Pretty easy question about squares of square roots

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The discussion centers on the mathematical relationship between the square root of the sum of squares and individual components under the condition that \(\sqrt{(a^2+b^2)} < \epsilon\). It is established that if this inequality holds, then both \(a < \epsilon\) and \(b < \epsilon\) can be derived by squaring the inequality and analyzing the resulting expressions. The participants emphasize the importance of understanding the behavior of square roots in the context of inequalities, particularly in the interval [0,1]. The discussion highlights common misconceptions when dealing with square roots and inequalities.

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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?
 
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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.
 
snipez90 said:
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.
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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