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proofs with real numbers

EqualElement

New member
May 8, 2020
4
Let m,n be real numbers. Prove that if n>m>0 , then (m+1)/(n+1) > m/n
I'm currently confuse in this one help will be very much needed
 

Cbarker1

Active member
Jan 8, 2013
236
How are you confused?
 

EqualElement

New member
May 8, 2020
4

Cbarker1

Active member
Jan 8, 2013
236
Let's start with several examples. What do you choose for n and m which m must be bigger than n and be positive for both (m and n)?
 
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EqualElement

New member
May 8, 2020
4
okay m=1 , n=2 would make it true.
 

Cbarker1

Active member
Jan 8, 2013
236
Any other example? Just keep making examples to see a pattern...
 

EqualElement

New member
May 8, 2020
4
m=2, n=3
m=3, n =4
m=12, n=69.
 

kaliprasad

Well-known member
Mar 31, 2013
1,283
let us calculate $\frac{m+1}{n+1}- \frac{m}{n}$
= $\frac{n(m+1) - m(n+1)}{m(n+1)}$
= $\frac{n - m}{m(n+1)}$
as n > m >0 so both numerator and denominator positive and hence

$\frac{m+1}{n+1}- \frac{m}{n}> 0$

or $\frac{m+1}{n+1}> \frac{m}{n}$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Since m and n are positive numbers, so are n and n+ 1 so you can eliminate the fractions by multiplying both sides by n and n+ 1 without changing the inequality.. That gives you n(m+1)> m(n+1) so that nm+ n> mn+ m. Can you finish?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
What I showed above was that "if $\frac{m+1}{n+1}> \frac{m}{m}$ then n> m. What you want to prove is the other way around- just reverse every step. From n> m, mn+ n> mn+ m.
n(m+1)> m(n+ 1). Dividing both sides by the positive number n and n+ 1, $\frac{m+1}{n+1}> \frac{m}{n}$.

It is often useful to see how to prove something by working backward, from the conclusion to the hypothesis. As long as every step is "reversible", it isn't necessary to actually show the reverse. That is called "synthetic proof".