anonymous
  • anonymous
Prove: if x>=0 and x<=epsilon for all epsilon>0, then x=0.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@wio would you be able to help with this?
anonymous
  • anonymous
Hmmm, well what are you supposed to use to do this proof?
anonymous
  • anonymous
Looks similar to squeeze theorem.

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anonymous
  • anonymous
im not sure what the squeeze theorem is. But I thought maybe you would just use contradiction and assume that either x>0 and x<0, and do both case so then by contradiction x>=0 and x<=0 so that means x=0. but Im really not sure.
anonymous
  • anonymous
Hmm
anonymous
  • anonymous
How about something like this... assume x>0. and we must show epsilson>0 such x>epsilon. Let epsilon=x/2. Since x>0, epsilson>0. then epsilson = x/2 = x/2< (x+x)/2 = x. so by contradiction x<=0. or something like that.
anonymous
  • anonymous
are you still need help?
anonymous
  • anonymous
yeah, i really dont know what I am doing for if I am even on the right track
anonymous
  • anonymous
ok,maybe i can help you;)
ybarrap
  • ybarrap
Seems like we need the completeness axiom
anonymous
  • anonymous
A = {x|x >= 0} B = {x|x <= (epsilon for all epsilon>0)} \[A \lceil \rceil B = {0}\] so, x = 0
anonymous
  • anonymous
ok?
anonymous
  • anonymous
are A and B like two different cases or are A and B like just different sets.
anonymous
  • anonymous
I can prove it in this way too: A = {x|x >= 0 , x <= (epsilon for all epsilon>0)} A = {0} so, x = 0
anonymous
  • anonymous
But if you are given that x>=0 and e>0. then all you are saying is that A=0 so x must equal 0. that is all you have to say??
anonymous
  • anonymous
yes ;)
anonymous
  • anonymous
easy enough. thank you.
anonymous
  • anonymous
your welcome :)
zzr0ck3r
  • zzr0ck3r
density of rational/irrationals..
zzr0ck3r
  • zzr0ck3r
assume \[x\ge0,x\le\epsilon\text{ for all }\epsilon>0\\\text{assume by contradiction }x\ne0\\\text{by density we have that there exists }\epsilon \le x\\a \space contradiction...\]

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