2004年9月22日 星期三

證明e是無理數的方法




  • [mathml]text(first)[/mathml]

  • [mathml]e = sum_(n=0)^oo 1/(n!) [/mathml]

  • [mathml]= 1 + 1 + 1/(2!) + 1/(3!)+... [/mathml]

  • [mathml]> 2[/mathml]

  • [mathml]text(and)[/mathml]

  • [mathml]e=1 + 1 + 1/(2!) + 1/(3!)+...[/mathml]

  • [mathml]
  • [mathml]= 1 + 1 + (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ....[/mathml]

  • [mathml]= 3[/mathml]

  • [mathml]text(implies) 2
  • [mathml]if e in RR, text(then) EE p, q in ZZ, text(s.t.) e=q/p, text(we claim we will find the conflict) [/mathml]

  • [mathml]text(because)[/mathml]

  • [mathml] 2 1[/mathml]

  • [mathml]text(so)[/mathml]

  • [mathml] p! * e=p! * (1+1+1/(2!)+1/(3!)+ ...)[/mathml]

  • [mathml] =p! * (1 + 1 + 1/(2!)+1/(3!) + ... + 1/(p!) + 1/(p+1)! + ...)[/mathml]

  • [mathml] = Integer + 1/(p+1)! +... [/mathml]

  • [mathml] = Integer + 1/(p+1)! + 1/(p+2)! + 1/(p+3)! + ...[/mathml]

  • [mathml]text(because p!e is a Integer, we can find the conflict from)[/mathml]

  • [mathml]1/(p+1)! + 1/(p+2)! + 1/(p+3)! + ...
  • [mathml]text(so)[/mathml]

  • [mathml]1/(p+1)! + 1/(p+2)! + 1/(p+3)! + ...[/mathml]

  • [mathml]
  • [mathml]
  • [mathml]= 1/{p}[/mathml]

  • [mathml]

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