Pick two nonzero integers $a$ and $b$, so $(a,b)\in (\mathbb{Z}\setminus\{0\})\times(\mathbb{Z}\setminus\{0\})$. We want to add the fractions $1/a$ and $1/b$ and use the standard algorithm. First carefully find the least common multiple of $a$ and $b$ (it is only well-defined up to a sign but that will not be important). Then convert each of the two fractions to get this common number as their denominators. Finally, add the two new numerators and keep the denominator.
For the scope of this question, let's call $(a,b)$ "easy" iff the resulting sum fraction is already in lowest terms (irreducible fraction).
Examples: If $a=12$ and $b=16$, then
$$\frac{1}{12}+\frac{1}{16}=\frac{4}{48}+\frac{3}{48}=\frac{7}{48}$$
so $(12,16)$ is "easy". On the other hand, since
$$\frac{1}{10}+\frac{1}{15}=\frac{3}{30}+\frac{2}{30}=\frac{5}{30}$$
the pair $(10,15)$ is not "easy".
The question: Is there an equivalent or simpler way to define this "easiness"? For example in terms of the signs and prime factorizations of $a$ and $b$? Does this property already have a conventional name?
Note that signs seem to matter since $(3,6)$ is not easy while $(3,-6)$ is.
Answer
Let $\gcd(a,b)=d$, and write $a=da', b=db'$, where $\gcd(a',b')=1$. Then you have $$\frac{1}{a}+\frac{1}{b}=\frac{1}{da'}+\frac{1}{db'}=\frac{b'+a'}{da'b'}$$
Now, $\gcd(a'+b',a'b')=1$ because if prime $p|\gcd(a'+b',a'b')$ then $p|a'b'$ then wlog $p|a'$; but then $p|(a'+b')$ and hence $p|b'$, a contradiction. Hence the sum is "easy" exactly when $$\gcd(a'+b',d)=1$$
In the example $a=10, b=15$, we have $a'=2, b'=3, d=5$ and $a'+b'=5$.
In the example $a=3, b=\pm 6$, we have $a'=1, b'=\pm 2, d=3$, and $a'+b'$ is $3$ or $-1$.
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