Q. \(\frac{2}{3} \times \frac{1}{2}\) as a fraction
Answer
Multiply numerators and denominators: \( \frac{2}{3}\times\frac{1}{2}=\frac{2\times1}{3\times2}=\frac{2}{6}=\frac{1}{3}\).
Detailed Explanation
Problem: Multiply the fractions \( \frac{2}{3} \times \frac{1}{2} \).
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Write the multiplication of fractions in terms of numerators and denominators.
The two fractions are \( \frac{2}{3} \) and \( \frac{1}{2} \). When multiplying fractions, multiply the numerators together and the denominators together. So we form
\[ \frac{2}{3} \times \frac{1}{2} = \frac{2 \times 1}{3 \times 2} \]
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Multiply the numerators.
Compute the product of the numerators: \( 2 \times 1 = 2 \). Thus the numerator of the result is 2.
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Multiply the denominators.
Compute the product of the denominators: \( 3 \times 2 = 6 \). Thus the denominator of the result is 6.
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Write the unsimplified result.
Putting the results of steps 2 and 3 together gives
\[ \frac{2 \times 1}{3 \times 2} = \frac{2}{6} \]
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Simplify the fraction by dividing numerator and denominator by their greatest common divisor (GCD).
The GCD of 2 and 6 is 2. Divide numerator and denominator by 2:
\[ \frac{2 \div 2}{6 \div 2} = \frac{1}{3} \]
Therefore the fraction simplifies to \( \frac{1}{3} \).
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Alternative (cancellation before multiplying):
Observe that the numerator 2 of the first fraction and the denominator 2 of the second fraction share a common factor 2. Canceling that common factor first yields
\[ \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \times \frac{1}{1} = \frac{1}{3} \]
This gives the same simplified result without computing the intermediate \( \frac{2}{6} \).
Final answer: \( \frac{1}{3} \).
FAQs
How do you multiply \( \frac{2}{3} \times \frac{1}{2} \)?
Can I simplify before multiplying?
Why do we multiply numerators and denominators?
What is \( \frac{1}{3} \) as a decimal?
How do I interpret \( \frac{2}{3}\times\frac{1}{2} \) in words?
What if one fraction is negative, e.g., \( -\frac{2}{3}\times\frac{1}{2} \)?
Can I multiply a fraction by a whole number?
Is fraction multiplication commutative and associative?
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