# in this epsilon delta continuity proof, why do inequalities change to equalities

i'm trying to follow this proof of continuity of the function $f(x)=3x+4$ using Delta epsilon.

On minute 5:29, written in green, an inequality suddenly changes to an equality and i'm not sure why.

It would seem that for the definition to hold, you would need $$|f(y)-f(x)|<\epsilon$$ not $$|f(y)-f(x)|=\epsilon$$

And even if that was valid i'm not sure how he got there.

Is this correct? if so why?

• From the picture, it looks like the author fixed some arbitrary $\varepsilon>0$ and defined $\delta := \frac{\varepsilon}{3}$. Hence the equality – joeb Mar 14 '17 at 1:17
• Hi, that's what he did, but i'm not sure why that leads to an equality, maybe i'm missing some concept in the definition? – Joaquin Brandan Mar 14 '17 at 1:19
• That $3\cdot\frac{\epsilon}{3}$ refers to $3\delta$. It does not mean that $<$ is replaced by $=$. – Juniven Mar 14 '17 at 1:19
• oh i dont mean that, i mean on the red arrows, the first one is an inequality, after that the author defines delta as $\epsilon / 3$, after doing that he replases delta with $\epsilon / 3$ on the second red arrow, and there he changes the inequality from the first red arrow into an equality. – Joaquin Brandan Mar 14 '17 at 1:21

What you are seeing is a long chain of relations which would be read left-to-right if space permitted: $$|f(y) - f(x)| = |(3y+4)-(3x+4)| = |3y-3x| = 3 |y-x| < 3 \delta = 3\cdot \frac{\epsilon}{3} = \epsilon$$ Each relation relates only the two expressions on either side. Since $\delta = \frac{\epsilon}{3}$, we know that $3 \delta = 3\cdot \frac{\epsilon}{3}$ at the second-to-last step.
But as they say, “a chain is only as strong as its weakest link.” So ignoring the intermediate steps by transitivity, we have $$|f(y) - f(x)| < \epsilon$$ and that's exactly what was to be shown.
• hi, thanks, why it it not written $|f(y)−f(x)|=|(3y+4)−(3x+4)|=|3y−3x|=|3(y−x)|<3 \delta <3⋅ \epsilon 3< \epsilon$ – Joaquin Brandan Mar 14 '17 at 1:24
• @JoaquinBrandan Because $\delta$ is not less than $\frac{\epsilon}{3}$; it's equal to it. And that's an em-dash (–) at the end which separates your comment from your signature, not an extraneous minus sign ($-$). – Matthew Leingang Mar 14 '17 at 1:25
• aha! got it!. Would this be correct then? again ignore the minus sign at the end. $|f(y)−f(x)|=|(3y+4)−(3x+4)|=|3y−3x|=|3(y−x)|<3 \delta$ where $3\delta = 3⋅ \epsilon /3 = \epsilon$. – Joaquin Brandan Mar 14 '17 at 1:27