Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-10: Express the number in the form \(a / b\), where \(a\) and \(b\) are integers. $$ (-0.008)^{2 / 3} $$

Step-by-Step Solution

Answer

The expression \((-0.008)^{2/3}\) is equal to \(\frac{1}{25}\).

Step 1: Understand the Expression

We need to express \((-0.008)^{2/3}\) in the form \(\frac{a}{b}\). This involves raising a number to a fractional power.

Step 2: Analyze the Base Number

The base in the expression is \(-0.008\). Convert \(-0.008\) to a fraction. This gives us \(-\frac{8}{1000}\), which can be simplified to \(-\frac{1}{125}\).

Step 3: Simplify the Exponentiation

The exponent \(\frac{2}{3}\) means we raise the fraction to the power of 2 and then take the cube root. So, we will calculate: \(\left(-\frac{1}{125}\right)^{2} = \frac{1}{15625}\).The cube root of \(\frac{1}{15625}\) is \(\frac{1}{25}\).

Step 4: Conclusion

After performing the calculations, we have the result \((-0.008)^{2/3} = \frac{1}{25}\). This is the simplified form of the expression where both \(a\) and \(b\) are integers.

Key Concepts

Negative Base

Working with negative bases in exponentiation can sometimes be tricky, especially when dealing with fractional exponents. A negative base refers to a number that is less than zero involved in a mathematical operation like raising to a power. When raising a negative number to a fractional exponent:

• If the denominator of the fraction is even, the result might involve complex numbers unless the negative sign is balanced out by the exponentiation.
• If the denominator is odd, the base retains its negative sign when raised to the power.

In our example of (-0.008)^{2/3}, the cube root (denominator is 3) ensures that the negative sign of -0.008 will result in a positive number since the expression requires squaring first, which is an even power, thus removing the negative sign before taking the cube root.

Fraction Simplification

Simplifying fractions is a crucial step when dealing with expressions like (-0.008)^{2/3}. Initially, we have the decimal -0.008, which we convert to a fraction. This is -rac{8}{1000}. Simplification involves finding the greatest common divisor (GCD) of the numerator and the denominator and dividing both by this number.

• Start by identifying common factors in both numbers. For 8 and 1000, the GCD is 8.
• Divide both the numerator and denominator by this GCD: -rac{8 ÷ 8}{1000 ÷ 8} = -rac{1}{125}.

By simplifying the fraction, -rac{1}{125} becomes easier to use in subsequent calculations, such as further exponentiation or root-taking.

Cube Root

The cube root is an inverse operation of cubing a number. It is a way to identify what number multiplied by itself three times yields the original number. In mathematical terms, the cube root of a number x is x^{1/3}. For fractional exponents like 2/3, the cube root acts as a part of the operation. Consider the expression (-rac{1}{125})^{2/3}. Performing the operations:

• First, raise the simplified fraction -rac{1}{125} to the power of 2: ( -rac{1}{125})^{2} = rac{1}{15625}. Notice how squaring a negative number results in a positive number.
• Next, take the cube root: the cube root of rac{1}{15625} is rac{1}{25}, as 25^{3} = 15625.

The cube root simplifies the exponentiation process by reducing a larger multiplication back down to a smaller base.

Exponentiation

Exponentiation refers to the process of raising a base number to a given power. For example, the expression a^b denotes the number a multiplied by itself b times. In the case of fractional exponents, this involves two sequential operations: a power and a root. When dealing with (-0.008)^{2/3}:

• The exponent is 2/3, which breaks down into a two-step operation: raising to a power and extracting a root.
• First, the base number is squared: (-rac{1}{125})^{2} = rac{1}{15625}. Squaring a negative fraction removes the negative sign.
• Then, the result is cube rooted, simplifying the expression to a smaller fraction, rac{1}{25}.

Understanding exponentiation with fractions is crucial for grasping the dual operations involved, especially when handling a negative base.