\:'p'\:and \:'q'\:\:are\:\:integers \:\:and\:\: q\neq 0]$

$\bf \rightarrow 3\sqrt{2}=\frac{p}{q}-7$

$\bf \rightarrow 3\sqrt{2}=\frac{p-7q}{q}$

$\bf \rightarrow \sqrt{2}=\frac{p-7q}{q} \times \frac{1}{3}$

$\bf \rightarrow \sqrt{2}=\frac{p-7q}{3q}$

Since,

(p), (q), (7), (3) are integers and q ≠ 0 so,

$\bf \frac{p-7q}{3q}$ is a rational number

Therefore, √2 is also a rational number which is not possible

√2 is an irrational number

∴ This contradiction arise due to our wrong assumption.

Hence,

7+3√2 cannot be a rational number,

∵ it is an irrational number.

Hence Proved