April 29th, 2014, 01:02 PM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  proooooooof
Proved that there is no two issues are natural different A, B so that: 1/A +1/B=1 
April 29th, 2014, 01:47 PM  #2 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Not fully sure if I understand your question correctly but let wlog, $\displaystyle a < b$. Then $\displaystyle a \ne 1$ since $\displaystyle 1/1+1/B \ne 1$. $\displaystyle 1/2 + 1/3 < 1$. As A or B increase, 1/A+1/B decrease so $\displaystyle 1/A+1/B \ne 1$. Only when A = B is allowed, A = B = 2 works. 
April 29th, 2014, 02:17 PM  #3 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
Sorry double post, I got redirected to posting the reply so I thought it hadn't posted. Could someone remove this one?

April 29th, 2014, 04:23 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
As I recall, no finite sum of fraction with coprime (natural) denominators can be equal to a natural number. That's slightly different, but has no restriction on the numerator.

April 29th, 2014, 04:59 PM  #5 
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 
$\displaystyle x,y \in\mathbb{N}^*, x\neq y\\\;\\\dfrac{1}{x}+\dfrac{1}{y}=1 \Rightarrow y=\dfrac{x}{x1}\\\;\\If\ x\geq3 \Rightarrow \dfrac{x}{x1}\ is\ irreducible.\\\;\\\dfrac{x}{x1} \not{\in}\ \mathbb{N}^*$

April 30th, 2014, 12:21 AM  #6 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
If either $A$ or $B$ is equal to $1$, then $\displaystyle \frac1A+\frac1B>1$. So for $\displaystyle \frac1A+\frac1B\leqslant1$ we must have $A,B\geqslant2$. Then if one of $A$ and $B$ is greater than $2$, $\displaystyle \frac1A+\frac1B < \frac12+\frac12=1$.
