Prove that the fraction $\dfrac{21n+4}{14n+3}$ is in lowest terms for any natural value of $n$.
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5Complex numbers are not numbers that are complicated! :-o – Brian Tung Jul 30 '18 at 17:09
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Any thoughts yourself? Why can't the numerator and denominator both be divided by $2$, $3$, $5$, $7$ or any other primes? – Henry Jul 30 '18 at 17:09
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3Anyway, use the Euclidean algorithm to find GCD of $21n+4$ and $14n+3$. What should the GCD be, if the fraction is always irreducible? – Brian Tung Jul 30 '18 at 17:10
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3@BrianTung You just made me choke. – Batominovski Jul 30 '18 at 17:11
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1@Batominovski: I do apologize! It appears your keyboard is unimpaired, though, thankfully. – Brian Tung Jul 30 '18 at 17:12
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5$$2(21n+4)-3(14n+3)=?$$ – lab bhattacharjee Jul 30 '18 at 17:12
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6For reference, this is IMO 1959, Problem 1---the first ever IMO problem. See https://artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/Problem_1. – Batominovski Jul 30 '18 at 17:13
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If $d$ is a common divisor of both numerator and denominator, then $d$ divides any linear combination of the two. In particular, $d$ divides $2(21n+4)-3(14n+3)=-1$. – Anurag A Jul 30 '18 at 17:14
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1Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Jul 30 '18 at 17:20
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1Duplicate of Prove the following fraction is irreducible – dxiv Jul 30 '18 at 17:41
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Dupe of this – Bill Dubuque Jan 27 '24 at 23:59
2 Answers
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Writing $(n,m)$ for the GCD of $m$ and $n$, immediately one has $(n,m)=(n,m-kn)$ for any $k\in\mathbb{Z}$. Then $$ (21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1. $$
Thus $21n+4$ and $14n+3$ are coprime, so their ratio is in lowest terms.
Ben West
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For starters such as yourself, you can begin by assuming otherwise. That is the fraction is reducible. So $\exists k \in \mathbb{N}, k > 1$ such that $k \mid 21n + 4, k \mid 14n+3\implies 21n+4 = ak, 14n+3 = bk$ for some natural numbers $a,b$. Thus: $42n+8 = 2ak, 42n+9 = 3bk\implies 3bk - 2ak = 1\implies k(3b-2a) = 1\implies k = 1$, contradicting the assumption that $k > 1$. This means $\text{gcd}(21n+4,14n+3) = 1$ or $\dfrac{21n+4}{14n+3}$ is irreducible.
DeepSea
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