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LostInParadise's avatar

Care for another elementary math problem?

Asked by LostInParadise (30071points) 1 month ago

For this you will need to remember how to add fractions and have a basic understanding of prime numbers.

We are given the denominators of four fractions as 4, 6, 9 and 10. All we know about the numerators is that the fractions are irreducible so that, for example, the numerator for 4 cannot be 2.

Find the prime factorization of the denominators to show that the least common denominator is 180. Then show that when the four fractions are added the resulting sum, with 180 as the denominator, is irreducible.

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4 Answers

LostInParadise's avatar

For some additional motivation, it is not always the case that use of least common denominator gives an answer that is irreducible. For example, for 1/6 + 1/10, the least common denominator is 30, giving 5/30 + 3/30 = 8/30, which can be reduced to 4/15.

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LostInParadise's avatar

I see I have no takers. Here is a simplified version. We have two fractions a/2 and b/3, both of which are irreducible. The least common denominator is 6, giving a/2 + b/3 = (3a+2b)/6. How do we know that this is irreducible?

LostInParadise's avatar

On the off chance someone cares to know how these problems are solved, here is the answer to the second one. The first is similar, but a little more elaborate. It is not necessary to use algebra to solve these problems, but it makes it easier to explain.

Since the denominator is 6=2×3, the final fraction is reducible if and only if (3a+2b) is divisible by 2 or 3. To show irreducibility, we need to show that 3a+2b is not divisible by 2 or 3.

Since a/2 is irreducible, a must be an odd number. Therefore 3a is an odd number. Since 2b is even, 3a+2b is odd and hence not divisible by 2.

Similarly, since b/3 is irreducible, b does not have 3 as one of its prime factors and therefore 2b is not divisible by 3, and so then neither is 3a+2b.

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