# proofs with real numbers

#### EqualElement

##### New member
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
How are you confused?

#### EqualElement

##### New member
How are you confused?
I understand the question but don't really know how to prove it.

#### Cbarker1

##### Active member
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
okay m=1 , n=2 would make it true.

#### Cbarker1

##### Active member
Any other example? Just keep making examples to see a pattern...

#### EqualElement

m=2, n=3
m=3, n =4
m=12, n=69.

##### Well-known member
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 $\frac{m+1}{n+1}> \frac{m}{n}$

• topsquark, Jameson and Ackbach

#### HallsofIvy

##### Well-known member
MHB Math Helper
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
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".