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If: 1^x=1^y and as: 1^2=1^99 then: 1=99

  1. Aug 1, 2005 #1
    if:
    1^x=1^y
    and as:
    1^2=1^99
    then:
    1=99
     
  2. jcsd
  3. Aug 1, 2005 #2

    arildno

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    Why do think this is true?

    Since 1*0=99*0, do you consider this as proof of your assertion as well?
     
    Last edited: Aug 1, 2005
  4. Aug 1, 2005 #3
    The log of your equation would beg to differ... EDIT: except you consistently used 1, not a variable. Duh! Mondays are rubbish. Anyway, shouldn't that read 2 = 99?
     
    Last edited: Aug 1, 2005
  5. Aug 1, 2005 #4

    VietDao29

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    I dunno if it's a joke or not.
    If you have a function f(x) such that [itex]f(x_0) \neq f(x_1)[/itex], for all [itex]x_0 \neq x_1[/itex]. Just in that case, you will have [itex]f(x) = f(y) \Leftrightarrow x = y[/itex]
    Some function like :f(x) = 0x, f(x) = 1 ^ x, f(x) = x ^ 0 ([itex]x \in \mathbb{R} - \{ 0 \}[/itex]). You cannot have [itex]f(x) = f(y) \Leftrightarrow x = y[/itex]. Why? Because in that 3 examples:
    [itex]\forall x, f(x) = const[/itex]
    Viet Dao,
     
    Last edited: Aug 1, 2005
  6. Aug 1, 2005 #5

    Alkatran

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    sin(0) = sin(2 pi)

    Oh no! 2 pi = 0, meaning pi = 0, which means circles don't exist!

    If only there was a flaw in the logic...
     
  7. Aug 1, 2005 #6

    Gokul43201

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    LHS = 1*1 = 1
    RHS = 1*1*1* ...(ninety five times) *1 = 1
    LHS = 1 = RHS

    How do you go from line (1) to line (2) ?

    nabodit : If you have a question to ask, ask it now.
     
  8. Aug 1, 2005 #7
    Reading this really brightened up my day. :rofl:
     
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