Question

show that 5-√3 is irrational

Answer 1

Let us assume that $5 - \sqrt{3}$ is rational.

Thus, $5 - \sqrt{3}$ = $\frac{a}{b}$, where a and b are co-primes.

$5 - \sqrt{3} = \frac{a}{b}$

- $\sqrt{3} = \frac{a}{b} - 5$

$-\sqrt{3} = \frac{a - 5b}{b} (LCM)$

$\sqrt{3} = - \left[\begin{array}{ccc} \frac{a - 5b}{b}\end{array}\right]$

Here, $\sqrt{3}$ is irrational, while $\frac{a - 5b}{b}$ is rational.

We know that irrational $\neq$ rational

Thus, it is a contradiction.

Due to this contradiction we can say that out assumption is incorrect.

Therefore, $5 - \sqrt{3}$ is irrational

Answer 2

Step-by-step explanation:

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