Question

How to prove that root 3 is an irrational number on a number line. ( NO LINKS).

Answer 1

Dear student
It is not possible to prove that $\sqrt{3}$ is an irrational number on number line.
But we can prove it to be an irrational number by the following method which is as follows:

$$\begin{gathered} \mathrm{Let}\mathrm{us}\mathrm{assume}\mathrm{on}\mathrm{the}\mathrm{contrary}\mathrm{that}\sqrt{3}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}. \\ \mathrm{Then},\mathrm{there}\mathrm{exist}\mathrm{co}-\mathrm{prime}\mathrm{positive}\mathrm{integers}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{such}\mathrm{that} \\ \sqrt{3}=\frac{\mathrm{a}}{\mathrm{b}},\mathrm{where}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{co}\mathrm{prime}\mathrm{i}.\mathrm{e}.\mathrm{their}\mathrm{HCF}\mathrm{is}1. \\ \mathrm{Now},\sqrt{3}=\frac{\mathrm{a}}{\mathrm{b}} \\ \Rightarrow 3=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2} \\ \Rightarrow 3{\mathrm{b}}^2={\mathrm{a}}^2 \\ \Rightarrow 3|{\mathrm{a}}^2\left[\mathrm{As}3|3{\mathrm{b}}^2\right] \\ \Rightarrow 3|\mathrm{a}...\left(1\right) \\ \Rightarrow \mathrm{a}=3\mathrm{c}\mathrm{for}\mathrm{some}\mathrm{integer}\mathrm{c} \\ \Rightarrow {\mathrm{a}}^2=9{\mathrm{c}}^2 \\ \Rightarrow 3{\mathrm{b}}^2=9{\mathrm{c}}^2\left[\mathrm{As}3{\mathrm{b}}^2={\mathrm{a}}^2\right] \\ \Rightarrow {\mathrm{b}}^2=3{\mathrm{c}}^2 \\ \Rightarrow 3|{\mathrm{b}}^2\left[\mathrm{As}\left[3|3{\mathrm{c}}^2\right]\right] \\ \Rightarrow 3|\mathrm{b}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{find}\mathrm{that}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{have}\mathrm{atleast}3\mathrm{as}\mathrm{a}\mathrm{common}\mathrm{factor}.\mathrm{This}\mathrm{contradicts} \\ \mathrm{the}\mathrm{fact}\mathrm{that}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{co}-\mathrm{prime}. \\ \mathrm{Hence},\sqrt{3}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$
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