How can I prove if $x>0$ then $-x<0$?












0














$$x>0implies-x<0$$



I thought about using the axioms of multiplication with $xcdot (-1) = -x$

but I am kind of stuck there.






inequality arithmetic







share|cite|improve this question




















  • 9




    How about adding $-x$ to both sides?
    – Lord Shark the Unknown
    Nov 25 at 14:00










  • I feel stupid. But thank you very much!
    – Another Noone
    Nov 25 at 14:02
















0














$$x>0implies-x<0$$



I thought about using the axioms of multiplication with $xcdot (-1) = -x$

but I am kind of stuck there.






inequality arithmetic







share|cite|improve this question




















  • 9




    How about adding $-x$ to both sides?
    – Lord Shark the Unknown
    Nov 25 at 14:00










  • I feel stupid. But thank you very much!
    – Another Noone
    Nov 25 at 14:02














0












0








0







$$x>0implies-x<0$$



I thought about using the axioms of multiplication with $xcdot (-1) = -x$

but I am kind of stuck there.






inequality arithmetic







share|cite|improve this question















$$x>0implies-x<0$$



I thought about using the axioms of multiplication with $xcdot (-1) = -x$

but I am kind of stuck there.






inequality arithmetic




inequality arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 14:07









amWhy

191k28224439




191k28224439










asked Nov 25 at 14:00









Another Noone

32




32








  • 9




    How about adding $-x$ to both sides?
    – Lord Shark the Unknown
    Nov 25 at 14:00










  • I feel stupid. But thank you very much!
    – Another Noone
    Nov 25 at 14:02














  • 9




    How about adding $-x$ to both sides?
    – Lord Shark the Unknown
    Nov 25 at 14:00










  • I feel stupid. But thank you very much!
    – Another Noone
    Nov 25 at 14:02








9




9




How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00




How about adding $-x$ to both sides?
– Lord Shark the Unknown
Nov 25 at 14:00












I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02




I feel stupid. But thank you very much!
– Another Noone
Nov 25 at 14:02










1 Answer
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If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0






share|cite|improve this answer





















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    1 Answer
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    If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0






    share|cite|improve this answer


























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      If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0






      share|cite|improve this answer
























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        If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0






        share|cite|improve this answer












        If x>0 then 0=-x+x>-x+0 so that -x<0. If x<0 then 0=-x+x<-x+0 so that -x>0







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        share|cite|improve this answer










        answered Nov 25 at 14:47









        John Nash

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