# Have I done this 'show that' problem correctly?

• Jul 8th 2010, 10:25 PM
Glitch
Have I done this 'show that' problem correctly?
The question:

Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

I did:

$\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$
$\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

Therefore:
$\displaystyle \sqrt{a} <= \sqrt{b}$

My textbook doesn't have answers for these types of questions, so I can't check.
• Jul 8th 2010, 10:43 PM
undefined
Quote:

Originally Posted by Glitch
The question:

Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

I did:

$\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$
$\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

Therefore:
$\displaystyle \sqrt{a} <= \sqrt{b}$

My textbook doesn't have answers for these types of questions, so I can't check.

Well, I think the proof is circular because it relies on being able to take the square root of both sides of an inequality, in other words: for $\displaystyle \displaystyle f(x)=\sqrt{x}$ and for non-negative reals $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b$ we have $\displaystyle \displaystyle a\le b \implies f(a) \le f(b)$, which is what you're trying to prove (and which is not true for general $\displaystyle \displaystyle f(x)$).

All we're really saying here is that $\displaystyle \displaystyle f(x)=\sqrt{x}$ is increasing. The "usual" way is to take the derivative and show that it's greater than equal to zero on that interval. This question is posted in the Pre-Calculus subforum; does it makes sense to take the derivative given where you are in the textbook?
• Jul 8th 2010, 10:45 PM
Glitch
Quote:

Originally Posted by undefined
This question is posted in the Pre-Calculus subforum; does it makes sense to take the derivative given where you are in the textbook?

Nope, this chapter is pre-calculus only, and hasn't mentioned derivatives yet.
• Jul 8th 2010, 10:47 PM
Also sprach Zarathustra
Hint:

$\displaystyle b-a=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})$
• Jul 9th 2010, 10:51 AM
earboth
Quote:

Originally Posted by Glitch
The question:

Show that if $\displaystyle 0 <= a <= b$, then $\displaystyle \sqrt{a} <= \sqrt{b}$

I did:

$\displaystyle \sqrt{0} <= \sqrt{a} <= \sqrt{b}$
$\displaystyle 0 <= \sqrt{a} <= \sqrt{b}$

Therefore:
$\displaystyle \sqrt{a} <= \sqrt{b}$

My textbook doesn't have answers for these types of questions, so I can't check.

I'm not sure if this approach is necessary any more, but nevertheless here it is:

If $\displaystyle a\leq b~\implies~\sqrt{a} \leq \sqrt{b}$

Substitute $\displaystyle b = a + k, k \geq 0$ and you'll get:

$\displaystyle \sqrt{a} \leq \sqrt{a + k} \leq \sqrt{a} + \sqrt{k}$

Finally you have $\displaystyle 0\leq \sqrt{k}$ which is true because the square-root of a positive number is positive too.
• Jul 10th 2010, 06:59 AM
HallsofIvy
It's not "unecessary", it is wrong. You want to prove that $\displaystyle \sqrt{a}\le \sqrt{b}$, yet the first line of your "proof" begins "$\displaystyle \sqrt{a}\le \sqrt{a+ k}$ which is exactly the same as what you want to prove. And, you are also assuming that $\displaystyle \sqrt{a+ k}\le \sqrt{a}+ \sqrt{k}$ which is a much harder theorem that was was asked here.