Problem 51
Question
Simplify each expression as much as possible. Show that multiplying 3 by 5 is the same as dividing 3 by \(\frac{1}{5}\)
Step-by-Step Solution
Verified Answer
Multiplying 3 by 5 equals 15, and dividing 3 by \(\frac{1}{5}\) also equals 15, confirming equivalence.
1Step 1: Understanding Multiplication
We begin by expressing the multiplication operation. We want to multiply 3 by 5, which is written as \(3 \times 5\). The result of this operation is straightforward: \(3 \times 5 = 15\).
2Step 2: Understanding Division by a Fraction
Next, we express the division operation. We are dividing 3 by \(\frac{1}{5}\). According to the rules of division involving fractions, dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, dividing by \(\frac{1}{5}\) is the same as multiplying by 5. This gives us: \(3 \div \frac{1}{5} = 3 \times 5\).
3Step 3: Evaluating Division by the Fraction
Now, we perform the operation from the divided expression. Since dividing by \(\frac{1}{5}\) is equivalent to multiplying by 5, we have \(3 \times 5\). We already calculated that \(3 \times 5 = 15\), confirming the equivalence.
Key Concepts
MultiplicationDivision of fractionsReciprocal
Multiplication
Multiplication is one of the four fundamental arithmetic operations. It involves adding a number to itself a certain number of times. For example, when you multiply 3 by 5, you add the number 3 together five times:
\(3 \times 5 = 15\).
This process of grouping numbers simplifies what could otherwise be a long addition process. It’s powerful for working with large numbers or repeated addition, making calculations faster and simpler.
When learning about multiplication, remember it's about finding how many instances you have of a particular number, grouped together.
- 3 + 3 + 3 + 3 + 3 = 15
\(3 \times 5 = 15\).
This process of grouping numbers simplifies what could otherwise be a long addition process. It’s powerful for working with large numbers or repeated addition, making calculations faster and simpler.
When learning about multiplication, remember it's about finding how many instances you have of a particular number, grouped together.
Division of fractions
The division of fractions might seem daunting at first, but it becomes simpler with the right approach. When you divide by a fraction, you're essentially finding out how many times a fraction fits into the whole.
Instead of dividing, you can multiply by the reciprocal of the fraction. This turns the division into a multiplication problem, which is often easier to solve. For example, dividing 3 by \(\frac{1}{5}\) becomes multiplying 3 by 5.
Thus, \[3 \div \frac{1}{5} \equiv 3 \times 5 = 15\] This equivalence shows how division by a fraction works perfectly with the rules of multiplication.
- For instance, dividing by \(\frac{1}{5}\) is asking, "How many fifths are in the number?"
Instead of dividing, you can multiply by the reciprocal of the fraction. This turns the division into a multiplication problem, which is often easier to solve. For example, dividing 3 by \(\frac{1}{5}\) becomes multiplying 3 by 5.
Thus, \[3 \div \frac{1}{5} \equiv 3 \times 5 = 15\] This equivalence shows how division by a fraction works perfectly with the rules of multiplication.
Reciprocal
A reciprocal is a special kind of number used in division and multiplication. To find the reciprocal of a fraction, you simply flip the numerator and the denominator. So, the reciprocal of \(\frac{1}{5}\) is 5, or \(\frac{5}{1}\), because flipping \(\frac{1}{5}\) results in \(\frac{5}{1}\).
- Reciprocals are vital in simplifying division with fractions.
- Knowing that any number times its reciprocal equals 1 (for example, \(\frac{5}{1} \times \frac{1}{5} = 1\)).
Other exercises in this chapter
Problem 51
Find the LCD for each of the following; then use the methods developed in this section to add or subtract as indicated. $$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\f
View solution Problem 51
Reduce each fraction to lowest terms. a. \(\frac{2}{90}\) b. \(\frac{3}{90}\) c. \(\frac{5}{90}\) d. \(\frac{6}{90}\) e. \(\frac{9}{90}\)
View solution Problem 51
Simplify by dividing the numerator by the denominator. $$\frac{37}{1}$$
View solution Problem 52
The following problems all involve the concept of borrowing. Subtract in case. \(7 \frac{1}{6}-6 \frac{5}{6}\)
View solution