Prove that numbers are coprime

Question

How to prove that if $a$ and $b$ are coprime then $a^k$ and $b^l$ are coprime? Please help me. I don't know how to do it.

Answer 1

For contradiction, assume $\gcd(a,b)=1$, but $\gcd\left(a^k,b^l\right)>1$. Let $p\mid \gcd\left(a^k,b^l\right)$ for some prime $p$. Then $p\mid a^k, b^l$. Also $p\mid a^k\implies p\mid a$ (by Euclid's Lemma) and similarly $p\mid b^l\implies p\mid b$. Then $p\mid a,b\implies \gcd(a,b)\ge p>1$, contradiction.

Answer 2

Suppose our numbers are $2$ and $3$. These numbers are coprime. The definition of coprime is that two numbers share no factors other than $1$.

Now, suppose we do $2^3$. This is $2 \times 2 \times 2$. Now, take $3^4$. This is $3 \times 3 \times 3 \times 3$. These two expansions show the prime factors of both numbers. Because their prime factorizations of the two numbers share no factors, they are coprime.

In other words, if you have two numbers that are coprime, and you multiply each of them by themselves a certain amount of times, they are still coprime, because it's mathematically impossible for them to share factors. You can't get new factors out of old factors.

Answer 3

Lemma 1: $\gcd(a,b)=1 \Rightarrow \forall n\in\mathbb{N}, \gcd(a^n,b)=1$

Proof: We prove by strong induction. Note that the base case is trivially true. Suppose that for each $k\in\mathbb{N}$ if $k<n$ then $\gcd(a^{k},b)=1$. By Bézout's identity there exists $x_k$ and $y_k$ in $\mathbb{Z}$ such that $x_k a^{k} + y_k b =1$. It follows that $x_{n-1}a^n + y_{n-1}ab=a$ and hence $x_1 (x_{n-1}a^n + y_{n-1}ab) + y_1b=1$ so that $x_1x_{n-1}a^n + (x_1y_{n-1}a + y_1)b=1$. It follows from Bézout's identity again that $\gcd(a^n,b)=1$ as required.

Using Lemma 1 twice we obtain $\gcd(a,b)=1 \Rightarrow \gcd(a^k,b)=1 \Rightarrow \gcd(a^k,b^l)=1$