Prove that for all $e \gt 0$, there exists $d \gt 0$ such that for all $x \gt 0$, if $x \lt d$, then ${x^2} \lt e$

Question

I tried to prove this using the contrapositive, but got stuck on $x^2\ge e \rightarrow x\ge d$. But how would I go from there?

Answer 1

Contrapositive of $P\to Q$ is $\lnot Q\to \lnot P$.

Contrapositive of $x>d \to x^2<e$ is $x^2\ge e \to x\le d$.

Since it is $\forall x>0$, one can let $x=d+e$, then: $x^2=(d+e)^2\ge e \not\to x=d+e\le d$.

Answer 2

It's been quite obvious... I think. If you fix such an $e$, then find an $x$ such that $x^2 < e$. You'll always find it. Fix that $x$. Then take a $d$ such that $d < x$. Again, you'll always find such $d$. I think from Archimedean property and density of $\mathbb R^+$, it follows.

Answer 3

• The contrapositive of $x > d \implies x^2 < e$ is $x^2 \ge e \implies x \le d$.

• The statement of the question states that for very large $x$, it is bounded from above which is clearly not true.

• For example, let $e=1$, for any positive $d$, let $x = \max(2d, e)$, then we have $x > d$ and $x^2 = \max(4d^2, e^2)= \max(4d^2, 1) \ge 1$

Answer 4

You first write down a number $d$ such that $d^2 < e$. Then If $0 < x < d$ you have $x^2 < d^2 < e$.

So try to find this number $d$. You might consider breaking into cases $e < 1$ and $e \geq 1$.