Suppose $k$ and $n$ are integers such that $1\leq k<n$ and $\gcd(n,k)=1$. Why is it that $\gcd(n-k,n)=1$ as well?
$\gcd(n,k)=1$ implies that there exist integers $a,b$ such that $an+bk=1$, but I can't see if/how this will end up being useful.
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$\gcd(n,k)=\gcd(n-k,n)$ for all pairs of integers $n$ and $k$. – Barry Cipra Feb 17 '19 at 00:32
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Say $d=\gcd(n-k,n)$. Then $d\,|\,n$ so we can write $n=da$. But $d\,|\,n-k$ so $n-k=db$ But then we have $$da-k=db\implies k=d(a-b)\implies d\,|\,k$$
Thus $d\,|\,\gcd(n,k)$ so $d=1$.
lulu
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The reason is because if $k$ and $n$ are coprimes then $k$ is invertible in $\frac{\mathbb{Z}}{n\mathbb{Z}}$, so there exist $a \in \{1,2,\ldots,n-1 \}$ such $ak \equiv 1 \; (mod \; n)$ and so $n-k \equiv -k \;(mod \; n)$ is invertible as his inverse will be $-a$, so $n-k$ and $n$ must be coprimes.
JoseSquare
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