1

If you cancel (incorrectly) the 6 in $\frac{26}{65}$, you get the (correct) equation $\frac{26}{65} = \frac{2}{5}$. What other fraction exhibits this property?

I tried considering only the simple case where the fraction is of the form $\frac{10a+n}{10n+b}$ (as in the example), then attempting to solve $\frac{a}{b} = \frac{10a+n}{10n+b}$ to get $n(10a-b) = 9ab$. I was thinking of using divisibility tricks, but am not getting anywhere.

The case of the forms $\frac{10n+a}{10n+b}$ and $\frac{10a+n}{10b+n}$ are easy; you just get $a = b$. I haven't attempted higher than two digits or non-rationals. It would be interesting to see a complicated formula not necessarily involving a ratio of integers which works out to a correct answer with incorrect cancellation.

NoChance
  • 6,427
user291130
  • 333
  • 1
    What would a "non-rational" example look like? – Travis Willse Jan 05 '16 at 23:56
  • @Travis I guess something like $ \sqrt{a+b} = \sqrt{a} + \sqrt{b}, (f \cdot g)' = f' \cdot g' $ and so on – Paolo Franchi Jan 06 '16 at 00:15
  • The classic I always thought was 16/64 = 1/4. – fleablood Jan 06 '16 at 00:43
  • Similar to: http://math.stackexchange.com/questions/49657/bad-fraction-reduction-that-actually-works – NoChance Jan 06 '16 at 01:38
  • @PaoloFranchi Square both sides of the radical equation and simplifying gives $2 \sqrt{a b} = 0$, so there are only solutions have one of $a, b$ zero. – Travis Willse Jan 06 '16 at 01:42
  • @PaoloFranchi Also, the ansatz $f(x) = e^{ax}, g(x) = e^{bx}$ leads to a one-parameter family of solutions to $(fg)' = f' g'$ given by the solutions to $a + b = a b$. – Travis Willse Jan 06 '16 at 01:44
  • @Travis I was just giving examples of "wrong operations" that could be solved as equations, since this would answer the final question of the OP. (Unfortunately,) $ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $ happens to have only trivial solutions – Paolo Franchi Jan 06 '16 at 01:52
  • 98/49=2. Cancel the 9's..... – DanielWainfleet Jan 06 '16 at 02:08
  • The Journal of Irreproducible Results, Vol. 35, No. 1 published a tongue-in-cheek article about this by Ben Ruekberg back in 1993 or so titled “Simplified Mathematics” that you can find at http://www.textfiles.com/humor/COMPUTER/maths.txt which included many examples of this sort of thing. – amd Jan 11 '16 at 08:49

4 Answers4

3

We'll solve $$\frac{10 a + n}{10 n + b} = \frac{a}{b},$$ or equivalently $n (10 a - b) = 9 a b$; to interpret the l.h.s. as a ratio of $2$-digit numbers, we restrict to solutions for which $1 \leq a, b, n \leq 9$.

Reducing modulo $9$ reduces $n(10 a - b) = 9 ab$ to $$n (a - b) \equiv 0 \pmod 9.$$ In particular $n$ and $(a - b)$ must have two factors of $3$ between them. This gives three cases:

CASE 1 $(3^2) \mid n$ : This forces $n = 9$, and canceling gives $10 a - b = ab$. Rearranging gives $(a + 1)(10 - b) = 10$, so since $1 \leq a, b \leq 10$ we must have either $a + 1 = 10$, or one of $a + 1$ and $10 - b$ is $2$ and the other is $5$. These cases respectively give $$\frac{99}{99}, \quad \frac{19}{95}, \quad \frac{49}{98}.$$

CASE 2 $3 \mid n$ and $3 \mid (a - b)$: We may as well assume $9 \nmid n$, as this is the content of case $1$, so, $n = 3$ or $n = 6$. If $n = 3$, simplifying gives $$10 a - b = 3 a b ,$$ and rearranging gives $(3 a + 1)(10 - 3 b) = 10$, which forces $a = b = 3$, giving $$\frac{33}{33} .$$ If $n = 6$, simplifying and rearranging gives $(3 a + 2)(20 - 3 b) = 40$. The factors of $40$ of the form $3 a + 2$ for $1 \leq a \leq 9$ are $5, 8, 20$, and these lead to $a = 1, b = 4$ and $a = 2, b = 5$, $a = 6, b = 6$, which respectively give $$\frac{16}{64}, \quad \frac{26}{65}, \quad \frac{66}{66}.$$

CASE 3 $9 \mid a - b$: Since $1 \leq a, b \leq 9$, we have $a = b$, and so $n (10 a - b) = 9 a b$ simplifies to $n = a$, giving the trivial cases $$\frac{11}{11}, \ldots, \frac{99}{99} .$$

In summary, the nontrivial solutions are $$\color{#bf0000}{\boxed{\frac{16}{64}, \qquad \frac{19}{95}, \qquad \frac{26}{65}, \qquad \frac{49}{98}}}. $$

Of course, we can also look for "false cancellations" of the form $$\frac{10 n + a}{10 b + n} = \frac{a}{b} ,$$ but this is exactly the equation formed by taking the reciprocals of both sides of the above equation and interchanging the roles of $a, b$, so this equation leads precisely to the four reciprocals, $\frac{64}{16}$, etc., of the above solutions.

(There are other two-digit "false cancellation" fractions, corresponding to other equations, but there are all trivial in some sense; they have the forms $\frac{d0}{e0}$ and $\frac{dd}{ee}$ for digits $d, e$.)

Travis Willse
  • 99,363
2

There are quite of few fractions that exhibit this property. They are also referred to as "lucky fractions".

Here is a paper on this topic that you might find useful: Lucky fractions

Mufasa
  • 5,434
0

One other springs to mind: $\frac{16}{64}$.

Senex Ægypti Parvi
  • 2,419
  • Actually it is an answer to the question "What other fraction exhibits this property?" in the first paragraph. To be sure it's not the only other such fraction. – Noam D. Elkies Jan 06 '16 at 00:38
0

The lucky fractions $ \frac{16}{64}, \frac{19}{95}, \frac{26}{65}, \frac{49}{98} $ and their inverses are the solutions to the first question.

About your final (vague) suggestion, I have tried some "wrong operations" concerning fractions, roots and powers, and it seems that the nicest results can be obtained as solutions of

$$ \frac{a+b}{c+d} = \frac{a}{c} + \frac{b}{d}. $$

For instance, we have $ \frac{2-8}{k+2k} = \frac{2}{k} + \frac{-8}{2k}$, which works for $k \neq 0$.

Reference: http://www.wolframalpha.com/input/?i=solve+%28a%2Bb%29%2F%28c%2Bd%29%3Da%2Fc+%2B+b%2Fd+over+the+integers

Paolo Franchi
  • 1,097