Proving that if $a$ and $b$ are two positive real numbers, then $a < b \iff \sqrt{a} < \sqrt{b} \iff \frac{1}{a} > \frac{1}{b}$

Question

I am so lost. We have been learning about set, sequences, infimum and supremum over the past week or two, and I have no idea how to apply it to my homework question. Here it is:

Suppose $a$ and $b$ are two positive real numbers. Prove that $$a < b \iff \sqrt{a} < \sqrt{b} \iff \frac{1}{a} > \frac{1}{b}$$

Answer 1

We have that

\begin{align} a<b&\iff b-a>0\\&\iff (\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})>0\\&\iff \sqrt{b}-\sqrt{a} >0\\&\iff \sqrt{a}<\sqrt{b}. \end{align}

In a similar way

\begin{align} \frac 1b<\frac 1a&\iff \frac 1a-\frac 1b>0\\&\iff ab\left(\frac 1a-\frac 1b\right)>0\\&\iff b-a>0 \\&\iff a<b. \end{align}

Answer 2

Let:

$$f_1:x \to \sqrt{x} \ \ \text{and} \ \ f_2:x \to \dfrac{1}{x}$$

(from $\mathbb{R}_+^*$ to itself).

The two properties to be established can be written in the following way:

$a<b \iff f_1(a)<f_1(b) \ \ $ (resp. $a<b \iff f_2(a)>f_2(b)),$

where we recognize the increasing (resp. decreasing) property for a bijective function $f$.

Thus we have to check that $f_1$ (resp. $f_2$) are bijective and increasing (resp. bijective and decreasing):

• bijectivity is checked by exhibiting an inverse of $f_1$ and $f_2$ (easy),

• increasing (resp. decreasing) property by proving that derivatives $f'_1$ (resp. $f'_2$) are always $>0$ (resp. $<0$) on $\mathbb{R}_+^*$ (easy as well).