Basic knowledge about fractions: Simplifying like factors is allowed. Simplifying like digits is not. And yet, there are fractions where this forbidden simplification seems to work. It's worth examining a particularly simple family of fractions more closely: fractions where the same digit appears at the end of the numerator and at the beginning of the denominator.
The numerator should now consist of a number \(a\) and an appended digit \(x\) , the denominator of the same digit \(x\) and an appended number \(c\):
$$\frac{\overline{a\,x}}{\overline{x\,c}}$$
Let the total number of digits in the numerator and denominator be \(n \ge 2\) .
Then \(a\) and \(c\) each have \(k=n-1\) decimal places. Furthermore, let \(x \in \{1,\dots,9\}\) .
In normal decimal notation, this means:
$$\overline{a\,x}\,=10a+x$$
and
$$\overline{x\,c}\,=x10^k+c.$$
The prohibited deletion, which is being examined here, would therefore be:
$$\frac{\overline{a\,x}}{\overline{x\,c}} \longmapsto \frac{a}{c}$$
We are looking for precisely those cases in which the value remains unchanged.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
The entire structure can now be derived from simple transformations. Cross-multiplication yields...:
$$c(10a+x)=a(x10^k+c)$$
When multiplied out, the:
$$10ac+cx=ax10^k+ac$$
If you combine the terms appropriately, it follows that:
$$9ac=x(a10^k-c)$$
Since \(a\) and \(c\) are each \(k\) digit positive integers, \(a10^kc>0\) . This is because \(a10^k\) is always greater than any \(k\) digit number \(c\) . Therefore, we can divide and obtain the central condition.:
$$\boxed{x=\frac{9ac}{a10^k-c}}$$
This formula describes exactly when apparent simplification works in this special form. It is not only necessary but also sufficient: If this equation holds, then all transformations can be reversed and lead back to the original result.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
The crucial point is that the expression \(\frac{9ac}{a10^kc}\) must ultimately result in a single decimal digit from \(\{1,\dots,9\}\) . Only then does such a fraction arise. For proper fractions \(a<c\) is additionally required. Then \(\frac{a}{c}<1\) is also true, and because of the equality of values, the original fraction is also proper.
The formula \(\displaystyle x=\frac{9ac}{a10^kc}\) is very convenient for the proof. However, a slightly rearranged form is more practical for actually finding such examples. From the equation \(\frac{10a+x}{x10^k+c}=\frac{a}{c}\) we already obtained \(\displaystyle 9ac=x(a10^kc)\) . Equivalently, we have \(\displaystyle c(9a+x)=xa10^k\) .
Now we factor the common divisor of \(a\) and \(x\) . Let \(\displaystyle g=\gcd(a,x)\) . Then there exist numbers \(b\) and \(y\) such that \(\displaystyle a=gb\) , \(\displaystyle x=gy\) , and \(\displaystyle \gcd(b,y)=1\) . Substituting this into \(\displaystyle c(9a+x)=xa10^k\) we get \(\displaystyle c(9b+y)=x b10^k\) . Since \(\displaystyle \gcd(9b+y,b)=\gcd(y,b)=1\) , the factor \(9b+y\) must completely divide the expression \(\displaystyle x10^k\) . If we set \(\displaystyle d=9b+y\) , then \(\displaystyle d\mid x10^k\) and also \(\displaystyle d\equiv y \pmod 9\) . Conversely, from such divisors we can directly obtain
$$\displaystyle a=g\frac{d-y}{9}$$
and
$$\displaystyle c=\frac{x10^k(d-y)}{9d}.$$
This means you no longer have to go blind. \(a\) and \(c\) Try them out. For each digit. \(x\in\{1,\dots,9\}\), each divider \(g\mid x\) and every suitable divider \(d\mid x10^k\) You receive candidates. All that remains is to check whether \(a\) and \(c\) really \(k\)Are -digits and, if you want real fractions, whether \(a\) and \(c\) really \(k\)Are -digits and, if you want real fractions, whether \(a<c\) applies.
Two examples:
$$\frac{16}{64}=\frac{1}{4}$$
Here, the digit \(6\) is deleted.
A significantly longer example with \(42\) places each and recursive cancellation is:
$$\frac{166666666666666666666666666666666666666666}{666666666666666666666666666666666666666664}=\frac{1}{4}$$
Here too, the same digit is removed: in the numerator the last \(6\) , in the denominator the first \(6\) .