Fred's Flummoxing Fractions

Algebra Level 3

Fred has two rational numbers ( $\frac{a}{b}$ and $\frac{c}{d}$ ), made of positive integers $a,b,c$ and $d$ .

Fred doesn't understand the rules for adding fractions, so he simply added the numerators together and added the denominators together. For example, Fred would get $\frac{5}{7} + \frac{12}{13} = \frac{5+12}{7+13} = \frac{17}{20}$ .

Can Fred ever get the correct answer, given two rational numbers?

Note: The fractions do not have to be in the simplest form and Fred would still be correct even if his answer wasn't in the simplest form

No, never is this possible Yes, there are infinitely many ways Yes, there are finitely many ways

2 solutions

Stephen Mellor
Dec 22, 2018

Performing the correct arithmetic, he would get $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}$

Performing the incorrect arithmetic, he would get $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d}$

Setting these equal, we can solve to find all possible solutions: $\dfrac{ad + bc}{bd} = \dfrac{a + c}{b + d}$

Cross-multiplying and simplifying we get: $ad^2 + cb^2 = 0$

Since $a,b,c,d$ are positive, the left side of this equation will be positive and hence cannot equal zero, meaning there are $\boxed{\text{Zero ways}}$ for Fred to be correct.

Note that if we set $a,b,c,d$ as "non-negative", we would get a whole set of trivial solutions such as $\dfrac{0}{13} + \dfrac{0}{16} = \dfrac{0}{29}$

Edwin Gray
Dec 24, 2018

Manipulation leads to cb^2 + ad^2 = 0, which is impossible in positive numbers. Ed Gray

except when a and c are exactly 0, which is possible as the question doesn't say they are non-zero

Skylar Deslypere - 2 years, 3 months ago

Fred's Flummoxing Fractions

2 solutions

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