Pretty easy question about squares of square roots

[AxiomOfChoice]

If you know \sqrt{(a^2+b^2)} < \epsilon, do you know a < \epsilon and b < \epsilon? If so, how?

 

[snipez90]

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 \leq e^2 and if a > 0, take square roots. If a < 0, then it's obvious. Same for b.

 

[AxiomOfChoice]

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 \leq 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 [0,1], the square root of a number is bigger than the number itself, which gets confusing.