Q. \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \)

Answer

\( \left(\frac{1}{2}\right)^6 = \frac{1^6}{2^6} = \frac{1}{64} \)

Detailed Explanation

Compute the product \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \).

  1. Recognize the repeated multiplication of the same factor. Six factors of \( \frac{1}{2} \) can be written with an exponent:

    \[ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \left(\frac{1}{2}\right)^6 \]

    Reason: multiplying the same number by itself \( n \) times equals that number to the \( n \)-th power.

  2. Use the power-of-a-fraction rule: for integers \( a \), \( b \) and positive integer \( n \), \( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \). Apply it here:

    \[ \left(\frac{1}{2}\right)^6 = \frac{1^6}{2^6} \]

  3. Compute numerator and denominator separately:

    • Numerator: \( 1^6 = 1 \) because any power of 1 is 1.

    • Denominator: compute powers of 2 step by step:
      \( 2^2 = 4 \),
      \( 2^3 = 8 \),
      \( 2^4 = 16 \),
      \( 2^5 = 32 \),
      \( 2^6 = 64 \).

  4. Combine the results:

    \[ \left(\frac{1}{2}\right)^6 = \frac{1}{64} \]

  5. Optional: express as a decimal if desired: \( \frac{1}{64} = 0.015625 \).

\( \frac{1}{64} \).

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FAQs

What is the value of \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}?

The product is \left(\frac{1}{2}\right)^6 = \frac{1}{64}.

How can I compute this quickly without multiplying six fractions one by one?

Use exponents: \left(\frac{1}{2}\right)^6. Compute numerator and denominator powers: 1^6 / 2^6 = 1 / 64.

What is the decimal form of \frac{1}{64}?

\frac{1}{64} = 0.015625.

How does this relate to negative exponents or reciprocals?

\left(\frac{1}{2}\right)^6 = 2^{-6} and its reciprocal is 64, since 2^6 = 64.

Can this be interpreted probabilistically?

Yes. Six independent fair coin flips all landing heads have probability \left(\frac{1}{2}\right)^6 = \frac{1}{64}.

How would you generalize multiplying the same fraction n times?

\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}. For a=\!1,\ b=\!2,\ n=\!6 you get \frac{1}{2^6}.

Does the order of multiplication matter here?

No. Multiplication is commutative, so any order of the six \frac{1}{2} factors gives the same result \frac{1}{64}.

If one factor were different, e.g. \frac{1}{2}^5 \times \frac{1}{4}, how to combine?

Write with common base: \left(\frac{1}{2}\right)^5 \times \frac{1}{4} = \frac{1}{32} \times \frac{1}{4} = \frac{1}{128}, or \left(\frac{1}{2}\right)^7 since \frac{1}{4} = \left(\frac{1}{2}\right)^2.
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