What is \(0.\overline {53} + 0.5\overline {3} \) equal to?
Solve Sum of Repeating Decimals: \(0.\overline {53} + 0.5\overline {3}\)
The question asks us to find the sum of two repeating decimals, \(0.\overline {53}\) and \(0.5\overline {3}\). To solve this, we will first convert each repeating decimal into a fraction. Then, we will add the fractions and finally convert the resulting sum back into a decimal form, specifically a repeating decimal if necessary.
Step-by-Step Solution
Converting \(0.\overline {53}\) to a Fraction
Let \(x = 0.\overline {53}\). This means \(x = 0.535353...\)
Since two digits are repeating, we multiply by \(100\):
\(100x = 53.535353...\)
Now, subtract the original equation (\(x = 0.535353...\)) from this new equation:
\(100x - x = 53.535353... - 0.535353...\)
\(99x = 53\)
Divide by 99 to find \(x\):
\(x = \frac{53}{99}\)
So, \(0.\overline {53} = \frac{53}{99}\).
Converting \(0.5\overline {3}\) to a Fraction
Let \(y = 0.5\overline {3}\). This means \(y = 0.533333...\)
First, multiply by \(10\) to move the non-repeating part (5) to the left of the decimal point:
\(10y = 5.333333...\)
Now, identify the repeating part, which is '3'. Multiply the \(10y\) equation by \(10\) (since one digit is repeating after the decimal in \(10y\)):
\(10 \times (10y) = 10 \times (5.333333...)\)
\(100y = 53.333333...\)
Subtract the equation \(10y = 5.333333...\) from the equation \(100y = 53.333333...\):
\(100y - 10y = 53.333333... - 5.333333...\)
\(90y = 48\)
Divide by 90 to find \(y\):
\(y = \frac{48}{90}\)
We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 6:
\(y = \frac{48 \div 6}{90 \div 6} = \frac{8}{15}\)
So, \(0.5\overline {3} = \frac{8}{15}\).
Adding the Fractions
Now we need to add \(\frac{53}{99}\) and \(\frac{8}{15}\). To add fractions, we need a common denominator.
The denominators are 99 and 15.
- Prime factors of 99: \(3 \times 3 \times 11\)
- Prime factors of 15: \(3 \times 5\)
The Least Common Multiple (LCM) of 99 and 15 is \(3 \times 3 \times 5 \times 11 = 9 \times 55 = 495\).
Convert each fraction to have the denominator 495:
\(\frac{53}{99} = \frac{53 \times 5}{99 \times 5} = \frac{265}{495}\)
\(\frac{8}{15} = \frac{8 \times 33}{15 \times 33} = \frac{264}{495}\)
Now, add the fractions:
Sum = \(\frac{265}{495} + \frac{264}{495} = \frac{265 + 264}{495} = \frac{529}{495}\)
Converting the Sum Back to a Decimal
We need to convert the fraction \(\frac{529}{495}\) back into a decimal form. We can perform division:
\(529 \div 495\)
\(529 \div 495 = 1\) with a remainder of \(529 - 495 = 34\).
So, \(\frac{529}{495} = 1 + \frac{34}{495}\).
Now, divide 34 by 495:
34.0000... \(\div\) 495
- \(340 \div 495 = 0\) (remainder 340). The decimal starts with 0.
- Bring down a 0: \(3400 \div 495\). \(495 \times 6 = 2970\). Remainder \(3400 - 2970 = 430\). The decimal is 0.06
- Bring down a 0: \(4300 \div 495\). \(495 \times 8 = 3960\). Remainder \(4300 - 3960 = 340\). The decimal is 0.068
- Bring down a 0: \(3400 \div 495\). \(495 \times 6 = 2970\). Remainder \(3400 - 2970 = 430\). The decimal is 0.0686
The remainder 340 has repeated, which means the sequence of digits '68' will repeat after the initial '0'.
So, \(\frac{34}{495} = 0.0686868... = 0.0\overline{68}\).
Therefore, \(\frac{529}{495} = 1 + 0.0\overline{68} = 1.0\overline{68}\).
Comparing with Options
Let's look at the options:
- Option 1: \(1.0\overline {68}\) means \(1.0686868...\)
- Option 2: \(1.06\bar 8\) means \(1.068888...\)
- Option 3: \(1.\overline {068}\) means \(1.068068068...\)
- Option 4: \(1.068\) means \(1.0680000...\)
Our calculated sum is \(1.0\overline{68}\), which matches Option 1.
Revision Table: Converting Repeating Decimals to Fractions
| Decimal Form | Fraction Form (General) | Example | Calculation Example |
|---|---|---|---|
| \(0.\overline{a}\) | \(\frac{a}{9}\) | \(0.\overline{7}\) | \(\frac{7}{9}\) |
| \(0.\overline{ab}\) | \(\frac{ab}{99}\) | \(0.\overline{23}\) | \(\frac{23}{99}\) |
| \(0.a\overline{b}\) | \(\frac{ab - a}{90}\) | \(0.5\overline{3}\) | \(\frac{53 - 5}{90} = \frac{48}{90} = \frac{8}{15}\) |
| \(0.ab\overline{c}\) | \(\frac{abc - ab}{900}\) | \(0.12\overline{3}\) | \(\frac{123 - 12}{900} = \frac{111}{900}\) |
Additional Information: Types of Decimals
Decimal numbers can be classified based on their digits after the decimal point:
- Terminating Decimals: These decimals have a finite number of digits after the decimal point. For example, 0.25, 1.7, 3.14159. These can always be expressed as a fraction where the denominator is a power of 10. For instance, \(0.25 = \frac{25}{100} = \frac{1}{4}\).
- Non-Terminating Decimals: These decimals have an infinite number of digits after the decimal point.
- Repeating Decimals (Recurring Decimals): In these decimals, a digit or a block of digits repeats infinitely. Examples include \(0.333... = 0.\overline{3}\), \(1.272727... = 1.\overline{27}\), and the decimals in this problem, \(0.\overline {53}\) and \(0.5\overline {3}\). Repeating decimals are rational numbers, meaning they can be expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
- Non-Repeating Decimals: In these decimals, there is an infinite sequence of digits, but no single digit or block of digits repeats in a pattern. These are irrational numbers. Examples include \(\pi \approx 3.14159265...\) and \(\sqrt{2} \approx 1.41421356...\)
The problem deals with adding two repeating decimals, which are both rational numbers. The sum of two rational numbers is always a rational number, which means the result will either be a terminating or a repeating decimal.
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