Math Help - analysis of a proof

1. Originally Posted by xalk
He is a bit confused but the problem is quite complicated
That is true only for people who do not understand the question.
Are you among them?

2. Originally Posted by Plato
That is true only for people who do not understand the question.
Are you among them?
I think considering this is the university help section, it is fair to say this proof is not complicated. It just relies on the properties of the real numbers.

$x\leq |x|$ follows from the fact that the reals are ordered.
$a \in \mathbb{R}, b \in \mathbb{R} \implies ab \in \mathbb{R}$ depends on how you look at R, easiest reasoning is because R is a field.

The rest follows from simple algebra.

3. Originally Posted by Plato
That is true only for people who do not understand the question.
Are you among them?
The statements of the proof are :

1) $ab\leq |ab|$

2) $2ab\leq 2|a||b|$

3) $(a+b)^2\leq (|a|+|b|)^2$

4) $|a+b|\leq |a|+|b|$

And the problem ask us :

a) To mention appropriate theorems or definitions that are responsible for those statements.

b) How are they involved .

For the (a) part the following tabular form shows clearly the theorems or definitions involved in the proof.

1) $ab\leq |ab|$.................................................. ............by using the theorem : for all ,x: $x\leq |x|$

2)|ab| = |a||b|............................................ ..................................by using the theorem : for all , x, y : |xy| = |x||y|

3) $ab\leq |a||b|$.................................................. .........by substituting (2) into (1)

4) $2ab\leq 2|a||b|$.................................................. .......by using the theorem : for all ,x,y,z : z>0 and $x\leq y\Longrightarrow xz\leq yz$

5) $a^2+b^2+2ab\leq 2|a||b|+a^2+b^2$.............................by using the theorem : for all ,x,y,z : $x\leq y\Longleftrightarrow x+z\leq y+z$

6) $a^2 = |a|^2$.................................................. ............by using the theorem :for all ,x : $x^2 = |x|^2$

7) $b^2 = |b|^2$.................................................. ............by using again the same theorem as in (6)

8) $a^2+b^2+2ab\leq 2|a||b|+|a|^2+|b|^2$.................................................. ..........by substituting (7) and (6) into (5)

9) $(a+b)^2 = a^2+b^2+2ab$............................................by using the identity (theorem) : for all , x,y : $(x+y)^2 = x^2+y^2+2xy$

10) $(|a|+|b|)^2 = |a|^2+|b|^2+2|a||b|$...........................by using the same theorem as in (9)

11) $(a+b)^2\leq (|a|+|b|)^2$............................................by substituting (10) and (9) into (8)

12) $\sqrt {(a+b)^2}\leq\sqrt {(|a|+|b|)^2}$........................by using the theorem : for all ,x,y : $x\geq 0$ and $y\geq 0\Longrightarrow (x\leq y\Longleftrightarrow\sqrt x\leq\sqrt y)$

13) $\sqrt{(a+b)^2} = |a+b|$..............................................by using the theorem : for all ,x : $\sqrt{ x^2} = |x|$

14) $\sqrt{(|a|+|b|)^2} = ||a|+|b||$.............................................by using the same theorem as in (13)

15) $|a+b|\leq ||a|+|b||$.................................................. ..........by substituting (14) and (13) into (12)

16) ||a|+|b|| = |a|+|b|........................................... .........................by using the definition of absolute value : for all , x : $x\geq 0\Longrightarrow |x| = x$

17) $|a+b|\leq |a|+|b|$.................................................. .by substituting (16) into (15).

Note: that in the above many other statements that are not mention in the proof are exposed . To include them all or part of them in the proof it is matter that depends on the writer's style .

NOW for part (b) .

Part (b) deals with the laws of logic ,the application of which, transforms the theorems involved in the proof into the statements of the proof.

That part i leave to Plato ,since he thinks that the whole problem is not complicated.

Of course if he wishes to do that

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