My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 29th, 2014, 02:02 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
mared is offline  
 
April 29th, 2014, 02:47 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.
Hoempa is offline  
April 29th, 2014, 03:17 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?
Hoempa is offline  
April 29th, 2014, 05:23 PM   #4
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,700
Thanks: 2682

Math Focus: Mainly analysis and algebra
As I recall, no finite sum of fraction with co-prime (natural) denominators can be equal to a natural number. That's slightly different, but has no restriction on the numerator.
v8archie is offline  
April 29th, 2014, 05:59 PM   #5
Senior Member
 
aurel5's Avatar
 
Joined: Apr 2014
From: Europa

Posts: 584
Thanks: 177

$\displaystyle x,y \in\mathbb{N}^*, x\neq y\\\;\\\dfrac{1}{x}+\dfrac{1}{y}=1 \Rightarrow y=\dfrac{x}{x-1}\\\;\\If\ x\geq3 \Rightarrow \dfrac{x}{x-1}\ is\ irreducible.\\\;\\\dfrac{x}{x-1} \not{\in}\ \mathbb{N}^*$
aurel5 is offline  
April 30th, 2014, 01:21 AM   #6
Senior Member
 
Olinguito's Avatar
 
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 $2$, $\displaystyle \frac1A+\frac1B < \frac12+\frac12=1$.
Olinguito is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
proooooooof



Thread Tools
Display Modes






Copyright © 2019 My Math Forum. All rights reserved.