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

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

The question:

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

I did:

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

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

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

2. Originally Posted by Glitch
The question:

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

I did:

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

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

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 $\displaystyle f(x)=\sqrt{x}$ and for non-negative reals $\displaystyle a$ and $\displaystyle b$ we have $\displaystyle a\le b \implies f(a) \le f(b)$, which is what you're trying to prove (and which is not true for general $\displaystyle f(x)$).

All we're really saying here is that $\displaystyle f(x)=\sqrt{x}$ 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?

3. 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.

4. Hint:

$b-a=(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})$

5. Originally Posted by Glitch
The question:

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

I did:

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

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

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

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

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

Finally you have $0\leq \sqrt{k}$ which is true because the square-root of a positive number is positive too.

6. It's not "unecessary", it is wrong. You want to prove that $\sqrt{a}\le \sqrt{b}$, yet the first line of your "proof" begins " $\sqrt{a}\le \sqrt{a+ k}$ which is exactly the same as what you want to prove. And, you are also assuming that $\sqrt{a+ k}\le \sqrt{a}+ \sqrt{k}$ which is a much harder theorem that was was asked here.