Squaring both sides in an epsilon-delta proof

Question

I am sorry if this is a silly question for all of you experienced mathematicians. I won't go through the whole problem since I am only concerned about one part. Why can't I do the following:

$ |\sqrt x-3| < \varepsilon \Longleftrightarrow |{(x-9)}| < {\varepsilon}^2 $

Maybe I have not thought it through enough, but will this work? Why or why not? Thanks

Answer 1

You are trying to square bot sides. You can do this without worrying about the inequality because both sides are already $> 0$.

But $$|\sqrt{x}-3|^2=(\sqrt{x}-3)^2=x-6\sqrt{x}+9\neq|x-9|$$

Answer 2

when you square , you have to square completely , i.e $$|\sqrt{x}-3|<\epsilon == |x+9-6\sqrt{x}|<\epsilon^2$$ moreover $$|{x}-9|=|\sqrt{x}-3||\sqrt{x}+3|<\epsilon|\sqrt{x}+3|$$

Answer 3

The two inequalities are not equivalent.

For instance

$$|\sqrt4-3|\stackrel?<2\text{ vs. }|4-9|\stackrel?<2^2.$$

Answer 4

Take a numerical example to show this is not true. For example, take $x=4$ and $\epsilon=1$, then $|2-3| \le 1$, but $|4-9|\not\le1$. If you want to square you have to square the whole term, that is $$ |\sqrt{x}-3|<\epsilon \iff |\sqrt{x}-3|^2 <\epsilon^2 \iff x - 6\sqrt{x} + 9 < \epsilon^2. $$

Answer 5

$(a+b)^2$ is not $a^2+b^2$.

It's not about remembering that it is like that. The point is that you must understand why. $(a+b)c = ac+bc$ for real numbers $a,b,c$ not because the police will arrest you if you don't accept it, but because real numbers are supposed to represent lengths, and the area of a rectangle is intuitively the product of the length and breadth, and putting an $a \times c$ rectangle and a $b \times c$ rectangle together in the right way yields an $(a+b) \times c$ rectangle, which ought to have the sum of the areas of the separate rectangles.

This in turn gives $(a+b)(a+b) = a(a+b)+b(a+b) = aa+ab+ba+bb = a^2 + 2ab + b^2$. No mathematical police!