let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

Hi, I have the following three problems, but I only want to know what the first one is like... I am sure I will then be able to figure out the other two.

The problem says:

"In the following, a, b are real numbers.

1) Let b>a+1. Show that there is an integer m such that a<m<b.

2) Let a,b > 0. Show that there is an integer n such that n*a > b.

3) Let a<b. Prove that there is an integer n such that n*b > n*a+1. Conclude that there is a rational number q such that a<q<b."

I would really appreciate it if you could give me a hand with the first one so that I can see how to solve the other 2 problems. Thanks a lot!

Re: let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

Let S be the set of all integers, x, with x< a. That set clearly is non-empty and has a as upper bound. By the "well ordered" property of the integers, any non-empty set of integers, having an upper bound, has a largest member. Let n be the largest member of this set. What can you say about n+1?

Re: let a<b with a, b being real numbers and b-a>1. Show that a<m<b for a random m

Quote:

Originally Posted by

**juanma101285** The problem says:

"In the following, a, b are real numbers.

1) Let b>a+1. Show that there is an integer m such that a<m<b.

I would really appreciate it if you could give me a hand with the first one so that I can see how to solve the other 2 problems.

**Theorm**: If $\displaystyle \alpha\in\mathbb{R}$ then $\displaystyle \left( {\exists j \in \mathbb{Z}} \right)\left[ {j \leqslant \alpha < j + 1} \right]$

The proof goes like this: Let $\displaystyle A=\{n\in\mathbb{Z}:n\le\alpha\}.$

That set exists because the integers are not bounded below.

Moreover the $\displaystyle \sup(A)\in\mathbb{Z}$. There is our $\displaystyle j$.

That proves that the greatest integer $\displaystyle \left\lfloor \alpha \right\rfloor $ exists.

Now for your question. Suppose that $\displaystyle b>a+1$.

The theorem tells us that $\displaystyle \left\lfloor a \right\rfloor\le a<\left\lfloor a \right\rfloor+1$.

So $\displaystyle a<\left\lfloor a \right\rfloor+1\le a+1<b$

**EDIT** this is a bit more. I did not see reply #2.