I am about teaching a course on inequalities to secondary school children and I find we'll have to use at some point that if $$0<a<b,$$ then $$\sqrt a<\sqrt b.$$
Therefore, I've been thinking of how one would go about explaining why this is true to these children, and the only way I know uses real analysis -- and they've definitely taken no calculus yet.
Hence, do you know of a way to prove this result using elementary means?
Thank you.
Let $0\leq a<b$.
Thus, $$\sqrt{b}-\sqrt{a}=\frac{b-a}{\sqrt{a}+\sqrt{b}}>0.$$
If $0 \le x \le y$ then $x^2 \le y^2$ since $y^2 - x^2 = (y - x)(y + x) \ge 0$.
Then assume that $\sqrt{b} \le \sqrt{a}$. Then applying above inequality (with $x = \sqrt{b}$ and $y = \sqrt{a}$) we get $b \le a$. This contradicts with $a < b$.
I'm not sure what you would mean by elementary. How "elementary" do you want?
Can you use the fact that "$(x- y)(x+ y)= x^2- y^2$" (which can, of course, be shown by just multiplying x- y and x+ y). Your students would also need to know that "positive times positive is positive", "negative times negative is positive", and "positive times negative is negative".
If 0< a< b, then b- a> 0. with $x= \sqrt{b}$ and $y= \sqrt{a}$, we can factor $b- a= (\sqrt{b}- \sqrt{a})(\sqrt{b}+ \sqrt{a})> 0$. Since $\sqrt{a}$ and $\sqrt{b}$ are positive by definition, $\sqrt{a}+ \sqrt{b}$ is positive and it follows that $\sqrt{b}- \sqrt{a}$ is positive so $\sqrt{b}> \sqrt{a}$.
Prove that the the function $f(x) = x^2$ on $[0,+\infty]$ is increasing.
Use some online tools to cut/paste/manipulate its graph and then display
Mention that using the paint tool eraser you can remove the red below the
'what was the $y\text{-axis}$ but is now the $x\text{-axis}$'.
Presumably you've already established that multiplication by a positive quantity preserves inequalities.
If $a < b$ then $$ aa < ab < bb . $$
What you want for square roots follows easily.
