Let's break this down step by step:

The given expression is:

$\frac{2^{-1} \times 4^{-1}}{2^{-2}}$

Step 1: Simplify the exponents

• $2^{-1} = \frac{1}{2}$
• $4^{-1} = \frac{1}{4}$, and since $4 = 2^2$, we can write $4^{-1} = (2^2)^{-1} = 2^{-2}$.
• $2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

So the expression now looks like:

$\frac{\left( \frac{1}{2} \times \frac{1}{4} \right)}{2^{-2}}$

Step 2: Multiply the terms in the numerator

$\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

Thus, the expression becomes:

$\frac{\frac{1}{8}}{2^{-2}} = \frac{\frac{1}{8}}{\frac{1}{4}}$

Step 3: Divide fractions

To divide fractions, multiply the numerator by the reciprocal of the denominator:

$\frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2}$

Final Answer:

$\boxed{\frac{1}{2}}$

Do you want any further details or have any questions?

Here are 5 related questions:

1. How do negative exponents affect the value of a number?
2. What is the rule for dividing exponents with the same base?
3. How can you convert powers of larger numbers like $4$ into powers of $2$?
4. How do you multiply and divide fractions in algebraic expressions?
5. What is the reciprocal of a fraction and when is it used?

Tip: When working with negative exponents, remember that $a^{-n} = \frac{1}{a^n}$, which can help simplify complex expressions.