February 4th, 2017, 02:28 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, 02:39 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,436 Thanks: 562 
4n=4n3+3. The rest is easy.

February 4th, 2017, 02:41 PM  #3 
Newbie Joined: Feb 2017 From: UK Posts: 2 Thanks: 0 
Cheers mate 
February 4th, 2017, 11:55 PM  #4 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176 
$\displaystyle n\in \mathbb{N} \\\;\\ \dfrac{4n}{4n3}=\dfrac{4n3+3}{4n3}=\dfrac{4n3}{4n3}+\dfrac{3}{4n3} = 1 +\dfrac{3}{4n3}$ 

Tags 
fractions, issue 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Need help on this issue  Frontier111  Physics  9  May 13th, 2016 04:18 AM 
Partial fractions issue  Chickwolf  Linear Algebra  9  November 12th, 2015 06:03 PM 
Tax issue  bigcity  Algebra  3  January 2nd, 2012 12:09 PM 
Simplifying fractions with variables in fractions  nova3421  Algebra  1  September 9th, 2009 11:03 AM 
complexity issue  baxy7  Applied Math  1  June 1st, 2009 06:43 AM 