Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 1752 by Rasheed Ahmad last updated on 14/Sep/15

Prove that: (−x)(−y)=xy

$${Prove}\:{that}: \\ $$$$\left(−{x}\right)\left(−{y}\right)={xy} \\ $$

Commented by 123456 last updated on 15/Sep/15

ax+bx=(a+b)x −x=−1×x −y=−1×y (−x)(−y)=(−1×x)(−1×y) =(−1)×(−1)xy=xy (−1)×(−1)+1×(−1)=(−1+1)(−1)=0 (−1)×(−1)=−1×(−1)=−−1=1

$${ax}+{bx}=\left({a}+{b}\right){x} \\ $$$$−{x}=−\mathrm{1}×{x} \\ $$$$−{y}=−\mathrm{1}×{y} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}×{x}\right)\left(−\mathrm{1}×{y}\right) \\ $$$$=\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right){xy}={xy} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)=\left(−\mathrm{1}+\mathrm{1}\right)\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)=−\mathrm{1}×\left(−\mathrm{1}\right)=−−\mathrm{1}=\mathrm{1} \\ $$

Answered by Rasheed Ahmad last updated on 16/Sep/15

Alternative Proof First of all we will prove: (−x)y=x(−y)=−xy We know that x+(−x)=0 (x+(−x))y=0y=0 So xy+(−x)y=0 Also xy+(−xy)=0 ∴ (−x)y=−xy Similarly x(−y)=−xy Hence (−x)y=x(−y)=−xy (−x)(−y)+(−xy) =(−x)(−y)+(−x)y=(−x){−y+y} =(−x)(0)=0 So (−x)(−y)+(−xy)=0 Also xy+(−xy)=0 ∴ (−x)(−y)=xy Rasheed Soomro

$${Alternative}\:{Proof} \\ $$$${First}\:{of}\:{all}\:{we}\:{will}\:{prove}: \\ $$$$\left(−{x}\right){y}={x}\left(−{y}\right)=−{xy} \\ $$$${We}\:{know}\:{that}\:\:{x}+\left(−{x}\right)=\mathrm{0} \\ $$$$\left({x}+\left(−{x}\right)\right){y}=\mathrm{0}{y}=\mathrm{0} \\ $$$${So}\:\:\:\:\:\:\:\:\:{xy}+\left(−{x}\right){y}=\mathrm{0} \\ $$$${Also}\:\:\:\:{xy}+\left(−{xy}\right)=\mathrm{0} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(−{x}\right){y}=−{xy} \\ $$$${Similarly}\:\:{x}\left(−{y}\right)=−{xy} \\ $$$${Hence}\:\:\:\left(−{x}\right){y}={x}\left(−{y}\right)=−{xy} \\ $$$$\:\:\:\left(−{x}\right)\left(−{y}\right)+\left(−{xy}\right) \\ $$$$\:\:\:=\left(−{x}\right)\left(−{y}\right)+\left(−{x}\right){y}=\left(−{x}\right)\left\{−{y}+{y}\right\} \\ $$$$=\left(−{x}\right)\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${So}\:\:\left(−{x}\right)\left(−{y}\right)+\left(−{xy}\right)=\mathrm{0} \\ $$$${Also}\:\:{xy}+\left(−{xy}\right)=\mathrm{0} \\ $$$$\therefore\:\left(−{x}\right)\left(−{y}\right)={xy} \\ $$$${Rasheed}\:{Soomro} \\ $$

Answered by 123456 last updated on 15/Sep/15

i dont know if it all right but is a try :D we have −x=−1×x=(−1)×x −y=−1×y=(−1)×y (−x)(−y)=(−1×x)(−1×y) by comugative (−x)(−y)=(−1)×(−1)xy we have by distributive that (−1)×(−1)+1×(−1)=(−1+1)×(−1) (−1)×(−1)+1×(−1)=0 (−1)×(−1)=−−1=1 so (−x)(−y)=(−1)×(−1)xy=xy

$$\mathrm{i}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{if}\:\mathrm{it}\:\mathrm{all}\:\mathrm{right}\:\mathrm{but}\:\mathrm{is}\:\mathrm{a}\:\mathrm{try}\::\mathrm{D} \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$−{x}=−\mathrm{1}×{x}=\left(−\mathrm{1}\right)×{x} \\ $$$$−{y}=−\mathrm{1}×{y}=\left(−\mathrm{1}\right)×{y} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}×{x}\right)\left(−\mathrm{1}×{y}\right) \\ $$$$\mathrm{by}\:\mathrm{comugative} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right){xy} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{by}\:\mathrm{distributive}\:\mathrm{that} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)=\left(−\mathrm{1}+\mathrm{1}\right)×\left(−\mathrm{1}\right) \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)+\mathrm{1}×\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right)=−−\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{so} \\ $$$$\left(−{x}\right)\left(−{y}\right)=\left(−\mathrm{1}\right)×\left(−\mathrm{1}\right){xy}={xy} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com