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Is this statement true?

If $ b $ does not divide $ a $ then $ gcd(a,b)=1 $ .

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    $\begingroup$ The statement can be made true by requiring $b$ to be prime. That is to say, for any $p$ prime one has $p\not\mid a\implies \gcd(a,p)=1$, otherwise if $p\mid a$ one has $\gcd(a,p)=p$ $\endgroup$ – JMoravitz Mar 19 '17 at 7:20
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Nope, because 4 doesn't divide 2, but their gcd is 2.

Suppose $a$ and $b$ are natural numbers. Then $a = 2^{a_1}3^{a_2}5^{a_3}...p_n^{a_n}$, and $b = 2^{b_1}3^{b_2}5^{b_3}...p_n^{b_n}$. Consider the specific case of $a = 12 = 2^2\times 3$ and $b = 18 = 2 \times 3^2$. Looking at the prime factorisation, because they have primes in common ($2$ and $3$) you know that their gcd is greater than 1. On the other hand, they can't divide each other because $a$ has more $2$s than $b$, and $b$ has more $3$s than $a$.

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  • $\begingroup$ A more interesting counterexample can be given by $6$ and $9$ as $3\vert 9$ and $3\vert 6$, so $\gcd(6,9)\geq3$. $\endgroup$ – Clayton Mar 19 '17 at 7:14
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    $\begingroup$ @RoshHashanah still no. Having played with a few randomly chosen examples should have been enough to convince yourself that the statement is false before needing to ask someone else. $\endgroup$ – JMoravitz Mar 19 '17 at 7:15
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Consider also that $4$ does not divide $6$ but $\gcd(4,6)=2$.

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No, since if a number doesn't divide other number it means that any factor(s) of denominator are not contained in numerator.

If $\gcd=1$ it means no factors are in common or the numbers are co-prime.

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