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Question Number 1752 by Rasheed Ahmad last updated on 14/Sep/15
Prove that: (−x)(−y)=xy
Provethat:(−x)(−y)=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=(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
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
AlternativeProofFirstofallwewillprove:(−x)y=x(−y)=−xyWeknowthatx+(−x)=0(x+(−x))y=0y=0Soxy+(−x)y=0Alsoxy+(−xy)=0∴(−x)y=−xySimilarlyx(−y)=−xyHence(−x)y=x(−y)=−xy(−x)(−y)+(−xy)=(−x)(−y)+(−x)y=(−x){−y+y}=(−x)(0)=0So(−x)(−y)+(−xy)=0Alsoxy+(−xy)=0∴(−x)(−y)=xyRasheedSoomro
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
idontknowifitallrightbutisatry:Dwehave−x=−1×x=(−1)×x−y=−1×y=(−1)×y(−x)(−y)=(−1×x)(−1×y)bycomugative(−x)(−y)=(−1)×(−1)xywehavebydistributivethat(−1)×(−1)+1×(−1)=(−1+1)×(−1)(−1)×(−1)+1×(−1)=0(−1)×(−1)=−−1=1so(−x)(−y)=(−1)×(−1)xy=xy

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