Question

Prove that $\left(4-\sqrt{3}\right)$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Answer 1

Let us assume that $\left(4-\sqrt{3}\right)$ is a rational number.

Thus, $\left(4-\sqrt{3}\right)$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} 4-\sqrt{3}=\frac{p}{q} \\ \Rightarrow -\sqrt{3}=\frac{p}{q}-4 \\ \Rightarrow -\sqrt{3}=\frac{p-4q}{q} \\ \Rightarrow \sqrt{3}=\frac{4q-p}{q} \\ \mathrm{Since},\frac{4q-p}{q}\mathrm{is}\mathrm{rational}\Rightarrow \sqrt{3}\mathrm{is}\mathrm{rational} \\ \mathrm{But},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\sqrt{3}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{our}\mathrm{assumption}\mathrm{is}\mathrm{wrong}. \\ \mathrm{Hence},4-\sqrt{3}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$