2

Given a rational number $a/b$ expressed in simplest terms (so $GCD(a,b)=1$), I want to raise it to an integer power $n$.

I think the result will always automatically be in simplest terms, but it's a long time since I was doing maths regularly, so I'm including my reasoning in an answer - please check it.

This question is related, and the answers seem particularly close, but I don't think it's a duplicate.

Community
  • 1

2 Answers2

3

As I say in the question, I think the answer is yes - the result is automatically in least terms.

If $a/b$ is in least terms, $a$ and $b$ don't share any prime factors. If they shared a prime factor $p$, you could reduce the fraction to $a'/b'$ where $a'=a/p$ and $b'=b/p$.

When you calculate $x'=x^n$ for any positive integer $n$, $x'$ has the exact same prime factors as $x$ - each prime factor just repeats $n$ times as often in $x'$.

So if $(a/b)^n = a^n/b^n = A/B$, then $A$ and $B$ share the same prime factors as $a$ and $b$ - just $n$ times as often. And since $a$ and $b$ don't share any prime factors, neither do $A$ and $B$.

  • 2
    Yes, if the unique ordered factorization (primes strictly ordered by index, positive exponents) of the reduced fraction is $$\frac{a}{b}=\frac{p_1^{n_1}p_2^{n_2}\cdots p_N^{n_N}}{q_1^{m_1}q_2^{m_2}\cdots q_M^{m_M}}$$ then $$\frac{a^k}{b^k}=\frac{p_1^{n_1k}p_2^{n_2k}\cdots p_N^{n_Nk}}{q_1^{m_1k}q_2^{m_2k}\cdots q_M^{m_Mk}}$$ which is still reduced as none of the primes $p_{n_i}$ coincide with any of the primes $q_{m_j}$ – MPW Dec 16 '14 at 01:32
  • 1
    Remark that you have implicitly used the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations). It is best to say that explicitly to make it clear that the argument is rigorous, e.g. "by FTA $x'$ has the exact same prime factors....". You might find it helpful to work through the details of that proof, using FTA. – Bill Dubuque Dec 18 '14 at 15:08
  • @Bill - there's simpler things I took as obvious too, such as dealing with negative integer powers. One thing that's been niggling me, though, is why my brain insists on prime factors. Any shared factor between numerator and denominator will do. I guess it's because you can't express a number as a product of all its factors, as MPW showed above, so although it's not necessary, it's a nice form. Also, all factors of a number are products of some subbag (if that's a word) of its prime factors anyway, so in a sense all factors are still covered. –  Dec 18 '14 at 15:57
  • @Steve314 The point is that we have much empirical intuition about the ways that integers can factor. But empirical inference is not rigorous. The FTA is the mathematical foundation supporting these properties of integers. It is important to understand the role it plays because unique factorization and related properties often fail in more general number systems, e.g. rings of quadratic integers. – Bill Dubuque Dec 18 '14 at 16:01
  • @Bill - OK, I hadn't thought about it that way. An obvious idea for an inductive proof (that might be more general) has been bugging me, but it seems on the one hand so trivial it's silly, on the other hand to be bugging me that somethings missing. Anyway, I should have time to work through some of these things in the next few days and it's so long since I made time for some math-related fun, so I'll get back to that. –  Dec 18 '14 at 16:09
  • @Steve It's worth emphasizing that these points are not pedantic. Historically it took an amazingly long time before a rigorous proof of FTA appeared (Gauss, Disq. Arith. circa 1800), and, furthermore, for the realization that a rigorous proof is required. The proof requires special properties of the ring of integers (that they have a Euclidean division algorithm), properties that may fail in more general rings of "integers". You might find it helpful to consider it algorithmically: if you and I have different algorithms for factoring integers into primes, why should they give equal results? – Bill Dubuque Dec 18 '14 at 16:18
  • @Bill - it turns out that Euclids Lemma was the bit I was missing for in that inductive proof, though it only really restates the shared-prime-factors proof anyway. –  Dec 18 '14 at 19:41
  • @Steve314 Yes, Euclid's Lemma is one way to go. It's essentally equivalent to uniqueness of prime factorizations, the nontrivial direction being that prime $,p\mid ab,\Rightarrow p\mid a,$ or $,p\mid b,$ allows us to "match up" then cancel the prime factors in any two prime factorizations of an integer. – Bill Dubuque Dec 18 '14 at 19:45
1

To add another proof to yours (which is perfectly correct), but avoiding relying on unique factorization, I would do it as follows:

First, prove that $\gcd(ab,c)$ divides $\gcd(a,c)\cdot\gcd(b,c)$. This can be done noting that, by the Euclidean algorithm, we can find integers $n_1,n_2,m_1,m_2$ such that: $$\gcd(a,c)=n_1a+n_2c$$ $$\gcd(b,c)=m_1b+m_2c.$$ Their product is thus $n_1m_1 ab + n_1m_2 ac + n_2m_1 bc + m_2n_2 c^2$. Each term of this sum is divisible by $ab$ or $c$, thus is divisible by $\gcd(ab,c)$.

Using this, is is clear that $\gcd(a^n,b^n)$ divides $\gcd(a,b)^{n^2}$. If the latter is $1$, so must be the former.

Milo Brandt
  • 60,888
  • Shouldn't $gcd(a,c)$ be smaller than both $a$ and $c$? I can see there should be a proof based on the greatest common divisor, if only because it's equal to the product of the shared prime factors, but something seems odd here. Should some of these GCDs be least common multiples? –  Dec 16 '14 at 02:13
  • 1
    @Steve Yes, $\gcd(a,c)$ is not larger than $a$ and $c$; the Euclidean algorithm does not contradict this: for instance, note that $\gcd(7,4) = -1\cdot 7 + 2\cdot 8 = 1$ - one of $n_1$ and $n_2$ will always non-positive to ensure $\gcd(a,c)$ is non-positive. In fact, you can define $\gcd$ as the smallest number writable as a sum (with integer coefficients) of its arguments. So, even though the expression on the left looks like it'd be bigger, it's not. – Milo Brandt Dec 16 '14 at 02:39
  • OK, that much makes sense, but I'm going to have to re-read this after some sleep. –  Dec 16 '14 at 02:49