Proof by Contradiction: Showing a ≤ b when a ≤ b1 for every b1 > b

[Mr Davis 97]

Homework Statement

Let $a,b \in \mathbb{R}$. Show if $a \le b_1$ for every $b_1 > b$, then $a \le b$.

Homework Equations

The Attempt at a Solution

We will proceed by contradiction. Suppose that $a \le b_1$ for every $b_1 > b$, and $a > b$. Let $b_1 = \frac{a+b}{2}$. We see that $b_1>b$: $a>b \implies \frac{a+b}{2}> b \implies b_1 >b$. Hence, by the hypothesis, it must be true that $a \le b_1$. But since $a \le b_1 \implies a \le \frac{a+b}{2} \implies a \le b$, we have reached a contradiction. We simultaneously have that $a \le b$ and $a >b$. Hence, it must be the case that the original statement is true, and we are done.

 

[fresh_42]

Homework Statement

Let $a,b \in \mathbb{R}$. Show if $a \le b_1$ for every $b_1 > b$, then $a \le b$.

Homework Equations

The Attempt at a Solution

We will proceed by contradiction. Suppose that $a \le b_1$ for every $b_1 > b$, and $a > b$. Let $b_1 = \frac{a+b}{2}$. We see that $b_1>b$: $a>b \implies \frac{a+b}{2}> b \implies b_1 >b$. Hence, by the hypothesis, it must be true that $a \le b_1$. But since $a \le b_1 \implies a \le \frac{a+b}{2} \implies a \le b$, we have reached a contradiction. We simultaneously have that $a \le b$ and $a >b$. Hence, it must be the case that the original statement is true, and we are done.

Is o.k.