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  1. J

    Solve 2^n+3=m^m

    If you are right that "m is of the form 4x+3 for n>1", then the only solution is the trivial solution.
  2. J

    Why is -1*-1=1

    dimitri151, why couldn't -1*-1 be some number other than -1 or 1? Why not 0 or 2?
  3. J

    Why is -1*-1=1

    Ah, looking at Mark44's post, there is a nicer way. First we prove 0*0 = 0. This is true because 0*0 = 0*(0+0) = 0*0 + 0*0. By uniqueness of the additive identity, 0*0 = 0. Also, 1 + -1 = 0, by definition. Thus (1 + -1)*(1 + -1) = 0. Distribution gives us (1*1 + -1*1) + (1*-1 + -1*-1) = 0...
  4. J

    Why is -1*-1=1

    If you are doing an axiomatic treatment of the integers, you define 1 to be the multiplicative identity. -1 is defined to be the additive inverse of 1 (so 1 + -1 = 0). -1*-1 = -1*(-1 + 0) [Use that 0 is the additive identity] -1*(-1 + 0) = -1*(-1 + (1 + -1)) [Use that 0 = 1 + -1] -1*(-1 + (1 +...
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