Fractions and Rational Numbers

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SUMMARY

The discussion focuses on the reduction of fractions and the identification of integers from a set of fractions formed by natural numbers 1-10. The fraction 3715/990 is correctly reduced to 743/198, confirming it is in its simplest form. Participants analyze the total number of fractions that reduce to integers, concluding that there are 27 valid fractions. Additionally, they determine that reducing these fractions yields 10 distinct numbers, all represented over 1.

PREREQUISITES
  • Understanding of fraction reduction techniques
  • Familiarity with natural numbers and their properties
  • Basic knowledge of integer identification from fractions
  • Ability to perform arithmetic operations with fractions
NEXT STEPS
  • Study the properties of rational numbers and their simplification
  • Learn about integer results from fraction operations
  • Explore the concept of equivalent fractions and their applications
  • Investigate the role of prime factorization in reducing fractions
USEFUL FOR

Students, educators, and anyone interested in mastering the concepts of fractions and rational numbers, particularly in mathematical problem-solving contexts.

sjaguar13
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I got a few questions. First of all, I reduced 3715/990 to 743/198. Is that reduced all the way?

Second, using the natural number, 1-10, as either the numerator or denominator of a fraction, there is 100 fractions, 1/1, 2/1,...,10/1, 1/2, 2/2,...10/10. How many of these reduce to integers. I say 27, all of the ones (10), evens over 2 (5), multiples of 3 over three (3), multiples of 4 over 4 (2), 5 and 10 over 5 (2), and then 6, 7, 8, 9, 10 over themselves (5)

If we reduce all of these fractions, how many different numbers do we get? I say 10. There was ten to begin with and they all appear over 1.
 
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I got a few questions. First of all, I reduced 3715/990 to 743/198. Is that reduced all the way?

...

If we reduce all of these fractions, how many different numbers do we get? I say 10. There was ten to begin with and they all appear over 1.

Yes and yes.
 
I motion to ban Mulliday...
 

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