# Prove that $0<a<b\implies \sqrt a < \sqrt b$

I'm working with a textbook that defines inequalities as:

$$a < b \implies b-a \in P$$, where P is the set of all positive numbers.

Also, if $$a \in P$$ and $$b \in P$$ then $$a+b \in P$$ and $$ab \in P$$

Using these properties, prove:

$$0

(taking positive roots)

The actual problem the book is asking to prove:

$$a < \sqrt{ab} < b$$

But getting the first part seems to be where the challenge lies.

• HINT: $b-a = (\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})$. – Arturo Magidin Apr 5 '19 at 4:02
• Thanks for the help, this helped me solve the problem. I have noticed that I'm very bad at finding useful factorizations, which has made me struggle with other problems as well. – David Davidson Apr 8 '19 at 21:35

Your textbook asks you to prove that $$a<\sqrt{ab} for $$a. This can be done in the following way: First we analyze $$a<\sqrt{ab}$$ and then $$\sqrt{ab}.

Part I: (Prove that $$a<\sqrt{ab}$$)

$$a^2 (square both sides)

Since $$a this is clearly true ($$a\times a)

Part II: (Prove that $$\sqrt{ab})

$$ab (square both sides)

Since $$a this is clearly true ($$a\times b < b\times b$$)

We conclude that $$a<\sqrt{ab}$$ and that $$\sqrt{ab} so that means $$a<\sqrt{ab}

Q.E.D.