February 4th, 2017, 01:28 PM  #1 
Newbie Joined: Feb 2017 From: UK Posts: 2 Thanks: 0  Fractions issue
Hi, I am sure I am missing a trick here, but could someone please help me understand why, $\displaystyle \frac{4n}{4n3}=1+\frac{3}{4n3}$ Thank you in advance, Matt 
February 4th, 2017, 01:39 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,497 Thanks: 580 
4n=4n3+3. The rest is easy.

February 4th, 2017, 01:41 PM  #3 
Newbie Joined: Feb 2017 From: UK Posts: 2 Thanks: 0 
Cheers mate 
February 4th, 2017, 10:55 PM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176 
$\displaystyle n\in \mathbb{N} \\\;\\ \dfrac{4n}{4n3}=\dfrac{4n3+3}{4n3}=\dfrac{4n3}{4n3}+\dfrac{3}{4n3} = 1 +\dfrac{3}{4n3}$ 

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