Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 183075 by Ml last updated on 19/Dec/22

Answered by BaliramKumar last updated on 19/Dec/22

let     n=2k       n^2  = (2k)^2  = 4k^2  = 2(2k^2 ) = 2(m)    even

$${let}\:\:\:\:\:{n}=\mathrm{2}{k} \\ $$$$\:\:\:\:\:{n}^{\mathrm{2}} \:=\:\left(\mathrm{2}{k}\right)^{\mathrm{2}} \:=\:\mathrm{4}{k}^{\mathrm{2}} \:=\:\mathrm{2}\left(\mathrm{2}{k}^{\mathrm{2}} \right)\:=\:\mathrm{2}\left({m}\right)\:\:\:\:{even} \\ $$

Answered by Rasheed.Sindhi last updated on 20/Dec/22

n≡0(mod 2)  (n)^2 ≡(0)^2 (mod 2)  n^2 ≡0 (mod 2)  ∴n∈E⇒ n^2 ∈E

$${n}\equiv\mathrm{0}\left({mod}\:\mathrm{2}\right) \\ $$$$\left({n}\right)^{\mathrm{2}} \equiv\left(\mathrm{0}\right)^{\mathrm{2}} \left({mod}\:\mathrm{2}\right) \\ $$$${n}^{\mathrm{2}} \equiv\mathrm{0}\:\left({mod}\:\mathrm{2}\right) \\ $$$$\therefore{n}\in\mathbb{E}\Rightarrow\:{n}^{\mathrm{2}} \in\mathbb{E} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com