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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:

$0<a<b \implies \sqrt a < \sqrt b$

(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.

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    $\begingroup$ HINT: $b-a = (\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})$. $\endgroup$ – Arturo Magidin Apr 5 '19 at 4:02
  • $\begingroup$ 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. $\endgroup$ – David Davidson Apr 8 '19 at 21:35
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Your textbook asks you to prove that $a<\sqrt{ab}<b$ for $a<b$. This can be done in the following way: First we analyze $a<\sqrt{ab}$ and then $\sqrt{ab}<b$.

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

$a^2<ab$ (square both sides)

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

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

$ab<b^2$ (square both sides)

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

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

Q.E.D.

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