Proof with inequalities

Solution 1:

Hint: Use contradiction. If $a>b$, then show $a$ is not less than $b+\frac{a-b}{2}$. This contradiction implies $a\leq b$.

Solution 2:

I'm a snob, so I like using contrapositive instead of contradiction whenever possible. The contrapositive would go like this: "if $a>b$, then there exists an $\epsilon>0$ such that $a\geq b+\epsilon$." Finding this epsilon should be easy with a number line.

Edit: the original poster figured it out in the comments, so here's the full argument: if $a>b$, then certainly $a-b>0$. Hence set $\epsilon=a-b$; then, since $b+\epsilon=b+a-b=a$, we certainly have $a\leq b+\epsilon=a$ (since $a=a$).

Solution 3:

I'm assuming this is from a first proof-based course, so apologies if this comes off as condescending.

Induction is harder to use in this case for precisely the reason you stated: what is the base case? We don't have any equivalent of some smallest nonzero number 1 with which to start like we do in $\mathbb{N}$. Maybe we could start with 0.0001? No, 0.00001 is smaller. And 0.000001 is smaller still. This sort of idea is known as the Well-Ordering Principle and is maybe not immediately clear, but it's worth thinking about. In particular, $\mathbb{R}$ (the real numbers) is not well-ordered, since there is no smallest element in $(0,1)$.

But in statements involving inequalities it's often a good idea to try to do things by contradiction. In that case, we'd assume $a > b$ and try to show that this is impossible. Well, if $a > b$ then $a - b > 0$. But we are given that $a < b + \epsilon$ for any $\epsilon > 0$. So what value should we choose for $\epsilon$ to find our contradiction?