Is there an elementary proof for the monotonicity of $\sqrt x$?

Question

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.

Answer 1

Let $0\leq a<b$.

Thus, $$\sqrt{b}-\sqrt{a}=\frac{b-a}{\sqrt{a}+\sqrt{b}}>0.$$

Answer 2

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 $a < b$.

Answer 3

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 $\sqrt{b}> \sqrt{a}$.

Answer 4

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 $x\text{-axis}$'.

Answer 5

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.