Note that $\displaystyle \displaystyle y - x > y - 2\sqrt{y}\sqrt{x} - x$.
So $\displaystyle \displaystyle \sqrt{y - x} > \sqrt{y - 2\sqrt{y}\sqrt{x} - x}$
$\displaystyle \displaystyle \sqrt{y - x} > \sqrt{(\sqrt{y} - \sqrt{x})^2}$
$\displaystyle \displaystyle \sqrt{y - x} > \sqrt{y} - \sqrt{x}$.
$\displaystyle \Leftrightarrow$ = if and only if
Considering the sentences p and q, $\displaystyle p\Leftrightarrow q$ if p and q have the same logical value (for example: if p is true and q is true, than $\displaystyle p\Leftrightarrow q$). So, what I wrote is correct.
You obviously didn't read what I wrote.
You can NOT start with the statement you are trying to prove. - Yes, you can. I'm not saying that is true, but supposing - if $\displaystyle \sqrt{y-x}>\sqrt{y}-\sqrt{x}$ is true, then $\displaystyle y-x>y+x-2\sqrt{xy}$ is true, then (...), then y>x is true. And y>x is true, so every expression that is forward y>x is true. You can't obtain something true starting with something fals.
He can start with what he wants to prove and do a series of logical steps with iff <==> , and if at the end he finishes with something which
is clearly true or that was proved before, then he can go back and show that what was to be proved is reachable.
What he did is correct if he: (1) states that it must be $\displaystyle y\geq x$ from the beginning (which isn't a tough
condition since otherwise the LHS of his inequality isn't defined), and (2) makes it clear to the checker that the abides
by the agreement that the square root of a positive real number is always positive.
Tonio
Actually you can. According to the rules of logical implication, going from something false to something true is considered to be true. This is why to prove something, you can not start with anything that you don't know the truth of, because there's no way of knowing the truth of the original statement.
"Actually you can." - No, you can't. I want an example.
"According to the rules of logical implication, going from something false to something true is considered to be true." - Fals. p <=> q only if v(p)=1 and v(q)=1.
Also we are not talking about implication, but equivalence. (implication: v(p => q)=0 only if v(p)=1 and v(q)=0; v(p => q)=1 if v(p)=1 and v(q)=1 // v(p)=1 and v(q)=0 // v(p)=0 and v(q)=0)
I think you're overseeing the iff (<==>) thing: of course, he can begin with something false and reach
something true, but NOT the other way around.
Thus, if I want to prove, as in this case, an inequality A > C, I can build the proof as follows:
$\displaystyle A> C\iff A_2\iff\ldots\iff B$ , and if $\displaystyle B$ is an obvious result (say, 0 = 0) or it's something
previously proved then A = C has been established.
The problem that usually can arise with a proof as above is that the double implications
not always are so, but if they are then the proof is sound.
Tonio
$\displaystyle \sqrt{y-x}+\sqrt{x}>0$
Square this
$\displaystyle y-x+2\sqrt{x}\sqrt{y-x}+x>0$
$\displaystyle y+2\sqrt{x}\sqrt{y-x}>0$
and since
$\displaystyle y+2\sqrt{x}\sqrt{y-x}>y$
$\displaystyle y-x+2\sqrt{x}\sqrt{y-x}+x>y$
Take square roots
$\displaystyle \sqrt{y-x}+\sqrt{x}>\sqrt{y}$
$\displaystyle \sqrt{y-x}>\sqrt{y}-\sqrt{x}$