Question

prove that 3-3√7 is an irrational number.​

Answer 1

Answer:

Step-by-step explanation:

Let it be assumed that 3-3√7 is rational.

Then 3(1-√7) = a, where a is rational.

Squaring both sides,

9(1+7-2√7) = a²

=> 8-2√7 = a²/9

=> -2√7 = (a²-72)/9

=> √7 = -(a²-72)/9

Here, in LHS the number is irrational whereas in RHS the number is rational, which contradicts our assumption. Hence, 3-3√7 is irrational.[proved]

Answer 2

Answer:

Step-by-step explanation:

Let it be assumed that 3-3√7 is rational.

Then 3(1-√7) = a, where a is rational.

Squaring both sides,

9(1+7-2√7) = a²

=> 8-2√7 = a²/9

=> -2√7 = (a²-72)/9

=> √7 = -(a²-72)/9

Here, in LHS the number is irrational whereas in RHS the number is rational, which contradicts our assumption. Hence, 3-3√7 is irrational.[proved]