1. ## Proof involving inequalities

Let $\displaystyle a,b \in \mathbb{R}$. If $\displaystyle a \le b_{1}$ for every $\displaystyle b_{1} > b$, then $\displaystyle a \le b$. I'm fairly certain I have to apply some of the ordered field properties but I am not sure how. Thanks!

2. Suppose $\displaystyle a>b$; let $\displaystyle a-b=\epsilon$. Let $\displaystyle b_1=b+\epsilon/2$. Then $\displaystyle b_1 >b$ so by assumption $\displaystyle a \leq b_1$. So $\displaystyle a \leq b+\epsilon/2 =b+\frac{a-b}{2} = \frac{a+b}{2} < a$, which is absurd.

3. So the contradiction you arrived at is $\displaystyle a > a$, if I'm not mistaken. Thanks!

4. That is correct!

There may be a more abstract approach but this is pretty straightforward.

5. Originally Posted by Pinkk
Let $\displaystyle a,b \in \mathbb{R}$. If $\displaystyle a \le b_{1}$ for every $\displaystyle b_{1} > b$, then $\displaystyle a \le b$. I'm fairly certain I have to apply some of the ordered field properties but I am not sure how. Thanks!
http://www.mathhelpforum.com/math-he...her-proof.html. Look familiar? Clearly, if it is true for every $\displaystyle b_1>b$ then it is true for the above post.