Question
True or false? 5 1/7 is a rational number between 5 and 5 1/4
Answers
GPT-4o mini
True.
To determine if \(5 \frac{1}{7}\) is a rational number between \(5\) and \(5 \frac{1}{4}\), we can first convert these numbers to improper fractions:
- \(5 = \frac{5 \times 7}{7} = \frac{35}{7}\)
- \(5 \frac{1}{4} = \frac{5 \times 4 + 1}{4} = \frac{21}{4}\)
Now, we convert \(5 \frac{1}{7}\) to an improper fraction as well:
- \(5 \frac{1}{7} = \frac{5 \times 7 + 1}{7} = \frac{36}{7}\)
Next, we need to compare \(5 \frac{1}{7} = \frac{36}{7}\) with \(5 = \frac{35}{7}\) and \(5 \frac{1}{4} = \frac{21}{4}\).
To compare \(\frac{36}{7}\) and \(\frac{21}{4}\), we find a common denominator. The least common multiple of \(7\) and \(4\) is \(28\):
\[
\frac{36}{7} = \frac{36 \times 4}{28} = \frac{144}{28}
\]
\[
\frac{21}{4} = \frac{21 \times 7}{28} = \frac{147}{28}
\]
Now we can see that:
\[
\frac{35}{7} = \frac{140}{28} < \frac{144}{28} < \frac{147}{28}
\]
Since \(\frac{36}{7} (5 \frac{1}{7})\) is greater than \(\frac{35}{7} (5)\) and less than \(\frac{21}{4} (5 \frac{1}{4})\), it is confirmed that \(5 \frac{1}{7}\) is indeed a rational number between \(5\) and \(5 \frac{1}{4}\).
Therefore, the statement is **true**.
To determine if \(5 \frac{1}{7}\) is a rational number between \(5\) and \(5 \frac{1}{4}\), we can first convert these numbers to improper fractions:
- \(5 = \frac{5 \times 7}{7} = \frac{35}{7}\)
- \(5 \frac{1}{4} = \frac{5 \times 4 + 1}{4} = \frac{21}{4}\)
Now, we convert \(5 \frac{1}{7}\) to an improper fraction as well:
- \(5 \frac{1}{7} = \frac{5 \times 7 + 1}{7} = \frac{36}{7}\)
Next, we need to compare \(5 \frac{1}{7} = \frac{36}{7}\) with \(5 = \frac{35}{7}\) and \(5 \frac{1}{4} = \frac{21}{4}\).
To compare \(\frac{36}{7}\) and \(\frac{21}{4}\), we find a common denominator. The least common multiple of \(7\) and \(4\) is \(28\):
\[
\frac{36}{7} = \frac{36 \times 4}{28} = \frac{144}{28}
\]
\[
\frac{21}{4} = \frac{21 \times 7}{28} = \frac{147}{28}
\]
Now we can see that:
\[
\frac{35}{7} = \frac{140}{28} < \frac{144}{28} < \frac{147}{28}
\]
Since \(\frac{36}{7} (5 \frac{1}{7})\) is greater than \(\frac{35}{7} (5)\) and less than \(\frac{21}{4} (5 \frac{1}{4})\), it is confirmed that \(5 \frac{1}{7}\) is indeed a rational number between \(5\) and \(5 \frac{1}{4}\).
Therefore, the statement is **true**.
are you 100% sure that is true?
GPT-4o mini
Let’s double-check the calculations step by step.
1. **Convert the mixed numbers to improper fractions**:
- \(5 = \frac{5 \times 7}{7} = \frac{35}{7}\)
- \(5 \frac{1}{4} = \frac{5 \times 4 + 1}{4} = \frac{20 + 1}{4} = \frac{21}{4}\)
2. **Convert \(5 \frac{1}{7}\) to an improper fraction**:
- \(5 \frac{1}{7} = \frac{5 \times 7 + 1}{7} = \frac{35 + 1}{7} = \frac{36}{7}\)
3. **Convert \(5\) and \(5 \frac{1}{4}\) to a common fraction to compare with \(5 \frac{1}{7}\)**:
- The denominators are \(7\) and \(4\).
- The least common multiple (LCM) of \(7\) and \(4\) is \(28\).
Now, convert each fraction:
- \(5 = \frac{35}{7} = \frac{140}{28}\)
- \(5 \frac{1}{4} = \frac{21}{4} = \frac{147}{28}\)
- \(5 \frac{1}{7} = \frac{36}{7} = \frac{144}{28}\)
4. **Compare the fractions** to check their order:
- \(\frac{140}{28} < \frac{144}{28} < \frac{147}{28}\)
This confirms that:
- \(5 < 5 \frac{1}{7} < 5 \frac{1}{4}\).
Based on this detailed review, my previous conclusion holds: ***the statement that \(5 \frac{1}{7}\) is a rational number between \(5\) and \(5 \frac{1}{4}\) is indeed true.***
1. **Convert the mixed numbers to improper fractions**:
- \(5 = \frac{5 \times 7}{7} = \frac{35}{7}\)
- \(5 \frac{1}{4} = \frac{5 \times 4 + 1}{4} = \frac{20 + 1}{4} = \frac{21}{4}\)
2. **Convert \(5 \frac{1}{7}\) to an improper fraction**:
- \(5 \frac{1}{7} = \frac{5 \times 7 + 1}{7} = \frac{35 + 1}{7} = \frac{36}{7}\)
3. **Convert \(5\) and \(5 \frac{1}{4}\) to a common fraction to compare with \(5 \frac{1}{7}\)**:
- The denominators are \(7\) and \(4\).
- The least common multiple (LCM) of \(7\) and \(4\) is \(28\).
Now, convert each fraction:
- \(5 = \frac{35}{7} = \frac{140}{28}\)
- \(5 \frac{1}{4} = \frac{21}{4} = \frac{147}{28}\)
- \(5 \frac{1}{7} = \frac{36}{7} = \frac{144}{28}\)
4. **Compare the fractions** to check their order:
- \(\frac{140}{28} < \frac{144}{28} < \frac{147}{28}\)
This confirms that:
- \(5 < 5 \frac{1}{7} < 5 \frac{1}{4}\).
Based on this detailed review, my previous conclusion holds: ***the statement that \(5 \frac{1}{7}\) is a rational number between \(5\) and \(5 \frac{1}{4}\) is indeed true.***