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Ptove that 21n+4/14n+3 is irreduciblefor every powitive integer n |
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Answer» Step-by-step explanation: Show that 21n+414n+3,n∈N cannot be reduced? Scan & PAY on Amazon. There are two ways that come to mind. One solution is mine, and the other I’ve taken from the IMO Aops website. For those interested, the problem can be found here: . Euclid’s Algorithm: This one is the one I’m proud of, since I came up with it. YES, I’m sure that everyone ELSE planned on using this one as well, but hey, a solution is a solution. Anyways, let’s get into it. 21n+414n+3 By definition, a quotient is reducible if both the numerator and denominator have a greatest COMMON factor that is greater than 1 . We can use Euclid’s method to determine the GCF between 21n+4 and 14n+3 . If the GCF is shown to be 1 , then we have our proof. So by the Division Theorem: 21n+4=(14n+3)q+r Now, apply Euclid’s Method. 21n+4=(14n+3)⋅1+(7n+1) 14n+3=(7n+1)⋅2+1 7n+1=1(7n+1)+0 So we see that gcf(21n+4,14n+3)=1 , MEANING that the quotient is irreducible this is the answer |
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