inequality proof

[noblegas]

I have a question about the direction I am going with a certain inequality proof . Here is the statement for the proof:

Suppose x>3 and y<2, then x^2-2*y>5. hint: used the theorem that if 0<a<b then a^2<b^2

Proof:

Assume x>3 and y<2 is true. When the inequality y<2 is multiply by -2, then y<2 ==> -2y>4. When the inequality x>3 is multiplied by the number 3, x>3 ==> 3*x>3*3==> 3*x>9. Similarly , when the inequality x>3 is multiplied by x, then x>3 ==> x^2>3*x. Since 3*x>9 and since x^2 >3*x, then x^2>9. Now I have -2*y>4 and x^2>9, adding the two inequalities together, I get x^2-2y>9-4 ==> x^2-2*y>5.


Is this proof right or am I AT LEAST going in the right direction with this proof. I would appreciate some feedback.

 

[pka]

I have a question about the direction I am going with a certain inequality proof . Here is the statement for the proof:

Suppose x>3 and y<2, then x^2-2*y>5. hint: used the theorem that if 0<a<b then a^2<b^2

Proof: Assume x>3 and y<2 is true. When the inequality y<2 is multiply by -2, then y<2 ==> -2y>4.

If $\displaystyle y<2$ then $\displaystyle -2y>-4$.

 

[stapel]

Suppose x>3 and y<2, then x^2-2*y>5. hint: used the theorem that if 0<a<b then a^2<b^2

The error in the first line has been corrected by another poster. Let's see if we can try something different, possibly that uses the "hint". (Those hints are generally very helpful!)

Let x > 3 and let 0 < y < 2. Then we have 0 < y < x, so y² < x². Since y < 2, then 2y < 2*2 = 4. Also, x > 3, so x² > 9. In any given subtraction, if we subtract something smaller, we get something larger. Then: [you do something with x² - 2y compared to 9 - 2y compared to 9 - 4].

On the other hand, let x > 3 and let y < 0. If y = 0, then x² - 2y = x² - 0 > 9 - 0 = 9 > 5. If y does not equal zero, then y < 0. Then [this should be an easy case].

Thus, in every case, we have x² - 2y > 5.
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Please write back if you get stuck. Thank you!

 

[noblegas]

If $\displaystyle y<2$ then $\displaystyle -2y>-4$.

Sorry, that was just a typo.

Let x > 3 and let 0 < y < 2.

but I am not given what y is. y could be zero. Besides the typo , what was wrong with the proof that I started with? I thought you could add two inequalities together?

 

[stapel]

I am not given what y is. y could be zero. Besides the typo , what was wrong with the proof that I started with? I thought you could add two inequalities together?

Try reading the rest of my post, where I show you how to handle y by "cases", being y > 0, y = 0, and y < 0.

 

[noblegas]

Try reading the rest of my post, where I show you how to handle y by "cases", being y > 0, y = 0, and y < 0.

I did and it seems way more complicated and more confusing to me than the proof technique I did. Was the proof technique I did too simplistic and incorrect?

 

[stapel]

Was the proof technique I did too simplistic and incorrect?

Your first line said that, if y < 2, then -2y > 4. The proof started wrong. Everything after that point must be ignored.

 

[Steven G]

I have a question about the direction I am going with a certain inequality proof . Here is the statement for the proof:

Suppose x>3 and y<2, then x^2-2*y>5. hint: used the theorem that if 0<a<b then a^2<b^2

Proof:

Assume x>3 and y<2 is true. When the inequality y<2 is multiply by -2, then y<2 ==> -2y>4. When the inequality x>3 is multiplied by the number 3, x>3 ==> 3*x>3*3==> 3*x>9. Similarly , when the inequality x>3 is multiplied by x, then x>3 ==> x^2>3*x. Since 3*x>9 and since x^2 >3*x, then x^2>9. Now I have -2*y>4 and x^2>9, adding the two inequalities together, I get x^2-2y>9-4 ==> x^2-2*y>5.


Is this proof right or am I AT LEAST going in the right direction with this proof. I would appreciate some feedback.

This proof is just fine other than your typo when you wrote -2y>4. I would have used the hint to get from x>3 to x^2>9 instead of your two step process. In my opinion a proof is never too simplistic.

 

[Ishuda]

I did and it seems way more complicated and more confusing to me than the proof technique I did. Was the proof technique I did too simplistic and incorrect?

Just some rambling which may (or may not) add to the discussion: As Jomo said, your proof (other than the typo) is correct. However some instructors might stop at the typo (thinking it was a mistake instead) and not follow the rest of your proof. That is, paraphrasing stapel, 'I won't look at the rest of the proof because it is based on a false premise'.

As to stapel's proof, I personally would think it would be good to study the 'cases proof' until you understand it. Many problems are solved easier this way because it allows different approaches for different conditions. The general idea behind this kind of proof is to divide all situations into (generally mutually exclusive) categories where, taking all categories together, all situations have been considered. In this particular example we have y is a real number. Therefore it is either greater than zero, equal to zero, or less than zero. Therefore showing the statement is true for all of these categories individually shows it is true for every possible y.