Proofs with real numbers

[EqualElement]

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]

How are you confused?

 

[EqualElement]

How are you confused?

I understand the question but don't really know how to prove it.

 

[cbarker1]

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)?

 

[EqualElement]

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

 

[cbarker1]

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

 

[EqualElement]

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

 

[kaliprasad]

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]

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?

 

[HOI]

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".