Question Number 1752 by Rasheed Ahmad last updated on 14/Sep/15
$${Prove}\:{that}: \\ $$$$\left(−{x}\right)\left(−{y}\right)={xy} \\ $$
Commented by 123456 last updated on 15/Sep/15
$${ax}+{bx}=\left({a}+{b}\right){x} \\ $$$$−{x}=−\mathrm{1}×{x} \\ $$$$−{y}=−\mathrm{1}×{y} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}×{x}\right)\left(−\mathrm{1}×{y}\right) \\ $$$$=\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right){xy}={xy} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)=\left(−\mathrm{1}+\mathrm{1}\right)\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)=−\mathrm{1}×\left(−\mathrm{1}\right)=−−\mathrm{1}=\mathrm{1} \\ $$
Answered by Rasheed Ahmad last updated on 16/Sep/15
$${Alternative}\:{Proof} \\ $$$${First}\:{of}\:{all}\:{we}\:{will}\:{prove}: \\ $$$$\left(−{x}\right){y}={x}\left(−{y}\right)=−{xy} \\ $$$${We}\:{know}\:{that}\:\:{x}+\left(−{x}\right)=\mathrm{0} \\ $$$$\left({x}+\left(−{x}\right)\right){y}=\mathrm{0}{y}=\mathrm{0} \\ $$$${So}\:\:\:\:\:\:\:\:\:{xy}+\left(−{x}\right){y}=\mathrm{0} \\ $$$${Also}\:\:\:\:{xy}+\left(−{xy}\right)=\mathrm{0} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−{x}\right){y}=−{xy} \\ $$$${Similarly}\:\:{x}\left(−{y}\right)=−{xy} \\ $$$${Hence}\:\:\:\left(−{x}\right){y}={x}\left(−{y}\right)=−{xy} \\ $$$$\:\:\:\left(−{x}\right)\left(−{y}\right)+\left(−{xy}\right) \\ $$$$\:\:\:=\left(−{x}\right)\left(−{y}\right)+\left(−{x}\right){y}=\left(−{x}\right)\left\{−{y}+{y}\right\} \\ $$$$=\left(−{x}\right)\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${So}\:\:\left(−{x}\right)\left(−{y}\right)+\left(−{xy}\right)=\mathrm{0} \\ $$$${Also}\:\:{xy}+\left(−{xy}\right)=\mathrm{0} \\ $$$$\therefore\:\left(−{x}\right)\left(−{y}\right)={xy} \\ $$$${Rasheed}\:{Soomro} \\ $$
Answered by 123456 last updated on 15/Sep/15
$$\mathrm{i}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{if}\:\mathrm{it}\:\mathrm{all}\:\mathrm{right}\:\mathrm{but}\:\mathrm{is}\:\mathrm{a}\:\mathrm{try}\::\mathrm{D} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$−{x}=−\mathrm{1}×{x}=\left(−\mathrm{1}\right)×{x} \\ $$$$−{y}=−\mathrm{1}×{y}=\left(−\mathrm{1}\right)×{y} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}×{x}\right)\left(−\mathrm{1}×{y}\right) \\ $$$$\mathrm{by}\:\mathrm{comugative} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right){xy} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{by}\:\mathrm{distributive}\:\mathrm{that} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)=\left(−\mathrm{1}+\mathrm{1}\right)×\left(−\mathrm{1}\right) \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)=−−\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{so} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right){xy}={xy} \\ $$