analysis of a proof

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November 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
November 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.
November 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
November 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.
November 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 : $ab\leq |ab|$ ??
November 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.
November 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 $\color{blue}ab \leqslant \left| ab \right|$
November 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

$-|ab|\leq ab\leq |ab|$
November 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
$-|ab|\leq ab\leq |ab|$

What is your point?
Surely you understand that $ab$ is a real number?
If not, why are you posting at this level?
November 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 ?
November 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.
$a^2=|a|^2,~b^2=|b|^2~\&~2ab\le 2|a||b|$ .

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

$(a+b)^2\le (|a|+|b|)^2$
November 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
November 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.
November 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?
November 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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