Although the definition of HCF for any given integers , on wikipedia clearly mentions it to be the largest positive integer which divides the given integers , I wanted to prove that even if we say HCF to be the largest positive real number which divides the given integers then also it has to be an integer always . For proving this I used the concept that if 'a' is divisible by 'b' (where 'a' and 'b' are any two real numbers and b is not 0 ) this means a/b is an integer . But I could not succeed in it . Please guide me by proving the above fact for any two positive integers 'a' and 'b' .
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3What does it mean for a real number to divide an integer? More specifically, when would it not divide an integer? – saulspatz Apr 16 '20 at 10:19
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HCF means gcd, right? – Wuestenfux Apr 16 '20 at 10:19
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Hint: If $\frac mn$ divides $a$ with $\gcd(m,n)=1$ then show that $m$ divides $a$. – lulu Apr 16 '20 at 10:19
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@Wuestenfux And I thought that gcd meant hcf. – Angina Seng Apr 16 '20 at 10:50
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Under the definition you use, if a real number $x$ divides an integer $a$ then $x$ must be rational (clearly).
The desired claim then follows quickly from the fact that, if $\frac mn$ is written in reduced form, then $$\frac mn\,|\,a\implies m\,|\,a$$
Pf: by your definition we know that there is an integer $k$ with $\frac mn\times k=a$ hence $$mk=an$$
Thus $m\,|\,an$. But $m$ and $n$ are relatively prime (by assumption), hence, by Bezout, $$m\,|\,an\implies m\,|\,a$$
lulu
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Integer gcds remain unchanged in the extension because each real divisor $\,\delta\,$ of an integer $\,k\,$ is $\rm\color{#0a0}{equivalent}$ to an integer divisor. Indeed, $\,\delta\mid k\,$ means $\,k/\delta = j\in\Bbb Z,\,$ hence $\, \delta\in\Bbb Q,\,$ say $\delta = n/m,\ (n,m)=1.\,$ Then $\,\delta\mid k\iff n\mid mk\iff \color{#c00}n\mid k\,$ by Euclid, so we have
$$\ \ \delta\mid k\color{#0a0}\iff \color{#c00}{n_\delta}\mid k,\ \ {\rm for\ some }\,\ n_\delta\in\Bbb Z$$
therefore gcds in $\,\Bbb Z\,$ persist: $\ \delta\mid j,k\color{#0a0}\iff n_\delta\mid j,k\iff n_\delta\mid (j,k)\color{#0a0}\iff \delta\mid (j,k)$
i.e. the gcd $\,(j,k)\in\Bbb Z\,$ remains the gcd of $\,j,k\,$ in $\Bbb R$ with said extension of integer divisibility.
The same holds true for analogous extensions of any gcd-domain (in place of $\Bbb Z$ above).
Bill Dubuque
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Alternatively, gcd persistence follows from the gcd Bezout equation, which works for any Bezout domain. – Bill Dubuque Apr 16 '20 at 17:54