Question

prove that 5root3÷11 is irrational

Answer 1

Let us assume that 5√3 ÷ 11 is a rational number.

Hence, there exist two integers 'a' and 'b', where b≠0 and

5√3 ÷ 11 = a ÷ b

5√3 = (a ÷ b) x 11

√3 = (a ÷ b) x (11 ÷ 5)

√3 = 11a ÷ 5b

Here, as 'a' and 'b' are real, rational integers, so 11a ÷ 5b is rational.

But √3 is an irrational number.

The derived condition states that an irrational number is equal to a rational number, which can never be true.

Hence, our whole assumption was wrong.

Hence, 5√3 ÷ 11 is not a rational number, that is,

5√3 ÷ 11 is an irrational number

... Hence Proved!

Answer 2

Assume that $\displaystyle \frac{5 \sqrt{3}}{11}$ is rational. Such that let it be indicated as x, where x is a rational number.

In the equation $\displaystyle x = \frac{5\sqrt{3}}{11}$, both LHS and RHS are rational. Then so will be the final result.

$\displaystyle \Longrightarrow \ \ x = \frac{5\sqrt{3}}{11} \\ \\ \\ \Longrightarrow\ \ x = \frac{5}{11}\sqrt{3} \\ \\ \\ \Longrightarrow\ \ x \times \frac{11}{5}=\sqrt{3} \\ \\ \\ \Longrightarrow\ \ \frac{11x}{5}=\sqrt{3}$

As we discussed earlier, if both sides of $\displaystyle x = \frac{5\sqrt{3}}{11}$ are rational, then so will be $\displaystyle \frac{11x}{5}=\sqrt{3}$. But this creates a contradiction that the LHS here is rational while the RHS is irrational.

Hence proved!!!