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Given an integer $n$, can the fraction $$\frac{2n}{4n-1}$$ be simplified?

I am not sure how I can show whether this is true or not or even begin to investigate it without writing out lots of terms.

Of course, the top is even and the denominator is odd but that's not sufficient to make any conclusion... I was thinking perhaps that $4n-1$ is not divisible by $2$ or by $n$ but then if $n$ is not prime, that argument doesn't work I don't think.

Anne Bauval
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PhysicsMathsLove
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    Hint: $\gcd(a,b)=\gcd(a-b,b)$. – Anne Bauval Jan 27 '24 at 22:53
  • For $~d \in \Bbb{Z_{\geq 2}},~$ if $~d~$ divides $~2n,~$ then $~d~$ divides $~4n,~$ which implies that $~d~$ does not divide $~4n-1.$ – user2661923 Jan 27 '24 at 23:00
  • $$2\cdot 2n - (4n-1) = 1$$ If $d$ divides both the numerator $2n$ and the denominator $(4n-1)$, then $d$ also divides the RHS $1$. So $d$ can only be $1$. – peterwhy Jan 27 '24 at 23:04
  • If $d\neq 1$ divides $2n$ then $d$ divides $4n$ but $d$ doesn't divide $-1$ therefore $d$ doesn't divide $4n-1$. – user Jan 27 '24 at 23:11
  • That is $\frac{2n}d =k \implies \frac{4n-1}d= \frac{4n}d-\frac 1d =2k-\frac 1d$. – user Jan 27 '24 at 23:26
  • By the linked dupe $,\gcd(a,ab-1) = \gcd(a,-1) = 1.\ $ OP is case $,a,b = 2n,2\ \ $ – Bill Dubuque Jan 27 '24 at 23:28
  • When you say "can be simplified" are you asking if $\frac {2n}{4n-1}$ is always in "lowest terms"? Or are you asking if we can rewrite $\frac {2n}{4n-1}$ as something else (maybe $\frac 2{4 -\frac 1n}$)? Or do you mean something else entirely. – fleablood Jan 27 '24 at 23:57
  • I suppose the easiest "explain as though I were 5" answer would be: Whatever (other than one) can divide $n$ will divide $4n$ and so will not divide $4n-1$. And $2$ doesn't divide $4n-1$. So (other than one) nothing that divides the numerator will divide the denominator. So the numerator and denominator have no factors (other than one) in common. – fleablood Jan 28 '24 at 00:02
  • "but then if n is not prime, that argument doesn't work I don't think." It doesn't work as easily but it does work. $n$, and all its divisors divide $4n$, so $n$ and all if its non-trivial (not $1$) divisors fail to divide $4n-1$. Argument works. – fleablood Jan 28 '24 at 00:07

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Suppose the numerator and denominator have a common factor $d$. Then $d$ divides both $2n$ and $4n-1$, and so $d$ also divides $$2\cdot(2n)-(4n-1)=1,$$ and so $d=\pm1$. This shows that the fraction cannot be simplified any further.

Servaes
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    Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 27 '24 at 23:29
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    @BillDubuque I disagree with the assessment that this is a duplicate. – Servaes Jan 27 '24 at 23:30
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Since $\text{gcd}(2n, 4n-1) = 1$, you cannot.

probafds123
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  • But why is this the case? – PhysicsMathsLove Jan 27 '24 at 22:54
  • Because if you want to simplify the fraction, you should divide the numerator and denominator with the same number. If a great common divisor is 1, that means that you can only divide the numerator and denominator simultaneously with 1. – probafds123 Jan 27 '24 at 22:56
  • Sorry I meant, why is $gcd(2n, 4n-1) = 1$? – PhysicsMathsLove Jan 27 '24 at 22:57
  • @PhysicsMathsLove Use twice my hint above: $\gcd(4n-1,2n)=\gcd(2n-1,2n)=\gcd(-1,2n)=1.$ – Anne Bauval Jan 27 '24 at 22:58
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    If $2n$ has a non-trivial (not equal to $1$) divisor, then $4n$ has the same divisor. But after subtracting $1$ we lose that divisor. – Aig Jan 27 '24 at 22:59
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    Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 27 '24 at 23:29