Proofs involving real numbers

[Max.Maxim]

Let m,n be real numbers. Prove that if n>m>0, then m+1/n+1>m/n
This is what I did but I'm stuck on how to finish.
Proof;
Let n, m, and 0 be real numbers.

 

[Dr.Peterson]

Did you mean (m+1)/(n+1) > m/n, that is, [MATH]\frac{m+1}{n+1}>\frac{m}{n}[/MATH], or what you wrote, which is [MATH]m + \frac{1}{n} + 1 > \frac{m}{n}[/MATH]?

One way to start would be to rewrite the goal as [MATH]\frac{m+1}{n+1}-\frac{m}{n}>0[/MATH].

 

[Steven G]

That really is a poor start--sorry.

You should at least had:

Proof: Given n and m are real numbers. Assume n>m>0.

Do you understand what this fact is saying? It says that given any positive fraction, if you add 1 to both the numerator and denominator then this new fraction will be larger than the original one. Do you believe that this is true? Why?

 

[lookagain]

Let m,n be real numbers. Prove that if n>m>0, then m+1/n+1>m/n
This is what I did but I'm stuck on how to finish.
Proof;
Let n, m, and 0 be real numbers.

Max.Maxim, you state that is what you did, but write down for us exactly
the statement of the problem (before you even starting working on it).

 

[JeffM]

Let m,n be real numbers. Prove that if n>m>0, then m+1/n+1>m/n
This is what I did but I'm stuck on how to finish.
Proof;
Let n, m, and 0 be real numbers.

Well you really did not do anything toward solving the problem given except repeat the givens.

One way to start is to check if what you want to prove is true.

[MATH]n = 2 \text { and } m = 1 \implies n > m > 0 \text { and } \dfrac{1 + 1}{2 + 1} = \dfrac{2}{3} > \dfrac{1}{2}[/MATH].

Sometimes doing that will give you a clue, and it will certainly stop you from wasting time on something that is blatantly untrue.

The standard way to do this sort of problem is

[MATH]n > m > 0 \text { and } x = \dfrac{m + 1}{n + 1} - \dfrac{m}{n}.[/MATH]
Now simplify and see what you can learn that may be helpful.

 

[Steven G]

I like to think of this problem the following way. Suppose you take a test where you got m questions correct out of n question (with m<n). Now your teacher gives you one more problem and you get 100% on that problem, that is you got 1/1. Since the rest of the test you had less than 100% and now you got 100% on this last unexpected problem your score had to increase. That is if 0<m<n, then m/n < (m+1)/(n+1)

 

[HallsofIvy]

Are m and n general real numbers, as you wrote, or are they required to be positive real numbers?

If they are allowed to be negative then the statement is not true! So your proof will have to use the fact that m and n are positive.