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    If you know that for all real numbers a, b that a > 0 and b < 0, then can you say that

    |a - b| = a - b always?

    (context is proving convergence of sequences, i.e. a = x_n and b = L)

    While I'm here, anyone know where I can find some worked examples of proving sequences converge using epsilon and n>N? My book doesn't cover this subject (so yes, I should get a different book, but just wondering).
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    Suppose x>0. Is |x|=x?
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    Yes, but I just feel kind of dirty doing it when there are two things in the modulus (and yes, I know, I could define c = a - b so that then there'd be just one thing :p:).

    Do you know any good resources?

    *Edit*

    Actually no, I don't. I wasn't thinking of it as a > b, so a - b > 0, so |a - b| = a - b, I was just thinking like "Well 3 - (-5) = 8, so yeah |a - b| must be a - b". Thanks.
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    (Original post by Swayum)
    Yes, but I just feel kind of dirty doing it when there are two things in the modulus (and yes, I know, I could define c = a - b so that then there'd be just one thing :p:).

    Do you know any good resources?

    *Edit*

    Actually no, I don't. I wasn't thinking of it as a > b, so a - b > 0, so |a - b| = a - b, I was just thinking like "Well 3 - (-5) = 8, so yeah |a - b| must be a - b". Thanks.
    Try Spivak's Calculus, it's bwilliant. A nice question:
    Prove that
    \displaystyle \lim_{n \to \infty} \sqrt{n+1} - \sqrt{n} = 0
 
 
 
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