# Thread: A square root inequality to prove

1. ## A square root inequality to prove

I'm having problems to prove it:

I am considering these propositions:

But up to now I am having no success.
Note: 0<x<y

2. Note that $\displaystyle \displaystyle y - x > y - 2\sqrt{y}\sqrt{x} - x$.

So $\displaystyle \displaystyle \sqrt{y - x} > \sqrt{y - 2\sqrt{y}\sqrt{x} - x}$

$\displaystyle \displaystyle \sqrt{y - x} > \sqrt{(\sqrt{y} - \sqrt{x})^2}$

$\displaystyle \displaystyle \sqrt{y - x} > \sqrt{y} - \sqrt{x}$.

3. Or (I think it is easier that way, even if it is almost the same thing):

$\displaystyle \sqrt{y-x}> \sqrt{y}-\sqrt{x}\Leftrightarrow y-x>y+x-2\sqrt{xy}\Leftrightarrow 2x<2\sqrt{xy}\Leftrightarrow x^{2}<xy\Leftrightarrow x<y$ Which is true.

4. Originally Posted by veileen
Or (I think it is easier that way, even if it is almost the same thing):

$\displaystyle \sqrt{y-x}> \sqrt{y}-\sqrt{x}\Leftrightarrow y-x>y+x-2\sqrt{xy}\Leftrightarrow 2x<2\sqrt{xy}\Leftrightarrow x^{2}<xy\Leftrightarrow x<y$ Which is true.
To prove a statement you need to start with something that is known to be true (an axiom) and use a series of logical steps to get to what you are trying to prove.

You can't do what you did because you started with something that you don't know is true...

5. $\displaystyle \Leftrightarrow$ = 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.

6. Originally Posted by veileen
$\displaystyle \Leftrightarrow$ = 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. You need to start with a statement that is already known to be true and follow logical steps to get to the statement you are trying to prove.

7. Originally Posted by Prove It
Note that $\displaystyle \displaystyle y - x > y - 2\sqrt{y}\sqrt{x} - x$.

So $\displaystyle \displaystyle \sqrt{y - x} > \sqrt{y - 2\sqrt{y}\sqrt{x} - x}$

$\displaystyle \displaystyle \sqrt{y - x} > \sqrt{(\sqrt{y} - \sqrt{x})^2}$

This is incorrect: $\displaystyle \left(\sqrt{y}-\sqrt{x}\right)^2=y-2\sqrt{y}\sqrt{x}+x$ , and not $\displaystyle -x$ as above.

Tonio

$\displaystyle \displaystyle \sqrt{y - x} > \sqrt{y} - \sqrt{x}$.
.

8. 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.

9. Originally Posted by Prove It
You obviously didn't read what I wrote. You can NOT start with the statement you are trying to prove. You need to start with a statement that is already known to be true and follow logical steps to get to the statement you are trying to prove.

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

10. Originally Posted by veileen
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.
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.

11. "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)

12. Originally Posted by Prove It
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.

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

13. Originally Posted by veileen
"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.

Not quite, because they both can be false. In fact, $\displaystyle p\iff q\mbox{ iff }v(p)=v(q)$

Tonio

Also we are not taking 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)
.

14. Originally Posted by jhonyGrund
I'm having problems to prove it:

I am considering these propositions:

But up to now I am having no success.
Note: 0<x<y
$\displaystyle \sqrt{y-x}+\sqrt{x}>0$

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

$\displaystyle \sqrt{y-x}+\sqrt{x}>\sqrt{y}$

$\displaystyle \sqrt{y-x}>\sqrt{y}-\sqrt{x}$

15. "Not quite, because they both can be false." Yup (I said that in the fifth post of this thread), but there is no math problem where you are asked to prove something that is false and we are refering to math problems, aren't we? ^^

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