# analysis of a proof

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• Nov 18th 2009, 04:02 PM
alexandros
analysis of a proof
for the following proof :

$ab\leq |ab|\Longrightarrow 2ab\leq 2|a||b|\Longrightarrow (a+b)^2\leq (|a|+|b|)^2$ $\Longrightarrow |a+b|\leq |a|+|b|$

Write the theorems involved in the proof and how are they involved
• Nov 18th 2009, 05:45 PM
Focus
Quote:

Originally Posted by alexandros
for the following proof :

$ab\leq |ab|\Longrightarrow 2ab\leq 2|a||b|\Longrightarrow (a+b)^2\leq (|a|+|b|)^2$ $\Longrightarrow |a+b|\leq |a|+|b|$

Write the theorems involved in the proof and how are they involved

I am not quite sure what you are asking. The first statement holds by definition, second one by Cauchy-Schwarz, third by squaring out and using the inequality, last one by taking square roots.
• Nov 19th 2009, 04:18 AM
alexandros
Quote:

Originally Posted by Focus
I am not quite sure what you are asking. The first statement holds by definition, second one by Cauchy-Schwarz, third by squaring out and using the inequality, last one by taking square roots.

Thank you.

The Nos are Real Nos and not vectors and i was asked to mention the theorems involved in each step of the proof ( like you did ) and then to show
how are they involved in the proof
• Nov 19th 2009, 07:56 AM
emakarov
(Read the note in the end.)

This seems like a nice exercise. Obviously, the reasoning displayed uses several things: (1) properties of addition, subtraction, multiplication and division; (2) properties of inequalities; (3) properties of the $|\cdot|$ function; and (4) possibly some other properties, like those of square root, etc.

Imagine that you are talking to a elementary-school student who is extremely smart and understands everything you say provided you explain every new concept. However, he/she knows only the properties of the four basic operations and inequalities, as well as the definition of the absolute value. As I understand, first you are supposed to write this reasoning in much finer detail, on the level something like: "Now add ... to both sides of inequality. Now move this term to the other side. Now use the formula for the expansion of the square of a sum". Then you have to note every place that the young student won't understand without more explanation. For example, you replace $|ab|$ with $|a|\cdot|b|$ because you claim they are equal. This is not a property of the basic operations or inequalities; it involves some reasoning (very simple) about $|\cdot|$. So you have to identify all such places and write the statements that you use, such as $|ab|=|a|\cdot|b|$.

Note: this is my interpretation of the problem. You may have recently studied concrete properties or theorems, to which you gave names, etc. Then maybe the professor wants you to identify those facts only.
• Nov 19th 2009, 02:43 PM
alexandros
Thank you.

I asked my professor and he said for the first step of the implication i.e

$ab\leq |ab|$ we use the theorem :

for all x : $-|x|\leq x\leq |x|$.

But how do we use that thyeorem to get : $ab\leq |ab|$??
• Nov 19th 2009, 02:55 PM
Focus
Quote:

Originally Posted by alexandros
Thank you.

I asked my professor and he said for the first step of the implication i.e

$ab\leq |ab|$ we use the theorem :

for all x : $-|x|\leq x\leq |x|$.

But how do we use that thyeorem to get : $ab\leq |ab|$??

Hint, $x\leq |x|$.

P.S. Cauchy-Schwarz still applies as multiplication in R^1 is an inner product.
• Nov 19th 2009, 02:55 PM
Plato
Quote:

Originally Posted by alexandros
Thank you.

I asked my professor and he said for the first step of the implication i.e

$ab\leq |ab|$ we use the theorem :

for all x : $-|x|\leq x\leq |x|$.

But how do we use that theorem to get : $ab\leq |ab|$??

If you know that $\left( {\forall x} \right)\left[ { - \left| x \right| \leqslant \color{blue}x \leqslant \left| x \right|} \right]$
then surely it follows that $\color{blue}ab \leqslant \left| ab \right|$
• Nov 19th 2009, 03:38 PM
xalk
Quote:

Originally Posted by Plato
If you know that $\left( {\forall x} \right)\left[ { - \left| x \right| \leqslant \color{blue}x \leqslant \left| x \right|} \right]$
then surely it follows that $\color{blue}ab \leqslant \left| ab \right|$

How ,by putting x= ab?

By what theorem are you allowed to do this?

And again by what theorem you get : $ab\leq |ab|$ from

$-|ab|\leq ab\leq |ab|$
• Nov 19th 2009, 04:16 PM
Plato
Quote:

Originally Posted by xalk
How ,by putting x= ab?
By what theorem are you allowed to do this?
And again by what theorem you get : $ab\leq |ab|$ from
$-|ab|\leq ab\leq |ab|$

Surely you understand that $ab$ is a real number?
If not, why are you posting at this level?
• Nov 20th 2009, 11:39 AM
alexandros
Quote:

Originally Posted by Plato
If you know that $\left( {\forall x} \right)\left[ { - \left| x \right| \leqslant \color{blue}x \leqslant \left| x \right|} \right]$
then surely it follows that $\color{blue}ab \leqslant \left| ab \right|$

Thank you ,can you help me with the rest of the problem ?
• Nov 20th 2009, 12:12 PM
Plato
Quote:

Originally Posted by alexandros
for the following proof :

$ab\leq |ab|\Longrightarrow 2ab\leq 2|a||b|\Longrightarrow (a+b)^2\leq (|a|+|b|)^2$ $\Longrightarrow |a+b|\leq |a|+|b|$

Write the theorems involved in the proof and how are they involved

Comming late to this thread, I don't know if this will help.
Here facts we have now.
$a^2=|a|^2,~b^2=|b|^2~\&~2ab\le 2|a||b|$.

$a^2+2ab+b^2\le |a|^2 +2|a||b|+|b|^2$

$(a+b)^2\le (|a|+|b|)^2$
• Nov 20th 2009, 02:05 PM
alexandros
Thank you ,first of all ,before we go any further, how do we know that this proof is correct
• Nov 20th 2009, 02:20 PM
Plato
Quote:

Originally Posted by alexandros
How do we know that this proof is correct?

Well it certainly is fairly standard.
What it does is to prove the triangle inequality.
We start very basic properties of the distance function (the metric, the absolute value) and then using basic properties of order we prove the triangle inequality.
This basic form of the proof is used in a variety of different courses.
• Nov 20th 2009, 02:22 PM
Drexel28
Quote:

Originally Posted by alexandros
Thank you ,first of all ,before we go any further, how do we know that this proof is correct

What class are you in?
• Nov 20th 2009, 03:03 PM
xalk
Quote:

Originally Posted by Drexel28
What class are you in?

He is a bit confused but the problem is quite complicated
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