Question Number 3262 by Rasheed Soomro last updated on 08/Dec/15
$$\mathcal{C}{ould}\:\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:\:{be}\:{drawn}\:{on}\:{numbered}\:{line}\:{with}\: \\ $$$${the}\:{help}\:\:{of}\:\:{ruler}\:{and}\:{compass}\:{only}? \\ $$
Commented by prakash jain last updated on 09/Dec/15
$$\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{using}\:\mathrm{a}\:\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}. \\ $$
Commented by Rasheed Soomro last updated on 10/Dec/15
$$\mathcal{C}{an}\:{this}\:{be}\:{proved}? \\ $$
Commented by 123456 last updated on 10/Dec/15
$$\mathrm{transcendental}\:\mathrm{number}? \\ $$
Commented by prakash jain last updated on 11/Dec/15
$$\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{transcendental}\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$$$\mathrm{of}\:{x}^{\mathrm{3}} −\mathrm{2}=\mathrm{0}. \\ $$$$\mathrm{In}\:\mathrm{2}\:\mathrm{D}\:\mathrm{with}\:\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}\:\mathrm{only}\:\sqrt{\mathrm{2}}\:\mathrm{and} \\ $$$$\mathrm{any}\:\mathrm{expression}\:\mathrm{inolving}\:\mathrm{square}\:\mathrm{roots}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{constructed}. \\ $$