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Help with a few questions about spivak calculus

  1. Sep 19, 2011 #1
    Question
    if a < b, then -b < -a

    proof
    if a < b then a-b<0 and b-a>0
    so a-b<0<b-a
    so -b<-a

    Question
    if a<b & c<0, then ab>bc

    proof
    if ac<ab then ab>bc
    then ac<ab>bc
    then ac>bc

    Question
    if a>1 then a^2>a

    proof
    a*a > 1*a
    lemma 1: a*a = a^2
    (a*a)*a^-1 = a^2 * a^-1
    a*(a*a^-1) = a
    a * 1 = a
    a = a
    so a*a>1*a
    a^2>a
     
  2. jcsd
  3. Sep 19, 2011 #2
    in the first proof,

    when you go from [itex]a-b<0[/itex] to [itex]b-a>0[/itex] , you have to do this first

    [tex]a-b<0 \Rightarrow \; -(b-a)<0 \;[/tex]

    but to go from here to [itex]b-a>0[/itex], you have to assume the thing which you set out to prove.

    in the last proof, when you go from [itex]\; a>1\;[/itex] to [itex]a*a > 1*a[/itex], you are
    assuming that [itex]1>0[/itex]. have you proved it before ?
     
  4. Sep 20, 2011 #3
    Thank you for replying and helping me out
    I have not proven 1 greater than 0 before
    but can you give me a pointer on how to prove that or another way of doing this proof
     
  5. Sep 25, 2011 #4
    order axioms say that between two real numbers a and b , three relations are possible.
    a>b or a=b or a<b. this is called property of trichotomy. to prove 1>0, you assume negation, assume [itex]1\ngtr 0 [/itex]. since one of the three possibilities between 1 and 0 is true and we have assumed [itex]1\ngtr 0 [/itex] , it must be true that
    [itex]1=0\;\mbox{or}\;1<0[/itex] . since the field axioms explicitly say that
    [itex]1\neq 0[/itex] , it must be true that [itex]1<0 [/itex] now work with axioms to find a contradiction.........

    are you studying analysis on your own or you are taking class ?
     
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