# Math Help - Proof involving inequalities

1. ## Proof involving inequalities

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

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

3. So the contradiction you arrived at is $a > a$, 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 $a,b \in \mathbb{R}$. If $a \le b_{1}$ for every $b_{1} > b$, then $a \le b$. 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 $b_1>b$ then it is true for the above post.