A common mistake students make when adding fractions is to add both the numerators and the denominators (I’ll use a special symbol to reinforce that this is ** not** proper addition): \[ \frac{2}{5} \heartsuit \frac{3}{7} = \frac{2+3}{5+7} = \frac{5}{12} \] The general approach is to tell students that that doesn’t normally work. However, while errors in process sometimes results in accidental correct answers, this incorrect method will

*never*result in a correct answer, at least not when both fractions are positive. Indeed, \(\frac{a}{b} \heartsuit \frac{c}{d}\) is always between \(\frac{a}{b}\) and \(\frac{c}{d}\).

To prove it, let’s consider what we’re working with: \[\frac{a}{b} \heartsuit \frac{c}{d} = \frac{a+c}{b+d}\] Write the three fractions to have a common denominator:\[\frac{a}{b} = \frac{a\cdot d\cdot (b + d)}{b\cdot d\cdot (b + d)} = \frac{abd + add}{bd(b + d)} \\ \frac{a+c}{b+d} = \frac{(a + c) \cdot b\cdot d}{(b + d) \cdot b\cdot d} = \frac{abd + bcd}{bd(b + d)} \\ \frac{c}{d} = \frac{c\cdot b\cdot (b + d)}{d\cdot b\cdot (b + d)} = \frac{bbc + bcd}{bd(b + d)}\]

Let us assume that \(\frac{a+c}{b+d}\) is not between \(\frac{a}{b}\) and \(\frac{c}{d}\). This would be the case if it were greater than both or less than both. In the case that it’s greater than both, we have: \[\frac{a+c}{b+d} > \frac{a}{b} \Rightarrow \frac{abd + bcd}{bd(b + d)} > \frac{abd + add}{bd(b + d)} \Rightarrow abd + bcd > abd + add \\ \Rightarrow bcd > add \Rightarrow bc > ad \] and \[\frac{a+c}{b+d} > \frac{c}{d} \Rightarrow \frac{abd + bcd}{bd(b + d)} > \frac{bbc + bcd}{bd(b + d)} \Rightarrow abd + bcd > bbc + bcd \\ \Rightarrow abd > bbc \Rightarrow ad > bc \]

But obviously it is a contradiction to say that \(bc > ad\) and \(ad > bc\). Assuming that the “sum” is less than both of the addend fractions leads to the same contradiction, so it must be the case that \(\frac{a+c}{b+d}\) is always between \(\frac{a}{b}\) and \(\frac{c}{d}\). Note that this means that if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a}{b} = \frac{c}{d} = \frac{a + b}{c + d}\).

We know, meanwhile, that adding two positive values always yields a greater positive value, that is, a value which is greater than either of the addends. Hence, when adding two positive fractions, “add the tops and add the bottoms” will *never* yield a correct answer.

### Extension: What about non-positives?

In the trivial case that both fractions are zero, this method does work by accident, since 0 + 0 is 0 and 0 is the only value “between” (inclusively speaking) 0 and 0. However, if only one fraction is 0 (over some arbitrary denominator), the method still fails. For example: \[\frac{0}{1} + \frac{1}{2} = \frac{1}{2} \text{ while } \frac{0}{1} \heartsuit \frac{1}{2} = \frac{1}{3} \]

If both fractions are negative, we have the same situation as when they’re both positive: The sum has to be less than both of the addends, while the misadded “sum” has to be between them.

That leaves the situation where one addend is positive and the other is negative There *are* cases here where the incorrect algorithm yields the correct answer, such as \(\frac{4}{6} + \frac{-1}{3} = \frac{3}{9}\) and \(\frac{9}{12} + \frac{-4}{8} = \frac{5}{20}\). Since the difference of two value can be between the two values, and since the misadded “sum” is always between the two addends, it’s not surprising that they can accidentally coincide.

Note that this does not occur when the fractions are actually being subtracted, since students making the error in question would subtract both the numerator and the denominator, rather than subtracting one and adding the other. So getting the correct answer through subtraction would rely on the student making two errors.

To find the specific pattern, we can perform both operations, set the values equal to each other, and then simplify as much as possible. Using the positive integers \(a, b, c,\) and \(d\): \[\frac{a}{c} \heartsuit \frac{-b}{d} = \frac{a-b}{c+d} \\ \frac{a}{c} + \frac{-b}{d} = \frac{ab-bc}{cd}\] We cross-multiply to get \[(ad – bc)(c + d) = (a – b)(cd)\] and then distribute and simplify \[acd + add – bcc – bcd = acd – bcd \\ \Rightarrow add – bcc = 0 \Rightarrow ad^2 = bc^2\]

However, in all cases where a misadded fraction sum yields the correct answer, at least one of the fractions can be simplified. It will never be the case, that is, that both \(a\) and \(c\) *and* \(b\) and \(d\) are coprime. We can see this fairly simply. Since \(bc^2 = ad^2\), either \(a\) or \(d\) must be a multiple of \(c\). Formally stated (where \(k\) is some integer): \[bc^2 = ad^2 \Rightarrow bc^2 = (kcd^2 \vee a(kc)^2)\] If \(a = kc\), then \(\frac{a}{c} = \frac{kc}{c}\), which can clearly be reduced. If \(d = kc\) then \(bc^2 = ak^2c^2\) and \(b = ak^2\) and so \(\frac{b}{d} = \frac{ak^2}{kc}\), which can clearly be reduced.

There is a third possibility, where \(c = mn\) and where \(a = km\) and \(d = ln\), but this leads to both fractions being reducible. In this case, \(bc^2 = ad^2 \Rightarrow bm^2n^2= kml^2n^2 \)\(\Rightarrow bm = kl^2 \Rightarrow b = kl^2 \div m\). This requires that: (1) \(k\) is a multiple of \(m\), (2) \(l\) is a multiple of \(m\), or (3) \(kl\) is a multiple of \(m\) (that is, all of the factors of \(k\) and \(l\), taken together, include all of the factors of \(m\).

In the first case, \(k = rm\) and so \(c = mn, a = rm^2, d = ln,\) and \(b = rl^2\). So \(\frac{a}{c} = \frac{rm^2}{mn}\) and \(\frac{b}{d} = \frac{rl^2}{ln}\), which can both be reduced.

In the second case, \(l = rm\) and so \(c = mn, a = km, d = rmn,\) and \(b = kr^2m\). So \(\frac{a}{c} = \frac{km}{mn}\) and \(\frac{b}{d} = \frac{kr^2m}{jmn}\), which can both be reduced.

In the third case, \(m = pq, k = rp,\) and \(l = sq\), and so \(c = pqn, a = rp^2q, d = sqn\) and \(b = rs^2q\), and so \(\frac{a}{c} = \frac{rp^2q}{pqn}\) and \(\frac{b}{d} = \frac{rs^2q}{sqn}\), which can both be reduced.

Thus, in all three cases, both fractions can be reduced.

For instance, \(\frac{9}{12} + \frac{-4}{8} = \frac{5}{20}\) is true with \((r, l, m, n) = (1, 2, 3, 4)\), so that \((a, b, c, d) = (rm^2, rl^2, mn, ln) \)\( = (9, 4, 12, 8)\).

As an example of the third case, consider \((p, q, r, s, n) = (2, 3, 5, 7, 4)\). This yields the sum \(\frac{60}{24}+\frac{-735}{84} = \frac{-675}{108}\), which is the same thing we get using the misadding algorithm. However, if we reduce the fractions, we get \(\frac{5}{2} + \frac{-35}{4} = \frac{-25}{4}\), while \(\frac{5}{2} \heartsuit \frac{-35}{4} = \frac{-30}{6} = -5\).

In conclusion, then, the only time that students would get the correct answer when using the strategy of “add the tops, add the bottoms” is in the case that the fractions have different signs *and* at least one of the fractions has not been simplified. This is a very unlikely scenario.