Write the number 7.5 23bar in the form p/q where p,q are integers in the lowest form

We are given the number \( 7.5\overline{23} \), which means \( 7.5232323\ldots \), where "23" repeats infinitely. We are tasked with expressing this number in the form \( \frac{p}{q} \) where \( p \) and \( q \) are integers and the fraction is in its lowest terms.

### Step 1: Let \( x = 7.5\overline{23} \).

So,
\[
x = 7.5232323\ldots
\]

### Step 2: Separate the integer and decimal part.
Let \( x = 7 + y \), where \( y = 0.5\overline{23} \).

Now, focus on \( y = 0.5\overline{23} \).

Let \( y = 0.5\overline{23} \), which means:
\[
y = 0.5232323\ldots
\]

### Step 3: Express \( y \) in the form of a fraction.
Let \( z = 0.\overline{23} \), which is the repeating decimal \( 0.232323\ldots \).

Thus, \( y = 0.5 + z \).

Now, we need to convert \( z = 0.\overline{23} \) into a fraction.

Let \( z = 0.\overline{23} \). To express this repeating decimal as a fraction, follow these steps:

1. Let \( z = 0.\overline{23} \).
2. Multiply both sides by 100 to shift the decimal two places:
   \[
   100z = 23.\overline{23}
   \]
3. Subtract the original equation \( z = 0.\overline{23} \) from the above equation:
   \[
   100z - z = 23.\overline{23} - 0.\overline{23}
   \]
   \[
   99z = 23
   \]
4. Solving for \( z \):
   \[
   z = \frac{23}{99}
   \]

Thus, \( z = \frac{23}{99} \).

### Step 4: Substitute back into \( y \).
Now, substitute \( z \) into the expression for \( y \):
\[
y = 0.5 + \frac{23}{99}
\]

Convert \( 0.5 \) to a fraction:
\[
0.5 = \frac{1}{2}
\]
So,
\[
y = \frac{1}{2} + \frac{23}{99}
\]

### Step 5: Find the common denominator and simplify.
The least common denominator of 2 and 99 is 198. So, rewrite both fractions with a denominator of 198:
\[
\frac{1}{2} = \frac{99}{198}, \quad \frac{23}{99} = \frac{46}{198}
\]

Now, add the fractions:
\[
y = \frac{99}{198} + \frac{46}{198} = \frac{145}{198}
\]

### Step 6: Add back the integer part.
Recall that \( x = 7 + y \). Thus,
\[
x = 7 + \frac{145}{198}
\]

Convert 7 to a fraction with the same denominator:
\[
7 = \frac{7 \times 198}{198} = \frac{1386}{198}
\]

Now, add the fractions:
\[
x = \frac{1386}{198} + \frac{145}{198} = \frac{1531}{198}
\]

Thus, the number \( 7.5\overline{23} \) in the form \( \frac{p}{q} \) is:
\[
\boxed{\frac{1531}{198}}
\]

This fraction is already in its lowest terms, so the final answer is \( \frac{1531}{198} \).

Would you like any further details or clarifications on this solution?

Here are 5 related questions for further exploration:
1. How do you convert a repeating decimal with three repeating digits into a fraction?
2. What is the process for simplifying fractions to their lowest terms?
3. How would you convert \( 0.\overline{456} \) to a fraction?
4. Can every repeating decimal be written as a fraction? Why or why not?
5. How does the process change if the repeating decimal starts after a non-repeating decimal sequence?

**Tip:** When dealing with repeating decimals, multiply the decimal by a power of 10 that matches the repeating length to simplify the subtraction step.

Learn how to convert the recurring decimal 7.5 recurring 23 to a fraction step-by-step. Understand the process of handling repeating decimals and expressing them as fractions. Suitable for high school students.

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