Prove ax+b is uniformly continuous on R

Question

Let f(x) = ax+ b Let any x,y ∈ R. Let ε > 0 be abritraty

|f(x) – f(y)| = |ax+b – (ay+b)|

= |ax-ay| < ε


= |a(x-y)| < ε 


≤ |a||x-y| < ε


If |a||x-y| < ε then |x-y| < ε / |a|

Let δ = ε / |a| > 0 |f(x)-f(y) < ε for |x-y|< δ where δ = ε / |a|

Answer 1

I agree with your answer. Here is how I would have formulated it.

We need to prove that $$ \forall \epsilon > 0 \ \exists \delta : \forall x,y \in \mathbb{R} \ |x-y| < \delta \Rightarrow |f(x) - f(y)|< \epsilon $$

Now let $\epsilon > 0$ and assume $a \not = 0$, choose $\delta = \frac{\epsilon}{|a|}$, then

$$ |x-y| < \delta = \frac{\epsilon}{|a|} \Leftrightarrow |a| |x - y | = | ax - ay| = |(ax - b) -(ay - b)| < \epsilon $$

Thus $\delta$ does not depend on $x,y$.

As pointed out in the comment under this answer, for $a = 0$, you can choose $\delta$, for example, to be $1$.