1. ## analysis midterm

Sorry for all the questions, but I have a midterm coming up and I'm having problems with these questions:

1. Prove that if a < b, then there is an irrational number x such that
a < x < b. Hint: first show that if q not equal to zero is rational, then q+ root(2) and q*root(2) are irrational.

2. Prove that if a > 0, there is an integer n such that 1/n < a < n.

3. For real numbers a and b, prove that if a <= b + 1/n for all positive integers n, then a <= b.

4. For each real number a, let the set S_a be the set of all rational numbers less than a. Prove that for each real number a, sup S_a = a.

Thanks.

2. Originally Posted by eigenvector11
1. Prove that if a < b, then there is an irrational number x such that
a < x < b. Hint: first show that if q not equal to zero is rational, then q+ root(2) and q*root(2) are irrational.
If $a so there exists rational $x$ so that $a-\sqrt{2} < x < b-\sqrt{2}$ thus $a < x +\sqrt{2} < b$ that means $x+\sqrt{2}$ is irrational. Q.E.D.

2. Prove that if a > 0, there is an integer n such that 1/n < a < n.
Given $a>0,1>0$ by Archimedean ordering there is nartual $n_1$ so that $a< n_1\cdot 1 = n_1$. Since $1/a>0,1>0$ by Archimedean ordering there is natural $n_2$ so that $1/a < n_2\cdot 1 = n_2$ thus $1/n_2 < a$. Let $n=\max\{n_1,n_2\}$ then $1/n \leq 1/n_2 < a < n_1 \leq n$. Q.E.D.

3. For real numbers a and b, prove that if a <= b + 1/n for all positive integers n, then a <= b.
Assume not, then $a > b$ that means $a-b>0$ (by definition). Thus by Archimedian ordering with $1/(a-b) > 0,1>0$ there is natural $n$ so that $1/(a-b) < n$ thus $a-b > 1/n \implies a > b+1/n$ which is a contradiction. Q.E.D.

4. For each real number a, let the set S_a be the set of all rational numbers less than a. Prove that for each real number a, sup S_a = a.
The set (non-empty) $S_a$ has an upper bound, i.e. $a$. Argue that this is the least upper bound.