Question

How to prove that root 5 is an irrational number on a number line. ( NO LINKS).

Answer 1

Dear student
It is not possible to prove that $\sqrt{5}$ is an irrational number on number line.
$$\begin{gathered} \mathrm{But}\mathrm{we}\mathrm{bcan}\mathrm{prove}\mathrm{it}\mathrm{to}\mathrm{be}\mathrm{an}\mathrm{irrational}\mathrm{number}\mathrm{by}\mathrm{the}\mathrm{following}\mathrm{method}\mathrm{which} \\ \mathrm{is}\mathrm{as}\mathrm{follows}: \\ \mathrm{Let}\mathrm{us}\mathrm{assume}\mathrm{on}\mathrm{the}\mathrm{contrary}\mathrm{that}\sqrt{5}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}.\mathrm{Then},\mathrm{there}\mathrm{exist} \\ \mathrm{co}-\mathrm{prime}\mathrm{positive}\mathrm{integers}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{such}\mathrm{that} \\ \sqrt{5}=\frac{\mathrm{a}}{\mathrm{b}} \\ \Rightarrow 5{\mathrm{b}}^2={\mathrm{a}}^2 \\ \Rightarrow 5|{\mathrm{a}}^2\left[\because 5|5{\mathrm{b}}^2\right] \\ \Rightarrow 5|\mathrm{a} \\ \Rightarrow \mathrm{a}=5\mathrm{c}\mathrm{for}\mathrm{some}\mathrm{positive}\mathrm{integer}\mathrm{c}....\left(1\right) \\ \Rightarrow {\mathrm{a}}^2=25{\mathrm{c}}^2 \\ \Rightarrow 5{\mathrm{b}}^2=25{\mathrm{c}}^2\left[\because {\mathrm{a}}^2=5{\mathrm{b}}^2\right] \\ \Rightarrow {\mathrm{b}}^2=5{\mathrm{c}}^2 \\ \Rightarrow 5|{\mathrm{b}}^2\left[\because 5|5{\mathrm{c}}^2\right] \\ \Rightarrow 5|\mathrm{b}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{find}\mathrm{that}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{have}\mathrm{atleast}5\mathrm{as}\mathrm{a}\mathrm{common}\mathrm{factor}.\mathrm{This}\mathrm{contradicts} \\ \mathrm{the}\mathrm{fact}\mathrm{that}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{co}-\mathrm{prime}. \\ \mathrm{Hence},\sqrt{5}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$
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