- #1
AxiomOfChoice
- 533
- 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?
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.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.
A square root is a number that, when multiplied by itself, gives the original number. For example, the square root of 25 is 5, because 5 multiplied by itself equals 25.
A square of a square root is the result of multiplying a square root by itself. For example, the square of the square root of 9 is 9, because the square root of 9 is 3 and 3 multiplied by itself is 9.
To find the square of a square root, simply multiply the square root by itself using a calculator or by hand. For example, to find the square of the square root of 16, multiply 4 (the square root of 16) by 4, which equals 16.
No, a square root of a negative number does not have a real value. This is because when a negative number is multiplied by itself, it results in a positive number. Therefore, there is no real number that when multiplied by itself will give a negative number.
Square roots and squares are inverse operations of each other. This means that taking the square root of a number and squaring a number will cancel each other out and give the original number. For example, the square root of 25 is 5, and 5 squared is 25.