Can Integers Be Found Between Scaled Real Numbers?

[saadsarfraz]

Homework Statement

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

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

Homework Equations

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

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[tex]\subset[/tex]Q and its safe to assume that Z[tex]\subset[/tex]R is always true but how do I exactly show that m lies in Z.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Mark44]

For a) you can't start off with

Since n(b-a) > 1

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.

 

[saadsarfraz]

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

 

[HallsofIvy]

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.