To solve for b in the equation b^2 = 64/9:
b^2 = 64/9
b = ±√(64/9)
b = ±(8/3)
So the solutions for b are b = 8/3 and b = -8/3.
Regarding the second part of the question:
1. If the square of a number is less than 1, it does not necessarily follow that the number must be less than 1. For example, if we take the number b = 1/2, then b^2 = (1/2)^2 = 1/4, which is indeed less than 1 but the number b = 1/2 is greater than 1.
2. If the square of a positive number is greater than 1, it does not necessarily follow that the number must be greater than 1. For example, if we take the number b = 2, then b^2 = 2^2 = 4, which is greater than 1 but the number b = 2 is in fact less than 1.
Therefore, the square of a number being less than 1 or greater than 1 does not dictate whether the number itself is less than or greater than 1.
Solve the equation. Use pencil and paper. If the square of a number is less than 1, does it follow that the number must be less than 1? If the square of a positive number is greater than 1, does it follow that the number must be greater than 1? Include examples to explain your reasoning.
b^2 = 64/9
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