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Pretty easy question about squares of square roots

  1. Sep 13, 2009 #1
    If you know [itex]\sqrt{(a^2+b^2)} < \epsilon[/itex], do you know [itex]a < \epsilon[/itex] and [itex]b < \epsilon[/itex]? If so, how?
  2. jcsd
  3. Sep 13, 2009 #2
    The left hand side of the first inequality is non-negative (I assume e > 0), so square both sides to get a^2 + b^2 < e^2. Now a^2 < e^2 - b^2 [itex]\leq[/itex] e^2 and if a > 0, take square roots. If a < 0, then it's obvious. Same for b.
  4. Sep 13, 2009 #3
    Thanks. That's pitifully easy. I don't know why, but I have trouble taking square roots when inequalities are involved. I guess I start thinking about how, if you're working in the interval [itex][0,1][/itex], the square root of a number is bigger than the number itself, which gets confusing.
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