Natural numbers, integars

1. Feb 21, 2009

1. The problem statement, all variables and given/known data

Let a,b $$\in$$R with a < b. and let n $$\in$$N where n(b-a) > 1.

a) How do you know that such an n must exist?
b) Show that there exists m $$\in$$Z where a < m/n < b
c) Show that there exists some irrational c where a < c < b (Hint:rational + irrational = irrational.)

2. Relevant equations

see above btw N is for natural numbers, Z for integers and R for real numbers

3. The attempt at a solution

a) Since n(b-a) > 1 , n > 1/(b-a) and since b does not equal a there is an n which exists OR can we say that nb > na and the n's cancel out to give us the condition given in the question.

b) we already know that n is natural and N$$\subset$$Q and its safe to assume that Z$$\subset$$R is always true but how do I exactly show that m lies in Z.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Feb 22, 2009
2. Feb 22, 2009

Staff: Mentor

For a) you can't start off with
You have to show that such an n exists to make that true.

For b, your task is not to show that "m lies in Z." You have to show that there exists an integer m such that m/n is between a and b.

3. Feb 22, 2009

so for part a) has something to do with the archimedean property?

Last edited: Feb 22, 2009
4. Feb 22, 2009

HallsofIvy

Staff Emeritus
Yes, Mark44's point is that you cannot start by asserting that n(b-a)> 1. What you can do is start with your second statement: since b-a> 0, 1/(b-a) is a positive real number and, by the Archimedean property, ...
For b) Show that there exists m Z where a < m/n < b
note that since n(b-a)= nb- na> 1, there must exist an integer between nb and na.